Let us now consider the case of an odd order of harmonic numbers. As usual we start from the integral representation of our sums . We have:
\begin{eqnarray}
&&{\bf H}^{(2q+1)}_n(t) - Li_{n+2q+1}(t) =\int\limits_0^t \frac{[\log(t/\xi)]^{n-1}}{(n-1)!} \cdot \frac{Li_{2q+1}(\xi)}{1-\xi} d\xi\\
&&= \sum\limits_{j=0}^q (-1)^{q+j} \left[\binom{q+j}{2 j}\frac{1}{2} + \binom{q+j}{2 j+1}\right]\cdot \int\limits_0^1 \frac{[\log(1/\xi)]^{n-2(j+1)}}{(n-2(j+1))!} \cdot \frac{[Li_{q+j+1}(t \xi)]^2}{\xi}d\xi\\
&&=\sum\limits_{l_1=0}^{2q+1}\left\{\sum\limits_{j=0 \vee (l_1-q-1)}^q (\binom{q+j}{2j+0}\frac{1}{2} + \binom{q+j}{2j+1})\binom{q+n-j-1-l_1}{n-2 j-2}\right\}(-1)^{1-l_1} \cdot \cdot (Li_{l_1}(t) 1_{l_1\ge 0} - \delta_{l_1,0}) \cdot Li_{2q+n+1-l_1}(t)+\sum\limits_{l_1=1}^{n-1} \left\{\sum\limits_{j=0 }^{q \wedge \lfloor \frac{n-1-l_1}{2}\rfloor } (\binom{q+j}{2j+0}\frac{1}{2} + \binom{q+j}{2j+1})\binom{q+n-j-1-l_1}{q+j}\right\}(-1)^1 \cdot {\bf H}^{(n+2q+1-l_1)}_{l_1}(t)
\end{eqnarray}
In the second line from the top we integrated by parts $(2q+2)$-times each time using the well known properties of the poly-logarithm. What we essentially did at each step was that we found the anti-derivatives of $Li_{\theta_1}(\xi) Li_{\theta_2}(\xi)/\xi$ for some integer values of $\theta_1$ and $\theta_2$. The result is a linear combination of products of pairs of poly-logs and a residual term which is either a half of a square of a poly-log or something else depending on whether $(n-p)$ is odd or even in the first and in the second case respectively.Since integration by parts produces surface terms we have to assume that $n\ge 2q+2$ for all those terms to disappear. In the subsequent line we just used Generalized definite dilogarithm integral. and we simplified the result.
The result constitutes a set of recurrence relations that entwine the harmonic sums. Here $q=0,1,2,\cdots$ and $n\ge 2q+2$ and $t\in (-1,1)$. In case $n=1,\cdots,2q+1$ we have to go back to the original integral representation and take into account the surface terms. We have:
\begin{eqnarray}
{\bf H}^{(2q+1)}_{2 n+1}(t) &=& \sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} \cdot Li_l(t) Li_{2q+2n+2-l}(t) (-1)^{l-(2n+1)} +\\
&&(-1)^{q+n} \frac{1}{2} \binom{q+n}{2 n} \cdot [Li_{q+n+1}(t)]^2+\\
&&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-1-2 j}}{(2n-1-2 j)!} \cdot \frac{[Li_{q+j+1}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+2}(t)\\
{\bf H}^{(2q+1)}_{2 n}(t) &=& \sum\limits_{l=2n}^{q+n} \binom{l-1}{2 n-1} \cdot Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n)} +\\
&&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-2-2 j}}{(2n-2-2 j)!} \cdot \frac{[Li_{q+j+1}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+1}(t)\\
%
{\bf H}^{(2q)}_{2 n}(t) &=& \sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2 n-1} \cdot Li_l(t) Li_{2q+2n-l}(t) (-1)^{l-(2n)} +\\
&&(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2 n-1} \cdot [Li_{q+n}(t)]^2+\\
&&\sum\limits_{j=0}^{n-1} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-1-2 j}}{(2n-1-2 j)!} \cdot \frac{[Li_{q+j}(\xi)]^2}{\xi}d\xi+Li_{2q+2n}(t)\\
%
{\bf H}^{(2q)}_{2 n+1}(t) &=& \sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} \cdot Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n+1)} +\\
&&\sum\limits_{j=0}^{n} (-1)^{q+j} \cdot \left(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j} \right)\cdot \int\limits_0^t \frac{[\log(t/\xi)]^{2n-2 j}}{(2n-2 j)!} \cdot \frac{[Li_{q+j}(\xi)]^2}{\xi}d\xi+Li_{2q+2n+1}(t)
\end{eqnarray}
both for $n\ge 0$ and for $q\ge 0$ in the two top cases above and for $n\ge 0$ and $q\ge 1$ in the two bottom cases above.
The integrals on the right hand side are evaluated in Generalized definite dilogarithm integral..
Bringing everything together we have:
\begin{eqnarray}
&&{\bf H}^{(2q+1)}_{2n+1}(t)=\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(t) Li_{2q+2n+2-l}(t) (-1)^{l-(2n+1)}+ (-1)^{q+n} \frac{1}{2} \binom{q+n}{2 n} [Li_{q+n+1}(t)]^2+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n}
\left\{\sum\limits_{j=0\vee (l-q-1)}^{n-1}(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-j-l}{2n-1-2j}\right\}(-1)^{1-l}
(Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+2-l}(t)+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n}
\left\{\sum\limits_{j=0}^{\lfloor n-\frac{l}{2}\rfloor}(\frac{1}{2}\binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n- j-l}{q+j}\right\} (-1)^1 {\bf H}^{(2q+2n+2-l)}_l(t)+Li_{2q+2n+2}(t)\\
%
&&{\bf H}^{(2q+1)}_{2n}(t)=\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n}^{q+n} \binom{l-1}{2 n-1} Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n)}+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n}
\left\{\sum\limits_{j=0\vee (l-q-1)}^{n-1}(\frac{1}{2} \binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-1-j-l}{2n-2-2j}\right\}(-1)^{1-l}
(Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+1-l}(t)+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n-1}
\left\{\sum\limits_{j=0}^{\lfloor n-\frac{1+l}{2}\rfloor}(\frac{1}{2}\binom{q+j}{2 j} + \binom{q+j}{2j+1})\binom{q+2n-1- j-l}{q+j}\right\} (-1)^1 {\bf H}^{(2q+2n+1-l)}_l(t)+Li_{2q+2n+1}(t)\\
%
&&{\bf H}^{(2q)}_{2n}(t)=\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2 n-1} Li_l(t) Li_{2q+2n-l}(t) (-1)^{l-(2n)}+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2n-1} [Li_{q+n}(t)]^2+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n-1}
\left\{\sum\limits_{j=0\vee (l-q)}^{n-1}(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-1-j-l}{2n-1-2j}\right\}(-1)^{l}
(Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n-l}(t)+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n}
\left\{\sum\limits_{j=0}^{\lfloor n-\frac{l}{2}\rfloor}(\frac{1}{2}\binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-1- j-l}{q+j-1}\right\} (-1)^0 {\bf H}^{(2q+2n-l)}_l(t)+Li_{2q+2n}(t)\\
%
&&{\bf H}^{(2q)}_{2n+1}(t)=\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(t) Li_{2q+2n+1-l}(t) (-1)^{l-(2n+1)}+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=0}^{q+n}
\left\{\sum\limits_{j=0\vee (l-q)}^{n}(\frac{1}{2} \binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n-j-l}{2n-2j}\right\}(-1)^{l}
(Li_{l}(t) 1_{l\ge 1} - \delta_{l,0})Li_{2q+2n+1-l}(t)+\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum\limits_{l=1}^{2 n+1}
\left\{\sum\limits_{j=0}^{\lfloor n+\frac{1-l}{2}\rfloor}(\frac{1}{2}\binom{q+j-1}{2 j-1} + \binom{q+j-1}{2j})\binom{q+2n- j-l}{q+j-1}\right\} (-1)^0 {\bf H}^{(2q+2n+1-l)}_l(t)+Li_{2q+2n+1}(t)
\end{eqnarray}
Note that as long as the first two equations are useful because the quantity in question only appears on the left hand side the last two equations are less useful because the quantity being searched for actually cancels on both sides of the equation. In general it turns out that the odd-odd quantities always reduce to even-even quantities and poly-logs. On the other hand the odd-even quantities always reduce to even-odd quantities and poly-logs. We have
\begin{eqnarray}
{\bf H}^{(1)}_1(t) &=& \frac{1}{2}\left( [\log(1-t)]^2 + 2 Li_2(t)\right)\\
{\bf H}^{(1)}_2(t) &=& \frac{1}{2}\left(-{\bf H}^{(2)}_1(t) - \log(1-t) Li_2(t) + 3 Li_3(t)\right)\\
{\bf H}^{(1)}_3(t) &=& \frac{1}{4} \left( -2 {\bf H}^{(2)}_2(t) + [Li_2(t)]^2 + 6 Li_4(t)\right)\\
{\bf H}^{(1)}_4(t) &=& \frac{1}{4}\left(-2 {\bf H}^{(2)}_3(t)+{\bf H}^{(4)}_1(t) + Li_2(t) Li_3(t) + \log(1-t) Li_4(t) + 5 Li_5(t) \right) \\
{\bf H}^{(1)}_5(t) &=& \frac{1}{4}\left( -2 {\bf H}^{(2)}_4(t) + {\bf H}^{(4)}_2(t) + [Li_3(t)]^2 - Li_2(t) Li_4(t) + 5Li_6(t)\right)\\
{\bf H}^{(1)}_6(t) &=&\frac{1}{4}\left(-2{\bf H}^{(2)}_5(t)+{\bf H}^{(4)}_3(t)-2{\bf H}^{(6)}_1(t) + Li_3(t) Li_4(t) - 2 Li_2(t)Li_5(t) - 2\log(1-t)Li_6(t) + 7 Li_7(t)\right)\\
{\bf H}^{(1)}_7(t) &=& \frac{1}{8}
\left( -4 {\bf H}^{(2)}_6(t)+ 2 {\bf H}^{(4)}_4(t) - 4 {\bf H}^{(6)}_2(t) + 5 [Li_4(t)]^2 - 8 Li_3(t) Li_5(t) + 4 Li_2(t) Li_6(t) + 14 Li_8(t)\right)\\
{\bf H}^{(1)}_8(t) &=& \frac{1}{8} (-4 {\bf H}^{(2)}_7(t)+2 {\bf H}^{(4)}_5(t)-4 {\bf H}^{(6)}_3(t)+17 {\bf H}^{(8)}_1(t)+5 \text{Li}_4(t)
\text{Li}_5(t)-13 \text{Li}_3(t) \text{Li}_6(t)+17 \text{Li}_2(t)
\text{Li}_7(t)-3 \text{Li}_9(t)+17 \text{Li}_8(t) \log (1-t))\\
{\bf H}^{(1)}_9(t) &=& \frac{1}{8} \left(-4 {\bf H}^{(2)}_8(t)+2 {\bf H}^{(4)}_6(t)-4 {\bf H}^{(6)}_4(t)+17 {\bf H}^{(8)}_2(t)+26
\text{Li}_5(t){}^2-47 \text{Li}_4(t) \text{Li}_6(t)+34 \text{Li}_3(t)
\text{Li}_7(t)-17 \text{Li}_2(t) \text{Li}_8(t)-3
\text{Li}_{10}(t)\right)\\
{\bf H}^{(1)}_{10}(t) &=& \frac{1}{8} (-4 {\bf H}^{(2)}_9(t)+2 {\bf H}^{(4)}_7(t)-4 {\bf H}^{(6)}_5(t)+17 {\bf H}^{(8)}_3(t)-124 {\bf H}^{(10)}_1(t)+26
\text{Li}_5(t) \text{Li}_6(t)-73 \text{Li}_4(t) \text{Li}_7(t)+107
\text{Li}_3(t) \text{Li}_8(t)-124 \text{Li}_2(t) \text{Li}_9(t)+121
\text{Li}_{11}(t)-124 \text{Li}_{10}(t) \log (1-t))\\
\end{eqnarray}
Likewise we have:
\begin{eqnarray}
{\bf H}^{(3)}_1(t) &=&\frac{1}{2} \left(-\text{Li}_2(t){}^2+2 \text{Li}_4(t)-2 \text{Li}_3(t) \log (1-t)\right)\\
{\bf H}^{(3)}_2(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_1(t)-\text{Li}_2(t) \text{Li}_3(t)+5 \text{Li}_5(t)-3 \text{Li}_4(t) \log (1-t))\\
{\bf H}^{(3)}_3(t) &=& \frac{1}{2}
\left(-3 {\bf H}^{(4)}_2-2 \text{Li}_3(t){}^2+3 \text{Li}_2(t) \text{Li}_4(t)+5 \text{Li}_6(t)\right)\\
{\bf H}^{(3)}_4(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_3(t)+5 {\bf H}^{(6)}_1(t)-2 \text{Li}_3(t) \text{Li}_4(t)+5 \text{Li}_2(t) \text{Li}_5(t)+5
\text{Li}_6(t) \log (1-t))\\
{\bf H}^{(3)}_5(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_4(t)+5 {\bf H}^{(6)}_2(t)-6 \text{Li}_4(t){}^2+10 \text{Li}_3(t) \text{Li}_5(t)-5 \text{Li}_2(t) \text{Li}_6(t)\right)\\
{\bf H}^{(3)}_6(t) &=& \frac{1}{2} (-3 {\bf H}^{(4)}_5(t)+5
{\bf H}^{(6)}_3(t)-21 {\bf H}^{(8)}_1(t)-6 \text{Li}_4(t) \text{Li}_5(t)+16 \text{Li}_3(t) \text{Li}_6(t)-21 \text{Li}_2(t) \text{Li}_7(t)+21 \text{Li}_9(t)-21 \text{Li}_8(t) \log (1-t))\\
{\bf H}^{(3)}_7(t) &=& \frac{1}{2} \left(-3 {\bf H}^{(4)}_6(t)+5
{\bf H}^{(6)}_4(t)-21 {\bf H}^{(8)}_2(t)-32 \text{Li}_5(t){}^2+58 \text{Li}_4(t) \text{Li}_6(t)-42 \text{Li}_3(t) \text{Li}_7(t)+21 \text{Li}_2(t) \text{Li}_8(t)+21 \text{Li}_{10}(t)\right)\\
\hline \\
{\bf H}^{(5)}_1(t)&=&\frac{1}{2} \left(\text{Li}_3(t){}^2-2 \text{Li}_2(t) \text{Li}_4(t)+2 \text{Li}_6(t)-2 \text{Li}_5(t) \log (1-t)\right)\\
{\bf H}^{(5)}_2(t)&=&\frac{1}{2} (-5 {\bf H}^{(6)}_1(t)+\text{Li}_3(t) \text{Li}_4(t)-3 \text{Li}_2(t) \text{Li}_5(t)+7 \text{Li}_7(t)-5 \text{Li}_6(t) \log (1-t))\\
{\bf H}^{(5)}_3(t)&=&\frac{1}{4} \left(-10 {\bf H}^{(6)}_2(t)+9\text{Li}_4(t){}^2-16 \text{Li}_3(t) \text{Li}_5(t)+10 \text{Li}_2(t) \text{Li}_6(t)+14 \text{Li}_8(t)\right)\\
{\bf H}^{(5)}_4(t)&=&\frac{1}{4} (-10 {\bf H}^{(6)}_3(t)+35 {\bf H}^{(8)}_1(t)+9
\text{Li}_4(t) \text{Li}_5(t)-25 \text{Li}_3(t) \text{Li}_6(t)+35 \text{Li}_2(t) \text{Li}_7(t)-21 \text{Li}_9(t)+35 \text{Li}_8(t) \log
(1-t))\\
{\bf H}^{(5)}_5(t)&=&
\frac{1}{4} \left(-10 {\bf H}^{(6)}_4(t)+35 {\bf H}^{(8)}_2(t)+52 \text{Li}_5(t){}^2-95 \text{Li}_4(t) \text{Li}_6(t)+70 \text{Li}_3(t) \text{Li}_7(t)-35
\text{Li}_2(t) \text{Li}_8(t)-21 \text{Li}_{10}(t)\right)\\
\hline \\
{\bf H}^{(7)}_1(t)&=&\frac{1}{2} \left(-\text{Li}_4(t){}^2+2 \text{Li}_3(t) \text{Li}_5(t)-2 \text{Li}_2(t) \text{Li}_6(t)+2 \text{Li}_8(t)-2 \text{Li}_7(t) \log
(1-t)\right)\\
{\bf H}^{(7)}_2(t)&=&\frac{1}{2} (-7 {\bf H}^{(8)}_1(t)-\text{Li}_4(t) \text{Li}_5(t)+3 \text{Li}_3(t) \text{Li}_6(t)-5 \text{Li}_2(t) \text{Li}_7(t)+9
\text{Li}_9(t)-7 \text{Li}_8(t) \log (1-t))\\
{\bf H}^{(7)}_3(t)&=&\frac{1}{2} \left(-7 {\bf H}^{(8)}_2(t)-8 \text{Li}_5(t){}^2+15 \text{Li}_4(t) \text{Li}_6(t)-12 \text{Li}_3(t)
\text{Li}_7(t)+7 \text{Li}_2(t) \text{Li}_8(t)+9 \text{Li}_{10}(t)\right)\\
\hline \\
{\bf H}^{(9)}_1(t)&=&\frac{1}{2} \left(\text{Li}_5(t){}^2-2 \text{Li}_4(t) \text{Li}_6(t)+2 \text{Li}_3(t) \text{Li}_7(t)-2 \text{Li}_2(t) \text{Li}_8(t)+2
\text{Li}_{10}(t)-2 \text{Li}_9(t) \log (1-t)\right)
\end{eqnarray}
Unfortunately both the even-even and the even-odd quantities cannot be worked out using the formalism above since the respective recurrence equations reduce to tautologies.
Update: Below we demonstrate that it is possible to get additional recurrence relations for both the even-odd and the even-even quantities provided $t=-1$. Let us start with the simplest possible example. Let us assume that $q\ge 1$ then we have:
\begin{eqnarray}
&&{\bf H}^{(2q)}_1(-1)= \sum\limits_{l=1}^q Li_l(-1) Li_{2q+1-l}(-1) (-1)^{l-1} + (-1)^q \underbrace{\int\limits_0^1 \frac{[Li_q(-\xi)]^2}{\xi} d\xi}_{{\mathcal A}^{(0,2)}_q(-1)}
+ Li_{2q+1}(-1)=\\
&&\frac{1}{4^q}\left(-1+(-2+4^q) q\right) \zeta(2q+1) - \log(2) \left(-1+\frac{1}{2^{2q-1}}\right) \zeta(2q)+\\
&&\sum\limits_{l=2}^q (-\frac{1}{2})^l \left(-2+2^{l-q}\right) \zeta(l) \zeta(2q+1-l)+\\
&&\sum\limits_{l=2}^{2q-1} \left(-\frac{1}{2} -2 (-1)^l +(-1)^l 2^{2-l} + \frac{1}{4^q} \right)\zeta(l) \zeta(2q+1-l)+\\
&&2{\bf H}^{(1)}_{2q}(-1)
\end{eqnarray}
In the top line we started from the integral representation which we integrated by parts $q$-times. In the bottom line we used the second answer to Generalized definite dilogarithm integral. to compute the integral on the right hand side. As a result we obtained a quite useful relation. Note that the harmonic sum on the left hand side is converging very slowly whereas the other sum on the right hand side converges quite fast. It is clear that this approach can be extended to more complicated cases. We have:
\begin{eqnarray}
&&{\bf H}^{(2q)}_{2n+1}(-1)=\\
&&\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(-1) Li_{2q+2n+1-l}(-1) (-1)^{l-(2n+1)}+\\
&&\sum\limits_{l=2}^{q+n} {\mathcal A}_0^{(n,l,q)} (-1)^l \frac{1-2^{1+l}+2^{2(n+q)}}{2^{2(n+q)}}\cdot \zeta(2n+2q+1-l) \zeta(l)+\\
&&\sum\limits_{l=2}^{2n+1} {\mathcal A}_1^{(n,l,q)} \frac{1-2^{1+l}+2^{2(n+q)}}{2^{2(n+q)}}\cdot \zeta(2n+2q+1-l) \zeta(l)+\\
&&\sum\limits_{l=1}^{2n+1} {\mathcal A}_1^{(n,l,q)}\left[(1-2^{1-2 n-2 q}) {\bf H}^{(l)}_{2n+2q+1-l}(+1) + 2 {\bf H}^{(l)}_{2n+2q+1-l}(-1)\right]+\\
&&Li_{2n+2 q+1}(-1)
\end{eqnarray}
for $n\ge 0$ and $q\ge 1$. Here the coefficients read:
\begin{eqnarray}
{\mathcal A}_0^{(n,l,q)}&:=& \sum\limits_{j=(l-q)\vee 0}^n (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j})
\binom{-j-l+2n+q}{2(n-j)}\\
{\mathcal A}_1^{(n,l,q)}&:=& \sum\limits_{j= 0}^{n-\lfloor \frac{l-1}{2}\rfloor} (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j})
\binom{-j-l+2n+q}{q+j-1}
\end{eqnarray}
In the even-even case we have:
\begin{eqnarray}
&&{\bf H}^{(2 q)}_{2 n}(-1)=\\
&&\sum\limits_{l=2n}^{q+n-1} \binom{l-1}{2n-1} Li_l(-1) Li_{2q+2n-l}(-1)(-1)^{l-2 n}+\\
&&(-1)^{q+n} \frac{1}{2} \binom{q+n-1}{2n-1} [Li_{q+n}(-1)]^2 +\\
&&\sum\limits_{l=2}^{q+n-1} {\mathcal A}_2^{(n,l,q)} \left( \frac{4-2^{2+l}+2^{2(n+q)}}{2^{2(n+q)}}\right)\cdot \zeta(2(n+q)-l)\zeta(l) (-1)^l+\\
&&\sum\limits_{l=2}^{2 n} {\mathcal A}_3^{(n,l,q)} \left( \frac{4-2^{2+l}+2^{2(n+q)}}{2^{2(n+q)}}\right)\cdot \zeta(2(n+q)-l)\zeta(l) +\\
&&\sum\limits_{l=1}^{2 n}{\mathcal A}_3^{(n,l,q)} \left((1-\frac{1}{2^{2(n+q-1)}}) {\bf H}^{(l)}_{2(n+q)-l}(+1) + 2 {\bf H}^{(l)}_{2(n+q)-l}(-1)\right)+\\
&&Li_{2n+2q}(-1)
\end{eqnarray}
where the coefficients read:
\begin{eqnarray}
{\mathcal A}_2^{(n,l,q)}&:=& \sum\limits_{j=(l-q)\vee 0}^n (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j})
\binom{-1-j-l+2n+q}{2(n-j)-1}\\
{\mathcal A}_3^{(n,l,q)}&:=& \sum\limits_{j= 0}^{n-\lfloor \frac{l}{2}\rfloor} (\frac{1}{2}\binom{q+j-1}{2j-1} + \binom{q+j-1}{2j})
\binom{-1-j-l+2n+q}{q+j-1}
\end{eqnarray}
where $n\ge 1$ and $q\ge 1$. As we can see from the above we also need the results for plus unity. They read:
\begin{eqnarray}
&&{\bf H}^{(2q)}_{2n+1}(+1)=\\
&&\sum\limits_{l=2n+1}^{q+n} \binom{l-1}{2 n} Li_l(1) Li_{2q+2n+1-l}(1) (-1)^{l-(2n+1)}+\\
&&\sum\limits_{l=2}^{q+n} {\mathcal A}_4^{(n,l,q)} \zeta(l) \zeta(1-l+2 n+2 q) (-1)^l+\\
&&\sum\limits_{l=2}^{2n+1} {\mathcal A}_5^{(n,l,q)} \zeta(l) \zeta(1-l+2 n+2 q) +\\
&&\sum\limits_{l=1}^{2n+1} {\mathcal A}_5^{(n,l,q)}(-1)^1 {\bf H}^{(l)}_{1+2n+2q-l}(+1)+\\
&&Li_{2n+2q+1}(+1)
\end{eqnarray}
where
\begin{eqnarray}
{\mathcal A}_4^{(n,l,q)}&:=& \sum\limits_{j=(l-q) \vee 0}^n \left(\frac{1}{2} \binom{q+j-1}{2j-1}+\binom{q+j-1}{2 j}\right)\binom{q-j+2n-l}{2n-2j}\\
{\mathcal A}_5^{(n,l,q)}&:=& \sum\limits_{j=0}^{n+\lfloor \frac{1-l}{2} \rfloor}\left(\frac{1}{2} \binom{q+j-1}{2j-1}+\binom{q+j-1}{2 j}\right)\binom{q-j+2n-l}{q+j-1}
\end{eqnarray}
It is clear that an analogous formula exists for the remaining even-even case at plus unity. We will write it down later on. Now I am going to argue that the last two formulae above along with the relations that combine the odd-odd and the odd-even cases with the even-even and the even-odd cases-- the relations that hold for arbitrary value of $t$ -- that those relations are sufficient in order to work out closed form solutions for all the harmonic sums at plus unity.
Indeed using this approach we found the following:
\begin{eqnarray}
{\bf H}^{(1)}_2(+1) &=& 2 \zeta(3)\\
\lim_{t\rightarrow 1} \left({\bf H}^{(2)}_1(t) + \log(1-t) Li_2(t)\right) &=& - \zeta(3)\\
\hline
{\bf H}^{(1)}_3(+1) &=& -\frac{1}{6} \zeta(2)^2+ \frac{5}{3} \zeta(4)\\
{\bf H}^{(2)}_2(+1) &=& +\frac{5}{6} \zeta(2)^2 - \frac{1}{3} \zeta(4) \\
\lim_{t \rightarrow 1} \left({\bf H}^{(3)}_1(t) + \log(1-t) Li_3(t)\right) &=& -\frac{1}{2} \zeta(2)^2 + \zeta(4) \\
\hline
{\bf H}^{(1)}_4(+1) &=& -\zeta(2)\zeta(3) + 3 \zeta(5)\\
{\bf H}^{(2)}_3(+1) &=&+3 \zeta(2) \zeta(3)-\frac{9}{2} \zeta(5)\\
{\bf H}^{(3)}_2(+1)&=& -2 \zeta(2) \zeta(3)+\frac{11}{2} \zeta(5)\\
\lim_{t\rightarrow 1} \left({\bf H}^{(4)}_1(t) + \log(1-t) Li_4(t)\right) &=&+1 \zeta(2) \zeta(3)-2 \zeta(5) \\
\hline
{\bf H}^{(1)}_5(+1) &=& -\frac{1}{2}\zeta(3)^2-\frac{1}{3}\zeta(2)\zeta(4) + \frac{7}{3} \zeta(6)\\
{\bf H}^{(2)}_4(+1) &=&+1 \zeta(3)^2+\frac{4}{3} \zeta(2) \zeta(4)-\frac{8}{3} \zeta(6)\\
{\bf H}^{(3)}_3(+1)&=& +\frac{1}{2} \zeta(3)^2-2 \zeta(2) \zeta(4)+4\zeta(6)\\
{\bf H}^{(4)}_2(+1)&=& -1 \zeta(3)^2+\frac{7}{3} \zeta(2) \zeta(4)-1\zeta(6)\\
\lim_{t\rightarrow 1} \left({\bf H}^{(5)}_1(t) + \log(1-t) Li_5(t)\right) &=&+\frac{1}{2} \zeta(3)^2- \zeta(2) \zeta(4)+ \zeta(6) \\
\hline
{\bf H}^{(1)}_6(+1) &=& -\zeta(3) \zeta(4)-\zeta(2) \zeta(5)+4 \zeta(7) \\
{\bf H}^{(2)}_5(+1) &=& +2\zeta(3) \zeta(4)+5\zeta(2) \zeta(5)-10 \zeta(7) \\
{\bf H}^{(3)}_4(+1) &=& +0\zeta(3) \zeta(4)-10\zeta(2) \zeta(5)+18 \zeta(7) \\
{\bf H}^{(4)}_3(+1) &=& +1\zeta(3) \zeta(4)+10\zeta(2) \zeta(5)-17 \zeta(7) \\
{\bf H}^{(5)}_2(+1) &=& -2\zeta(3) \zeta(4)-4\zeta(2) \zeta(5)+11 \zeta(7) \\
\lim_{t\rightarrow 1} \left({\bf H}^{(6)}_1(t) + \log(1-t) Li_6(t)\right) &=& +1 \zeta(3) \zeta(4)+ \zeta(2) \zeta(5) - 3 \zeta(7) \\
\hline \\
{\bf H}^{(1)}_7(+1) &=& \frac{9 \zeta(8)}{4}-\zeta (3) \zeta (5) \\
{\bf H}^{(2)}_6(+1) &=& \\
{\bf H}^{(3)}_5(+1) &=& -\frac{5}{2} {\bf H}^{(2)}_6(+1)-\frac{21 \zeta(8)}{8}+5 \zeta (3) \zeta (5) \\
{\bf H}^{(4)}_4(+1) &=& \frac{13 \zeta(8)}{12} \\
{\bf H}^{(5)}_3(+1) &=& \frac{5}{2} {\bf H}^{(2)}_6(+1)+\frac{29 \zeta(8)}{8}-4 \zeta
(3) \zeta (5) \\
{\bf H}^{(6)}_2(+1) &=& \frac{8 \zeta(8)}{3}-{\bf H}^{(2)}_6(+1) \\
\lim_{t\rightarrow 1} {\bf H}^{(7)}_1(t) + \log(1-t) Li_7(t) &=& \zeta (3) \zeta (5)-\frac{5 \zeta(8)}{4}\\
\hline\\
{\bf H}^{(1)}_8(+1)&=&\frac{-2 \pi ^6 \zeta(3)-21 \pi ^4 \zeta(5)-315 \pi ^2 \zeta(7)+9450 \zeta(9)}{1890} \\
{\bf H}^{(2)}_7(+1)&=& \frac{2}{945} \pi ^6 \zeta(3)+\frac{2}{45} \pi ^4 \zeta(5)+\frac{7}{6} \pi ^2
\zeta(7)-\frac{35 \zeta(9)}{2} \\
{\bf H}^{(3)}_6(+1)&=& -\frac{1}{15} \pi ^4 \zeta(5)-\frac{7}{2} \pi ^2 \zeta(7)+\frac{85 \zeta(9)}{2} \\
{\bf H}^{(4)}_5(+1)&=& \frac{1}{18} \pi ^4 \zeta(5)+\frac{35}{6} \pi ^2
\zeta(7)-\frac{125 \zeta(9)}{2} \\
{\bf H}^{(5)}_4(+1)&=& -\frac{2}{45} \pi ^4 \zeta(5)-\frac{35}{6} \pi ^2 \zeta(7)+\frac{127 \zeta(9)}{2} \\
{\bf H}^{(6)}_3(+1)&=& \frac{1}{945} \pi ^6 \zeta(3)+\frac{1}{15} \pi ^4
\zeta(5)+\frac{7}{2} \pi ^2 \zeta(7)-\frac{83 \zeta(9)}{2} \\
{\bf H}^{(7)}_2(+1)&=& -\frac{2}{945} \pi ^6 \zeta(3)-\frac{2}{45} \pi ^4 \zeta(5)-\pi ^2 \zeta(7)+\frac{37 \zeta(9)}{2} \\
\lim_{t \rightarrow 1}\left( {\bf H}^{(8)}_1(t) + \log(1-t) Li_8(t) \right)&=& \frac{1}{945} \pi ^6 \zeta(3)+\frac{1}{90} \pi ^4 \zeta(5)+\frac{1}{6} \pi ^2 \zeta(7)-4 \zeta(9) \\
\hline \\
{\bf H}^{(1)}_9(+1)&=& \frac{\pi ^{10}}{34020}-\frac{\zeta (5)^2}{2}-\zeta (3) \zeta (7) \\
{\bf H}^{(2)}_8(+1)&=& {\bf H}^{(2)}_8(+1) \\
{\bf H}^{(3)}_7(+1)&=& -\frac{7}{2} {\bf H}^{(2)}_8(+1)+7 \zeta (3) \zeta (7)+4 \zeta (5)^2-\frac{\pi ^{10}}{11340} \\
{\bf H}^{(4)}_6(+1)&=& \frac{7}{2} {\bf H}^{(2)}_8(+1)-7 \zeta (3) \zeta (7)-5 \zeta
(5)^2+\frac{227 \pi ^{10}}{1871100} \\
{\bf H}^{(5)}_5(+1)&=& \frac{\pi ^{10}}{187110}+\frac{\zeta (5)^2}{2} \\
{\bf H}^{(6)}_4(+1)&=& -\frac{7}{2} {\bf H}^{(2)}_8(+1)+7 \zeta (3) \zeta (7)+5 \zeta (5)^2-\frac{37 \pi ^{10}}{374220} \\
{\bf H}^{(7)}_3(+1)&=& \frac{7}{2} {\bf H}^{(2)}_8(+1)-6 \zeta (3) \zeta (7)-4
\zeta (5)^2+\frac{37 \pi ^{10}}{374220} \\
{\bf H}^{(8)}_2(+1)&=& \frac{53 \pi ^{10}}{1871100}-{\bf H}^{(2)}_8(+1) \\
\lim_{t \rightarrow 1} \left( {\bf H}^{(9)}_1(t) + \log(1-t) Li_9(t) \right) &=& -\frac{\pi ^{10}}{53460}+\frac{\zeta (5)^2}{2}+\zeta (3) \zeta (7) \\
\hline\\
\vdots\\
\hline\\
{\bf H}^{(1)}_{11}(+1)&=& -\zeta (5) \zeta (7)-\zeta (3) \zeta (9)+\frac{691 \pi ^{12}}{196465500}\\
{\bf H}^{(3)}_9(+1)&=& \frac{428652000 \zeta (5) \zeta (7)+321489000 \zeta (3) \zeta (9)-691 \pi ^{12}}{35721000}-\frac{9}{2}
{\bf H}^{(2)}_{10}(+1)\\
{\bf H}^{(4)}_8(+1)&=& 8 {\bf H}^{(2)}_{10}(+1)-16 \zeta (3) \zeta (9)-28 \zeta (5) \zeta (7)+\frac{86096 \pi ^{12}}{1915538625}\\
{\bf H}^{(5)}_7(+1)&=& -7 {\bf H}^{(2)}_{10}(+1)+14 \zeta (3) \zeta (9)+28 \zeta (5) \zeta (7)-\frac{316027 \pi
^{12}}{7662154500}\\
{\bf H}^{(6)}_6(+1)&=& \frac{703 \pi ^{12}}{638512875}\\
{\bf H}^{(7)}_5(+1)&=& 7 {\bf H}^{(2)}_{10}(+1)-14 \zeta (3) \zeta (9)-27 \zeta (5) \zeta (7)+\frac{324319 \pi ^{12}}{7662154500}\\
{\bf H}^{(8)}_4(+1)&=& -8 {\bf H}^{(2)}_{10}(+1)+16 \zeta (3) \zeta (9)+28 \zeta (5)
\zeta (7)-\frac{327083 \pi ^{12}}{7662154500}\\
{\bf H}^{(9)}_3(+1)&=& \frac{9}{2} {\bf H}^{(2)}_{10}(+1)-8 \zeta (3) \zeta (9)-12 \zeta (5) \zeta (7)+\frac{104341 \pi ^{12}}{5108103000}\\
{\bf H}^{(10)}_2(+1)&=& \frac{1219 \pi ^{12}}{425675250}-{\bf H}^{(2)}_{10}(+1)\\
\lim_{t\rightarrow 1} \left({\bf H}^{(11)}_1(t)+\log(1-t) Li_{11}(t) \right)&=&
\frac{283783500 \zeta (5) \zeta (7)+283783500 \zeta (3) \zeta (9)-691 \pi ^{12}}{283783500}
\end{eqnarray}
On the face of it is seemed that all harmonic sums at plus unity are functions of zeta values at positive integers only. However when the weight became strictly bigger than seven something new happened. One of the equations appeared to be linearly dependent on the others which rendered it impossible to evaluate one of the sums. Now to the case of minus unity.
In the even-even and the even-odd cases we will be using the relations for minus unity
whereas in the odd-odd and in the odd-even cases we will be using the relations that are valid for arbitrary $t$.
\begin{eqnarray}
{\bf H}^{(1)}_1(-1) &=& \frac{1}{2} [\log(2)]^2 - \frac{1}{2} \zeta(2)\\
\hline\\
{\bf H}^{(1)}_2(-1) &=& - \frac{5}{8} \zeta(3)\\
{\bf H}^{(2)}_1(-1) &=& \frac{1}{2} [\log(2)] \zeta(2) - \zeta(3)\\
\hline\\
{\bf H}^{(1)}_3(-1) &=& \frac{1}{360} \left(30 \left(24 \text{Li}_4\left(\frac{1}{2}\right)+21 \zeta (3) \log (2)+\log ^4(2)\right)-11 \pi ^4-30 \pi ^2 \log ^2(2)\right)\\
{\bf H}^{(2)}_2(-1) &=& -4\text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{17 \pi ^4}{480}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)\\
{\bf H}^{(3)}_1(-1) &=& \frac{1080 \zeta (3) \log (2)-19 \pi
^4}{1440}\\
\hline\\
{\bf H}^{(1)}_4(-1) &=& \frac{1}{96} \left(8 \pi ^2 \zeta (3)-177 \zeta (5)\right)\\
{\bf H}^{(2)}_3(-1) &=& \frac{11 \zeta (5)}{32}-\frac{5 \pi ^2 \zeta (3)}{48} \\
{\bf H}^{(3)}_2(-1) &=& \frac{21 \zeta (5)}{32}-\frac{\pi ^2 \zeta (3)}{8} \\
{\bf H}^{(4)}_1(-1) &=& \frac{\pi ^2 \zeta (3)}{16}-2 \zeta (5)+\frac{7}{720} \pi ^4 \log (2)\\
\hline \\
{\bf H}^{(1)}_5(-1) &=& \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) \\
{\bf H}^{(2)}_4(-1) &=& \frac{1}{2} \int\limits_0^1 \frac{[log(1/\xi)]^1}{1!} \cdot \frac{Li_2(-\xi)^2}{\xi} d\xi
- \int\limits_0^1 \frac{[log(1/\xi)]^3}{3!} \cdot \frac{[\log(1+\xi)]^2}{\xi} d\xi + Li_6(-1) \\
{\bf H}^{(3)}_3(-1) &=& -6 {\bf H}^{(1)}_5(-1)-3 {\bf H}^{(2)}_4(-1)+\frac{1701 \zeta (3)^2-62 \pi ^6}{6048} \\
{\bf H}^{(4)}_2(-1) &=& 4 {\bf H}^{(1)}_5(-1)+2 {\bf H}^{(2)}_4(-1)-\frac{9 \zeta (3)^2}{16}+\frac{359 \pi ^6}{60480} \\
{\bf H}^{(5)}_1(-1) &=& \frac{5670 \zeta (3)^2+18900 \zeta (5) \log (2)-37 \pi^6}{20160}\\
\hline\\
{\bf H}^{(1)}_6(-1)&=&+\frac{56 \pi ^4 \zeta (3)+480 \pi ^2 \zeta (5)-16965 \zeta (7)}{5760}\\
{\bf H}^{(2)}_5(-1)&=&+\frac{249 \zeta (7)}{64}-\frac{49 \pi ^2 \zeta (5)}{192}-\frac{7 \pi ^4 \zeta (3)}{360}\\
{\bf H}^{(3)}_4(-1)&=&-\frac{363 \zeta(7)}{128}+\frac{3 \pi ^2 \zeta (5)}{16}\\
{\bf H}^{(4)}_3(-1)&=&-\frac{199 \zeta (7)}{64}+\frac{13 \pi ^2 \zeta (5)}{96}+\frac{7 \pi ^4 \zeta (3)}{960}\\
{\bf H}^{(5)}_2(-1)&=&+\frac{519 \zeta (7)}{128}-\frac{5 \pi ^2 \zeta (5)}{16}-\frac{7 \pi ^4\zeta (3)}{480}\\
{\bf H}^{(6)}_1(-1)&=&-3 \zeta (7)+\frac{5 \pi ^2 \zeta (5)}{64}+\frac{7 \pi ^4 \zeta (3)}{960}+\frac{31 \pi ^6 \log (2)}{30240} \\
\hline\\
{\bf H}^{(1)}_7(-1) &=& {\bf H}^{(1)}_7(-1)\\
{\bf H}^{(2)}_6(-1) &=& {\bf H}^{(2)}_6(-1)\\
{\bf H}^{(3)}_5(-1) &=& -9 {\bf H}^{(1)}_7(-1)-4 {\bf H}^{(2)}_6(-1)-\frac{63}{128} {\bf H}^{(2)}_6(+1)+\frac{123 \zeta (3) \zeta (5)}{64}-\frac{127 \pi ^8}{76800}\\
{\bf H}^{(4)}_4(-1) &=& 16 {\bf H}^{(1)}_7(-1)+6 {\bf H}^{(2)}_6(-1)+\frac{63}{32} {\bf H}^{(2)}_6(+1)-\frac{123 \zeta (3) \zeta (5)}{16}+\frac{3097 \pi ^8}{1036800}\\
{\bf H}^{(5)}_3(-1) &=& -15 {\bf H}^{(1)}_7(-1)-5 {\bf H}^{(2)}_6(-1)-\frac{315}{128} {\bf H}^{(2)}_6(+1)+\frac{165 \zeta (3) \zeta (5)}{16}-\frac{2257 \pi ^8}{691200}\\
{\bf H}^{(6)}_2(-1) &=& 6 {\bf H}^{(1)}_7(-1)+2 {\bf H}^{(2)}_6(-1)+\frac{63}{64} {\bf H}^{(2)}_6(+1)-\frac{21 \zeta (3) \zeta (5)}{4}+\frac{193 \pi ^8}{145152}\\
{\bf H}^{(7)}_1(-1) &=& \frac{45 \zeta (3) \zeta (5)}{64}+\frac{63}{64} \zeta (7) \log (2)-\frac{23 \pi ^8}{96768}\\
\hline\\
{\bf H}^{(1)}_8(-1)&=&+\frac{496 \pi ^6 \zeta (3)+4704 \pi ^4 \zeta (5)+40320 \pi ^2 \zeta (7)-1926855 \zeta (9)}{483840}\\
{\bf H}^{(2)}_7(-1)&=&+\frac{4837 \zeta (9)}{512}-\frac{107 \pi ^2 \zeta (7)}{256}-\frac{7 \pi ^4 \zeta
(5)}{180}-\frac{31 \pi ^6 \zeta (3)}{15120}\\
{\bf H}^{(3)}_6(-1)&=&-\frac{7367 \zeta (9)}{512}+\frac{97 \pi ^2 \zeta (7)}{128}+\frac{7 \pi ^4 \zeta (5)}{120}\\
{\bf H}^{(4)}_5(-1)&=&+\frac{3259 \zeta (9)}{512}-\frac{335 \pi ^2 \zeta
(7)}{768}-\frac{343 \pi ^4 \zeta (5)}{11520}\\
{\bf H}^{(5)}_4(-1)&=&+\frac{3385 \zeta (9)}{512}-\frac{25 \pi ^2 \zeta (7)}{64}-\frac{7 \pi ^4 \zeta (5)}{192}\\
{\bf H}^{(6)}_3(-1)&=&-\frac{7451 \zeta (9)}{512}+\frac{187 \pi ^2 \zeta
(7)}{256}+\frac{7 \pi ^4 \zeta (5)}{128}+\frac{31 \pi ^6 \zeta (3)}{40320}\\
{\bf H}^{(7)}_2(-1)&=&+\frac{4873 \zeta (9)}{512}-\frac{63 \pi ^2 \zeta (7)}{128}-\frac{7 \pi ^4 \zeta (5)}{192}-\frac{31 \pi ^6 \zeta
(3)}{20160}\\
{\bf H}^{(8)}_1(-1)&=&-4 \zeta (9)+\frac{21 \pi ^2 \zeta (7)}{256}+\frac{7 \pi ^4 \zeta (5)}{768}+\frac{31 \pi ^6 \zeta (3)}{40320}+\frac{127 \pi ^8 \log (2)}{1209600}\\
\hline \\
{\bf H}^{(9)}_1(-1)&=&+\frac{189 \zeta (3) \zeta (7)}{256}+\frac{225 \zeta (5)^2}{512}+\frac{255}{256} \zeta (9) \log (2)-\frac{563 \pi ^{10}}{19160064}\\
{\bf H}^{(8)}_2(-1)&=&+8 {\bf H}^{(1)}_9(-1)+2 {\bf H}^{(2)}_8(-1)+\frac{255}{256} {\bf H}^{(2)}_8(+1)-\frac{237 \zeta(3) \zeta (7)}{32}-\frac{15 \zeta (5)^2}{4}+\frac{36067 \pi ^{10}}{159667200}\\
{\bf H}^{(7)}_3(-1)&=&-28 {\bf H}^{(1)}_9(-1)-7 {\bf H}^{(2)}_8(-1)-\frac{1785}{512} {\bf H}^{(2)}_8(+1)+\frac{2751 \zeta (3) \zeta (7)}{128}+\frac{615 \zeta (5)^2}{64}-\frac{223 \pi^{10}}{304128}\\
{\bf H}^{(6)}_4(-1)&=&+24 {\bf H}^{(1)}_9(-1)+3 {\bf H}^{(2)}_8(-1)+\frac{2295}{512} {\bf H}^{(2)}_8(+1)-2 {\bf H}^{(3)}_7(-1)-\frac{6831 \zeta (3) \zeta (7)}{256}-\frac{2745 \zeta (5)^2}{256}+\frac{64811 \pi ^{10}}{95800320}\\
{\bf H}^{(5)}_5(-1)&=&+10 {\bf H}^{(1)}_9(-1)+10{\bf H}^{(2)}_8(-1)-\frac{1275}{512} {\bf H}^{(2)}_8(+1)+5 {\bf H}^{(3)}_7(-1)+\frac{3795 \zeta (3) \zeta (7)}{256}+\frac{2775 \zeta (5)^2}{512}+\frac{893 \pi ^{10}}{31933440}\\
{\bf H}^{(4)}_6(-1)&=&-16 {\bf H}^{(1)}_9(-1)-9 {\bf H}^{(2)}_8(-1)+\frac{255}{512} {\bf H}^{(2)}_8(+1)-4{\bf H}^{(3)}_7(-1)-\frac{759 \zeta (3) \zeta (7)}{256}-\frac{255 \zeta (5)^2}{256}-\frac{43817 \pi ^{10}}{159667200}
\end{eqnarray}
Different approach to compute our main sum $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}$.
From here we have
$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}\tag{1}$$
Using the identity
$$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\tag{2}$$
Multiply both sides of (2) by $\frac{\ln^2x}{1-x}$ then integrate from $x=0$ to $1$ we have
\begin{align}
I&=\int_0^1\frac{\ln^2x\ln^2(1+x)}{1-x}\ dx=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1\frac{x^n\ln^2x}{1-x}\ dx\\
&=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(2\zeta(3)-2H_n^{(3)}\right)\\
&=4\zeta(3)\underbrace{\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)}_{\text{use (2) where}\ x=1}+4\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}\\
&=2\ln^22\zeta(3)+4\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}\tag{3}
\end{align}
On the other hand:
\begin{align}
I&=\small{\int_0^1\frac{\ln^2x\ln^2(1+x)}{1-x}\ dx\overset{x\mapsto 1-x}=\int_0^1\frac{\ln^2(1-x)\ln^2(2-x)}{x}=\int_0^1\frac{\ln^2(1-x)}{x}\left(\ln2+\ln\left(1-\frac x2\right)\right)^2\ dx}\\
&=\small{\ln^22\int_0^1\frac{\ln^2(1-x)}{x}\ dx+2\ln2\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ln\left(1-\frac x2\right)\ dx}_{x\mapsto 1-x}+\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ln^2\left(1-\frac x2\right)\ dx}_{\text{use (2)}}}\\
&=\small{2\ln^22\zeta(3)+2\ln2\underbrace{\int_0^1\frac{\ln^2x}{1-x}\ln\left(\frac{1+x}{2}\right)\ dx}_{\text{use (1)}}+2\sum_{n=1}^\infty\frac1{2^n}\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx}\\
&=2\ln^22\zeta(3)+4\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}+2\sum_{n=1}^\infty\frac1{2^n}\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\
&=\small{2\ln^22\zeta(3)+4\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}+2\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}+2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}}\quad \quad \quad \quad \text{(4)}
\end{align}
From (3) and (4) we conclude that
$$\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}=\\
\small{\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}-\frac12\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}-\frac12\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}
+\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}+\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}}\tag{5}$$
We have the following results:
$$S_1=\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$
$$S_2=\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$
$$S_3=\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52$$
$$S_4=\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52$$
$$S_5=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$
$$S_6=\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{64}\zeta(5)-\frac{37}{16}\ln2\zeta(4)-\frac{9}{16}\zeta(2)\zeta(3)\\+\frac{7}{16}\ln^22\zeta(3)+\frac1{12}\ln^32\zeta(2)-\frac{1}{40}\ln^52$$
By substituting these results in (5) we get
$$\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}=4 \operatorname{Li}_5\left(\frac{1}{2}\right)+2\ln2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{167}{32}\zeta(5)+\frac{49}{16}\ln2\zeta(4)-\frac{3}{8}\ln^22\zeta(3)\\-\frac{1}{6}\ln^32\zeta(2)+\frac{1}{16}\zeta(2)\zeta(3)+\frac{1}{20}\ln^52$$
NOTE:
$S_1$ and $S_2$ can be found here, $S_3$ and $S_4$ can be found here and $S_5$ and $S_6$ can be found here.
Best Answer
We proved in this solution that
$$\mathcal{I}=\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2(2)\tag1$$
and we proved here that
$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag2$$
By $(2)$ we get
$$\mathcal{I}=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1\frac{ x^{2n}}{1+x}dx$$
Using the identity $$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
it follows that
$$\mathcal{I}=-2\ln(2)\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}+2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{2}}{2n+1}$$
$$=-2\ln(2)\mathcal{S}_1-2\mathcal{S}_2+2\mathcal{S}_3\tag3$$
For $\mathcal{S}_1$ and $\mathcal{S}_3$, we use the classical identity:
$$\sum_{n=1}^\infty(-1)^n f(2n)=\Re\sum_{n=1}^\infty i^n f(n)$$
Therefore
$$\mathcal{S}_1=\Re\sum_{n=1}^\infty i^n\frac{H_n}{n+1}=\Re\left\{\frac{\ln^2(1-i)}{i}\right\}=-\frac{\pi}{8}\ln(2)\tag4$$ where we used $\sum_{n=1}^\infty x^n\frac{H_n}{n+1}=\frac{\ln^2(1-x)}{x}$ which follows from integrating $\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$.
Similarly,
$$\mathcal{S}_3=\Re\sum_{n=1}^\infty i^n\frac{H_n^2}{n+1}$$
Using the generating function
$$\sum_{n=1}^\infty x^{n}\frac{ H_n^{2}}{n+1}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}$$
it follows that
$$\mathcal{S}_3=\Re\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3i}\right\}$$
$$=\Im\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3}\right\}$$
$$=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{16}\ln^2(2)+\frac{5}{96}\pi^3\tag5$$
Plug the results of $(4)$ and $(5)$ in $(3)$ we get
$$\mathcal{I}=4\Im\{\operatorname{Li}_3(1-i)\}+\ln(2)\ G+\frac{5\pi}{8}\ln^2(2)+\frac{5}{48}\pi^3-2\mathcal{S}_2\tag6$$
By $(1)$ and $(6)$ we get