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
$$\mathcal{S}_2=\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{8}\ln^2(2)+\frac{3}{64}\pi^3$$
Considering the algebraic identity
\begin{align*}
&(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\
&\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b
\end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that
\begin{align*}
2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =& - \frac 1 2\int_0^1 {\ln^4(1+x)\over x}d x \\
&-\frac 12 \int_0^1 \frac{\ln^4(1+x) + 6\ln^2(1-x)\ln^2(1+x)}{x}dx\\
&+3\int_0^1 \frac{\ln^3(1-x)\ln(1+x) + \ln(1-x)\ln^3(1+x)}{x}dx\\
&- \int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}dx\\
=:& -I_1 - I_2 + I_3 -I_4.
\end{align*}
For $I_1$, make substitution $y = \frac x {1+x}$ to get:
\begin{align*}
I_1 =& \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{y(1-y)} dy \\
=& \frac 1 2\underbrace{ \int_0^{\frac 12} \frac{\ln^4(1-y)}{y} dy}_{z=1-y}+ \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{1-y} dy\\
=& \frac 1 2 \int_{\frac 1 2 }^1 \frac{\ln^4 z} {1-z} dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \int_{\frac 1 2}^1 z^{n-1}\ln^4 z\ dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \frac{\partial^4}{\partial n^4}\left[\frac 1 n - \frac 1 {n2^n}\right] + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \left[\frac{24}{n^5} - \frac {24}{n^52^n} - \frac{24 \ln 2}{n^42^n}-\frac{12\ln^2 2}{n^3 2^n}-\frac{4\ln^3 2}{n^2 2^n} - \frac{\ln^4 2}{n2^n}\right] + \frac {\ln^5 2}{10}\\
=&12\zeta(5) - 12\text{Li}_5(1/2) - 12\ln 2 \text{Li}_4(1/2) -6\ln^2 2 \text{Li}_3(1/2) -2\ln^3 2\text{Li}_2(1/2)-\frac {2}{5}\ln^5 2\\
=&\boxed{-12\Big(\text{Li}_5(1/2) + \ln 2\text{Li}_4(1/2)-\zeta(5)\Big)-{21 \over 4}\zeta(3)\ln^2 2 +{1\over 3} \pi^2 \ln^3 2-{2 \over 5} \ln^5 2}
\end{align*} where the well-known values
\begin{align*}\text{Li}_2(1/2) = {\pi^2 \over 12}-{\ln^2 2\over 2} , \qquad \text{Li}_3(1/2) ={7\zeta(3) \over 8} -{\pi^2 \ln 2\over 12} + {\ln^3 2 \over 6}
\end{align*} are used.
Actually, $I_2$ was already evaluated by the OP here using the algebraic identity $$b^4 + 6a^2b^2 = \frac {(a-b)^4} 2+\frac{(a+b)^4}{2} -a^4.$$
It holds that
$$
\boxed{I_2 = \frac {21}{8} \zeta(5).}
$$
In fact, the value of $I_3$ can also be found in the previous answer of @Przemo's. For $I_3$, one can use the algebraic relation $3(a^3b + ab^3) =\frac 3 8 \left[ (a+b)^4 - (a-b)^4\right]$.
This gives
\begin{align*}
I_3=& \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4(1-x^2)}{x} dx}_{x^2 = y} - \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x} dx}_{\frac{1-x}{1+x} = y}\\
=&\frac 3 {16}\underbrace{\int_0^1 \frac{\ln^4(1-y)}{y} dy }_{1-y\mapsto y}- \frac 3 4 \int_0^1 \frac{\ln^4 y}{1-y^2} dy\\
=&\frac 3 {16}\int_0^1 \frac{\ln^4 y}{1-y} dy - \frac 3 4 \sum_{n=0}^\infty \int_0^1 y^{2n} \ln^4 y \ dy\\
=&\frac 3 {16}\sum_{n=1}^\infty \int_0^1 y^{n-1}\ln^4 y \ dy - \frac 3 4 \sum_{n=0}^\infty \frac {24}{(2n+1)^5}\\
=&\frac 3 {16}\sum_{n=1}^\infty \frac{24}{n^5} - 18 \sum_{n=0}^\infty \frac {1}{(2n+1)^5}\\
=&\frac {9}{2} \zeta(5)- 18\cdot \frac {31}{32}\zeta(5)\\
=&\boxed{-\frac{207}{16}\zeta(5)}
\end{align*} as can be found in @Przemo's answer.
For $I_4$, make substitution $ \frac{1-x}{1+x}\mapsto x$ to get
\begin{align*} I_4 = &2\int_0^1 \frac{\ln^3 x \ln\left(\frac 2 {1+x}\right)}{1-x^2} dx \\
=&2\ln 2 \int_0^1 \frac{\ln^3 x}{1-x^2} dx - \underbrace{2\int_0^1\frac{\ln^3 x \ln(1+x)}{1-x^2} dx }_{=:J}\\
=& 2\ln 2\sum_{n=0}^\infty \int_0^1 x^{2n} \ln^3 x\ dx - J\\
=& - 12\ln 2 \underbrace{\sum_{n=0}^\infty \frac 1 {(2n+1)^4}}_{\frac{15}{16}\zeta(4) = \frac{\pi^4}{96}} - J \\
=& -\frac{\pi^4 \ln 2}{8} - J.
\end{align*}
\begin{align*}
J = &\int_0^1\frac{2\ln^3 x \ln(1+x)}{1-x^2} dx \\
=& \underbrace{\int_0^1 \frac{\ln^3 x \ln(1+x)}{1+x}dx}_{=:A} + \int_0^1 \frac{\ln^3 x \ln(1+x)}{1-x}dx\\
=& A + \int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \int_0^1 \frac{(1+x)\ln^3 x \ln(1-x^2)}{1-x^2}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \underbrace{\int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x^2}dx }_{=:B}+\underbrace{\int_0^1 \frac{x\ln^3 x \ln(1-x^2)}{1-x^2}dx}_{x^2 \mapsto x}-\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + B - \underbrace{\frac {15}{16} \int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx}_{=:C}\\
=&A + B - C.
\end{align*}
For $A$, we can use the McLaurin series of
$$
\frac{\ln (1+x)}{1+x} = \sum_{n=0}^\infty (-1)^{n-1}H_n x^n
$$ ($H_0= 0$) to get
\begin{align*}
A = & \sum_{n=0}^\infty (-1)^{n-1}H_n \int_0^1 x^n\ln^3 x \ dx \\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_n}{(n+1)^4}\\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_{n+1}}{(n+1)^4} - 6\sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^5}\\
=&6 \sum_{n=1}^\infty \frac{(-1)^{n-1}H_{n}}{n^4} - 6\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^5}\\
=& 6\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-6\cdot \frac{15}{16}\zeta(5)\\
=& \frac{87}{16}\zeta(5) - \frac{\pi^2 \zeta(3)}{2}.
\end{align*}
Here, the known value of $ \sum_{n=1}^\infty (-1)^{n-1}{H_n \over n^4}$ is used.
For $B$, make substitution $u = x^2$ to get
\begin{align*}
B =& \frac 1 {16} \int_0^1 \frac{\ln^3 u \ln(1-u)}{\sqrt u (1-u)} du \\
=& \frac 1 {16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=\frac 1 2, y = 0^+}
\end{align*} where $\text{B}(\cdot,\cdot)$ is Euler's Beta function. We can use the fact that
\begin{align*}
\lim_{y\to 0^+}\frac{\partial^2}{\partial x\partial y} \text{B}(x,y) = -\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]
\end{align*} to get
\begin{align*}
B =& \frac 1 {16}\frac{d^2}{dx^2}\left[-\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]\right]_{x=\frac 1 2}\\
=&\frac 1 {16} \left[-\frac 1 2 \psi''''(1/2) + \psi'''(1/2)\big[\psi(1/2) + \gamma\big] + 3\psi'(1/2)\psi''(1/2)\right]\\
=& \frac 1 {16}\left[-21\pi^2 \zeta(3) + 372\zeta(5) - 2\pi^4 \ln 2\right]
\end{align*} which can be evaluated using the series representations of polygamma functions $$\psi(x) +\gamma = - \frac 1 x +\sum_{n=1}^\infty \frac 1 n - \frac 1 { n+x},\\
\psi'(x) = \sum_{n=0}^\infty \frac 1 {(n+x)^2}$$ and the derived fact that $\psi(\tfrac 1 2 )+\gamma = -2\ln 2$ and $\psi^{(k)}(\tfrac 1 2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$ for $k\ge 1$.
For $C$, we can use the same method as used in the evaluation of $B$. It holds that
\begin{align*}
C =& \frac {15}{16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=1, y = 0^+}\\
=&\frac {15} {16}\left[-\frac 1 2 \psi''''(1) + \psi'''(1)\big[\psi(1) + \gamma\big] + 3\psi'(1)\psi''(1)\right]\\
=&\frac{15}{16}\left[12\zeta(5) -6\zeta(2)\zeta(3)\right]\\
=&\frac {45}{4}\zeta(5) -\frac {15\pi^2 \zeta(3)}{16}
\end{align*} where $\psi(1) +\gamma = 0$, $\psi'(1) = \zeta(2)$, $\psi''(1) = -2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Combining $A,B,C$, we have that $$J =A+B-C= \frac{279}{16}\zeta(5) -\frac{7\pi^2\zeta(3)}{8} - \frac{\pi^4 \ln 2}{8}$$ and
$$
\boxed{I_4 = -\frac{\pi^4 \ln 2}{8} - J = -\frac{279}{16}\zeta(5)+\frac{7\pi^2\zeta(3)}{8}}
$$
Finally, these evaluate $\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =\frac 1 2\big[-I_1-I_2+I_3-I_4\big]$ as follows.
\begin{align*}
\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&\ 6\text{Li}_5(1/2) + 6\ln 2\ \text{Li}_4(1/2)-\frac{81}{16}\zeta(5)-{7\pi^2 \over 16}\zeta(3)\\
&+\frac{21\ln^2 2}{8}\zeta(3)- \frac{1}{6}\pi^2\ln^3 2+\frac{1}{5}\ln^5 2.
\end{align*}
Using the identity given in the OP, we get the desired integral $I$
\begin{align*}
\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\
&-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}.
\end{align*}
Best Answer
Using the generating function of $\displaystyle\{H_k^2\}_{k=1}^\infty$: $$ \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} = \sum_{k=1}^\infty H_k^2 x^k $$ we can observe that \begin{align*} S =& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1{k^2}\\ =& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1 2\int_0^1 x^{k-1}\ln^2 x\ dx\\ =&\frac 1 2 \int_0^1 \left(\sum_{k=1}^\infty H_k^2\left(\frac x 2\right)^k\right)\frac{\ln^2 x}{x} dx\\ =& \frac 1 2 \int_0^{\frac 1 2}\left(\sum_{k=1}^\infty H_k^2 x ^k\right) \frac{\ln^2 (2x)}{x} dx \\ =& \frac 1 2 \int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln^2 x}{x} dx \\ &+ \ln 2\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln x}{x} dx \\ &+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{dx}{x}\\ =&: I_1 + I_2 + I_3. \end{align*}
For $I_1$, we have \begin{align*} I_1=&\frac 1 2 \int_0^{\frac 12 } \frac{\big[\text{Li}_2(x)+\ln^2(1-x)\big]\ln^2 x}{x(1-x)}dx \\ =& \frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{x}dx+\frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{1-x}dx +\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2 (1-x)\ln^2 x}{x(1-x)} dx\\ =&:I_1'+I_1''+I_1'''. \end{align*}
For $I_1'$, we integrate by parts twice to get \begin{align*} I_1' \underset{\text{IBP}}{=}& \frac 1 2\left[ \text{Li}_3(x)\ln^2 x\right]^{1/2}_0 -\int_0^{\frac 1 2}\frac{\text{Li}_3(x) \ln x}{x} dx\\ \underset{\text{IBP}}{=}&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 - \left[\text{Li}_4(x)\ln x\right]^{1/2}_0 +\int_0^{\frac 1 2 }\frac{\text{Li}_4(x)}x dx\\ =&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 + \ln 2\ \text{Li}_4(1/2)+\text{Li}_5(1/2)\\ =&\boxed{\text{Li}_5(1/2)+\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{16}\zeta(3)-\frac {\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{12}} \end{align*} where the well-known value of $ \text{Li}_3(1/2) = \frac 78 \zeta(3) -\frac{\pi^2\ln 2}{12}+\frac{\ln^3 2}{6} $ is used to simplify.
For $I_1''$, by integrating by parts, \begin{align*} I_1'' \underset{\text{IBP}}{=}& \frac 1 2 \int_0^{\frac 1 2} \ln(1-x)\left[\frac{2\ln x\text{Li}_2(x)}{x} - \frac{\ln(1-x)\ln^2 x}{x}\right]dx +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\ =& {\int_0^{\frac 1 2} \ln x\frac{\ln(1-x)\text{Li}_2(x)}{x} dx}-\underbrace{\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln^2 x}{x}dx}_{=:J} +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\ \underset{\text{IBP}}{=}&\frac{\ln 2}2\text{Li}_2^2(1/2) +{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\ =&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J. \end{align*} The well-known value of $\text{Li}_2(1/2) = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$ is used to simplify. In fact, the integral ${\int_0^{1/2}\frac{\text{Li}_2^2(x)}{x} dx}$ was already evaluated in my previous answer here: \begin{align*} {\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx} = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\ &-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}. \end{align*}
For $J$, we make substitution $y= \frac{x}{1-x}$ to get \begin{align*} J=&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y(1+y)}dy\\ =&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy-\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\ =&:K-L. \end{align*}
For $K$, expanding $\ln^2\left(\frac y {1+y}\right)=\big[\ln y -\ln(1+y)\big]^2$ and integrating by parts we get \begin{align*} K =&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy\\ =& \frac 12{ \int_0^1\frac{\ln^2 y\ln^2(1+y)}{y} dy}-{\int_0^1 \frac{\ln y\ln^3(1+y)}{y} dy}+\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy\\ \underset{\text{IBP}}{=}&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \frac 3 2\int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy \end{align*} Doing the same thing for $L$, \begin{align*} L = & \frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\ =& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy- {\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy} +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\ \underset{\text{IBP}}{=}& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy+\frac 1 4 \int_0^1 \frac{\ln^4(1+y)}y dy +\frac{\ln^5 2}{10}. \end{align*} This gives that \begin{align*} J=&K-L\\ =&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy -\frac{\ln^5 2}{10}\\ =&:-V_1+V_2+V_3 -\frac{\ln^5 2}{10}. \end{align*}
For $V_1$, we can use the Maclaurin series of $\frac{\ln (1+y)}{1+y} = \sum_{k=0}^\infty (-1)^{k-1} H_k y^k$ to get \begin{align*} V_1=&\frac 1 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy \\ =& \frac 1 3\sum_{k=0}^\infty (-1)^{k-1}H_k {\int_0^1 y^k\ln^3 y\ dy} \\ =& \frac{-6}{3}\sum_{k=0}^\infty \frac{(-1)^{k-1}H_k}{(k+1)^4}\\ =&2\sum_{k=0}^\infty \frac{(-1)^k \left(H_{k+1}-\frac 1{k+1}\right)}{(k+1)^4} \\ =&2 \sum_{k=1}^\infty \frac{(-1)^{k-1}H_k}{k^4} -2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}\tag{$k+1\mapsto k$}\\ =&2\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-2\cdot \frac{15}{16}\zeta(5)\\ =&\frac{29}{16}\zeta(5) - \frac{\pi^2}{6}\zeta(3) \end{align*} where the known value of alternating Euler sum $\sum_{k=1}^\infty \tfrac{(-1)^{k-1}H_k}{k^4}$ is used.
For $V_2$, we consider the algebraic identity $$ 6a^2b^2 = (a-b)^4 - a^4 +4a^3b +4ab^3 -b^4 $$ with $a=\ln y$ and $b = \ln(1+y)$ to get \begin{align*} V_2 =&\frac 1 6{\int_0^1 \frac{\ln^4\left(\frac y{1+y}\right)}{1+y} dy}-\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +\underbrace{\frac 2 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy}_{=2V_1}\\ &+\frac 2 3\underbrace{\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy}_{=- V_3\text{ by IBP}} -\frac 1 6\int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\ =&\frac 1 6 \int_0^{\frac 1 2} \frac{\ln^4 x}{1-x} dx -\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +2V_1-\frac 2 3 V_3 -\frac{\ln^5 2}{30}.\tag{$\tfrac y{1+y}= x$} \end{align*} For the first integral, we have \begin{align*} W:=&\frac 1 6{\int_0^{\frac 1 2 }\frac{\ln^4 x}{1-x} dx}\tag{$2x\mapsto x$} \\ =& \frac 1 6\int_0^1 \frac{\ln^4(\tfrac x 2)}{2-x}dx\\ =&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\int_0^1 x^{k-1}\Big[\ln^4 x -4\ln 2 \ln^3 x + 6\ln^2 2\ln^2 x - 4\ln^3 2 \ln x + \ln^4 2\Big]dx\\ =&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\left[\frac{24}{k^5} + \frac{24\ln 2}{k^4} +\frac {12\ln^2 2}{k^3} +\frac{4\ln^3 2}{k^2} +\frac{\ln^4 2}{k}\right]\\ =&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + 2\ln^2 2\text{Li}_3(1/2) + \frac{2\ln^3 2}{3}\text{Li}_2(1/2) + \frac{\ln^5 2}6\\ =&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + \frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^2\ln^3 2}{9}+ \frac{\ln^5 2}6. \end{align*}
For the second integral, we have \begin{align*} \frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy =& \frac 1 6\sum_{k=1}^\infty (-1)^{k-1} \int_0^1 y^{k-1}\ln^4 y \ dy \\ =& \frac 1 6 \sum_{k=1}^\infty(-1)^{k-1} \frac{24}{k^5}\\ =&\frac{15}{4}\zeta(5). \end{align*}
This gives $$ V_2 = W +2V_1-\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}. $$
For $V_3$ we have \begin{align*} V_3=&\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy \tag{$y\mapsto y+1$}\\ =& \frac 1 {4}{ \int_1^2 \frac{\ln^4 y }{y-1} dy} \tag{$\tfrac 1 y\mapsto y$}\\ =&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y(1-y)} dy\\ =&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y} dy+\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{1-y} dy\\ =&\frac{\ln^5 2}{20} + \frac 1 {4} \int_0^1 \frac{\ln^4 y}{1-y} dy -\frac 1 {4} \underbrace{\int_0^{\frac 1 2}\frac{\ln^4 y}{1-y}dy}_{=6 W}\\ =&\frac{\ln^5 2}{20} +\frac 1 {4} \sum_{k=1}^\infty \int_0^1 y^{k-1}\ln^4 y\ dy- \frac 3 2 W\\ =&\frac{\ln^5 2}{20} +6\zeta(5)- \frac 3 2 W. \end{align*}
Combining $V_1$, $V_2$ and $V_3$, we get \begin{align*} J = & V_2 -V_1+V_3 -\frac{\ln^5 2}{10}\\ =& \left[W+2V_1 -\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}\right]-V_1+V_3 -\frac{\ln^5 2}{10}\\ =& W+V_1+\frac 1 3 V_3-\frac{15}{4}\zeta(5)-\frac{2\ln^5 2}{15}\\ =&\frac 1 2 W+V_1 -\frac 7 4\zeta(5) -\frac{7\ln^5 2}{60}\\ =&2\text{Li}_5(1/2) +2\ln 2\ \text{Li}_4(1/2) +\frac 1 {16}\zeta(5) -\frac{\pi^2}6 \zeta(3) +\frac {7\ln^2 2}{8} \zeta(3) -\frac{\ln^2 2\pi^3}{18}-\frac{\ln^5 2}{30}. \end{align*} This gives \begin{align*} I_1'' =&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx-J\\ =&\boxed{\small -3\text{Li}_5(1/2) -3\ln 2\text{Li}_4(1/2) +\frac{23}{64}\zeta(5) +\frac {23\pi^2}{96}\zeta(3) -\frac {21\ln^2 2}{16}\zeta(3) +\frac{7\pi^2\ln^3 2}{72} - \frac{3\ln^5 2}{20}.} \end{align*}
For $I_1'''$, we exploit the symmetric nature of the integrand to write \begin{align*} I_1''' :=& \frac 1 2\int_0^{\frac 1 2} \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx\\ =& \frac 1 4\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx \\ =& \frac 1 4\underbrace{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x} dx}_{1-x\mapsto x}+\frac 1 4{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx}\\ =&\frac 1 2\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx\\ =&\frac 1 2 \left[\frac{\partial^4}{\partial x^2 \partial y^2 } \text{B}(x,y)\right]_{x=1,y=0^+} \end{align*} where $\text{B}(x,y)=\tfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is Euler's Beta function. Now we can use the fact that \begin{align*} \lim_{y\to 0^+}\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,y) =&-\frac 1 3\psi'''(x)+\psi''(x)\Big[\psi(x) +\gamma\Big] + \psi'(x)\Big[\psi'(x)-\zeta(2) - \big[\psi(x) + \gamma\big]^2\Big] \end{align*} to obtain \begin{align*} I_1'''=& \frac 1 2\frac d{dx}\left[\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,0^+)\right]_{x=1} \\ =& -\frac 1 6 \psi''''(1) +\psi'(1)\psi''(1) \\ =&\boxed{4\zeta(5) -\frac{\pi^2}3 \zeta(3)} \end{align*} where the values of $\psi(1) +\gamma = 0$, $\psi'(1) =\zeta(2)$, $\psi''(1) =-2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Finally, from $I_1 = I_1'+I_1''+I_1'''$ we get
For $I_2$, we observe that \begin{align*}\require{cancel} I_2 =& \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x(1-x)} dx \\ =& \ln 2 {\int_0^{\frac 12} \frac{\text{Li}_2(x) \ln x}{1-x} dx}+\ln 2\int_0^{\frac 12} \frac{\ln^2(1-x)\ln x}{1-x} dx+ \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x} dx\\ \underset{\text{IBP}}{=}&-\ln^3 2\ \text{Li}_2(1/2)+\ln 2 \int_0^{\frac 12} \ln(1-x)\frac{-\cancel{\ln(1-x)\ln x}+\text{Li}_2(x)}{x} dx \\ &+\ln2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln x}{1-x} dx + \ln 2{ \int_0^{\frac 12} \frac{\big[\text{Li}_2(x)+\cancel{ \ln^2(1-x)}\big] \ln x}{x} dx}\\ \underset{\text{IBP}}{=}&\small-\ln^3 2 \text{Li}_2(1/2)-\tfrac{\ln 2}{2} \left[\text{Li}^2_2(x)\right]^{1/2}_0-\frac{\ln^5 2}3+{\frac{\ln 2}3{\int_0^{\frac 1 2} \frac{\ln^3(1-x)}{x} dx}} -\ln^2 2\ \text{Li}_3(1/2)-\ln 2\int_0^{\frac 1 2} \frac{\text{Li}_3(x)}{x} dx\normalsize\\ =&-\ln^3 2\ \text{Li}_2(1/2) -\tfrac{\ln 2}{2} \text{Li}^2_2(1/2)-\frac{\ln^5 2}3+ \small\underbrace{\frac{\ln 2}3{\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx}}_{1-x\mapsto x, \ =:I_2'}\normalsize-\ln^2 2\ \text{Li}_3(1/2)-\ln 2\ \text{Li}_4(1/2)\\ =&-\ln 2\ \text{Li}_4(1/2) -\frac{7\ln^2 2}8 \zeta(3) -\frac{\pi^4 \ln 2}{288}+\frac {\pi^2\ln^3 2}{24} -\frac{\ln^5 2}{8} + I_2'. \end{align*}
For $I_2'$, by integrating by parts, we have \begin{align*} I_2' =& \frac{\ln 2}3\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx\\ =&\frac{\ln 2}3\int_{0}^1 \frac{\ln^3 x}{1-x} dx -\frac{\ln 2}3{\int_{0}^{\frac 12} \frac{\ln^3 x}{1-x} dx}\tag{$x=\tfrac y 2$}\\ =&\frac{\ln 2}3\sum_{k=1}^\infty {\int_0^1 x^{k-1}\ln^3 x\ dx}-\underbrace{\frac{\ln 2}3\int_{0}^{1} \frac{\ln^3 (\tfrac y 2)}{2-y} dy}_{=:A}\\ =&-\frac{\pi^4\ln 2}{45}-A. \end{align*} \begin{align*} A=&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\ln^3 (\tfrac y 2) dy\\ =&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\left[\ln^3 y - 3\ln 2\ln^2 y +3\ln^2 2\ln y -\ln^3 2\right]dy\\ =&-\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\left[\frac 6 {k^4}+\frac{6\ln 2}{k^3} +\frac{3\ln^2 2}{k^2} +\frac{\ln^3 2}{k}\right]\\ =& -2\ln 2\ \text{Li}_4(1/2) - 2\ln^2 2\ \text{Li}_3(1/2)-\ln^3 2\ \text{Li}_2(1/2)-\frac{\ln^5 2}3\\ =&-2\ln 2\ \text{Li}_4(1/2)-\frac{7\ln^2 2}{4}\zeta(3)+\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}. \end{align*}
This gives $$ I_2'= 2\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^4\ln 2}{45}-\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6} $$ and
For $I_3$, we have \begin{align*} I_3=&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)+\ln^2(1-x)}{x(1-x)}dx\\ =&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)}{x}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} {\left[\frac{\ln^2(1-x)}{x}+\frac{\text{Li}_2(x)}{1-x}\right]}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\ln^2(1-x)}{1-x}dx\\ =&\frac{\ln^2 2}{2}\text{Li}_3(1/2) +\frac{\ln^2 2}{2}\big[-\ln(1-x)\text{Li}_2(x)\big]^{1/2}_0+\frac{\ln^5 2}6. \end{align*} Using the well-known values of $\text{Li}_3(1/2)$ and $\text{Li}_2(1/2)$, this simplifies to
From $S = I_1+I_2 + I_3$, we finally get
We can observe that the values of $I_2$ and $I_3$ can be used to evaluate sums of lower order in a similar way: