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*}
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
\begin{align*}
I_1 =& -2\text{Li}_5(1/2) - 2\ln 2\text{Li}_4(1/2) + \frac {279}{64}\zeta(5) -\frac {3\pi^2}{32}\zeta(3)-\frac {7\ln^2 2}{8} \zeta(3)+\frac {\pi^2\ln^3 2}{18}-\frac{\ln^5 2}{15}.
\end{align*}
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
\begin{align*}
I_2=\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{8}\zeta(3)-\frac{37\pi^4\ln 2}{1440}-\frac{\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{24}.
\end{align*}
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
$$
I_3 =\frac {7\ln^2 2}{16}\zeta(3).
$$
From $S = I_1+I_2 + I_3$, we finally get
\begin{align*}
\sum_{k=1}^\infty \frac{H_k^2}{k^32^k} =& -2\text{Li}_5(1/2) -\ln 2\ \text{Li}_4(1/2) + \frac{279}{64}\zeta(5) - \frac{3\pi^2}{32}\zeta(3) + \frac{7\ln^2 2}{16}\zeta(3) - \frac{37\pi^4 \ln 2}{1440}\\
& + \frac{\pi^2 \ln^3 2}{72} - \frac{\ln^5 2}{40}.
\end{align*}
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:
\begin{eqnarray*}
&\sum_{k=1}^\infty \frac{H_k^2}{k^2 2^k} = -\frac 1{\ln 2} I_2 - \frac{2}{\ln 2} I_3=-\text{Li}_4(1/2) -\frac{7\ln 2}4\zeta(3) +\frac{37 \pi^4}{1440} +\frac{\pi^2\ln^2 2}{24} -\frac{\ln^4 2}{24},\\
&\sum_{k=1}^\infty \frac{H_k^2}{k 2^k} = \frac{2}{\ln^2 2} I_2 = \frac{7}{8}\zeta(3).
\end{eqnarray*}
Best Answer
\begin{align}J&=\int_0^1\frac{\text{Li}_4(x)\ln x}{1-x}dx\\ &=-\frac{1}{6}\int_0^1\int_0^1\frac{x\ln^3 t\ln x}{(1-x)(1-tx)}dtdx\\ &=-\frac{1}{6}\int_0^1\int_0^1\left(\frac{\ln^3 t\ln x}{(1-x)(1-t)}-\frac{\ln^3 t\ln x}{(1-tx)(1-t)}\right)dtdx\\ &=-\zeta(2)\zeta(4)+\frac{1}{6}\int_0^1\int_0^1\frac{\ln^3 t\Big(\ln(t x)-\ln t\Big)}{(1-tx)(1-t)}dtdx\\ &=-\zeta(2)\zeta(4)+\frac{1}{6}\int_0^1 \frac{\ln^3 t}{t(1-t)}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt-\frac{1}{6}\int_0^1 \frac{\ln^4 t}{t(1-t)}\left(\int_0^t\frac{1}{1-x}dx\right)dt\\ &\overset{\text{IBP}}=-\zeta(2)\zeta(4)-\frac{1}{24}\int_0^1\frac{\ln^5 t}{1-t}dt+\frac{1}{6}\int_0^1\frac{\ln^3 t}{1-t}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt+\frac{1}{30}\int_0^1\frac{\ln^5 t}{1-t}dt+\\ &\frac{1}{6}\int_0^1\frac{\ln^4 t\ln(1-t)}{1-t}dt\\ &=\zeta(6)-\zeta(2)\zeta(4)+\frac{1}{6}\underbrace{\int_0^1\frac{\ln^3 t}{1-t}\left(\int_0^t\frac{\ln x}{1-x}dx \right)dt}_{=A}+\frac{1}{6}\int_0^1\frac{\ln^4 t\ln(1-t)}{1-t}dt\\ A&=\int_0^1\int_0^1\frac{t\ln(tx)\ln^3 t}{(1-t)(1-tx)}dtdx=\int_0^1\int_0^1\left(\frac{\ln(tx)\ln^3 t}{(1-t)(1-x)}-\frac{\ln(tx)\ln^3 t}{(1-x)(1-tx)}\right)dtdx\\ &\overset{\text{Fubini}}=6\zeta(2)\zeta(4)+\int_0^1\int_0^1\left(\frac{\ln^4 t}{(1-t)(1-x)}-\frac{\ln(tx)\Big(\ln(tx)-\ln x\Big)^3}{(1-x)(1-tx)}\right)dtdx\\ &=6\zeta(2)\zeta(4)+\int_0^1\int_0^1\left(\frac{\ln^4 t}{(1-t)(1-x)}-\frac{\ln^4(tx)}{(1-x)(1-tx)}\right)dtdx+\\&3\!\int_0^1\!\!\int_0^1\frac{\ln^3(tx)\ln x}{(1-x)(1-tx)}\!dtdx-3\int_0^1\int_0^1\frac{\ln^2(tx)\ln^2 x}{(1-x)(1-tx)}dtdx+\\&\int_0^1\int_0^1\frac{\ln(tx)\ln^3 x}{(1-x)(1-tx)}dtdx\\ &=6\zeta(2)\zeta(4)+\int_0^1\frac{1}{1-x}\left(\int_0^1\frac{\ln^4 t}{1-t}dt-\frac{1}{x}\int_0^x\frac{\ln^4 t}{1-t}dt\right)dx+\\&3\int_0^1\frac{\ln x}{(1-x)x}\left(\int_0^x \frac{\ln^3 t}{1-t}dt\right)dx-3\int_0^1\frac{\ln^2 x}{(1-x)x}\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)dx+\\&\int_0^1\frac{\ln^3 x}{(1-x)x}\left(\int_0^x \frac{\ln t}{1-t}dt\right)dx\\ &\overset{\text{IBP}}=6\zeta(2)\zeta(4)-30\zeta(6)-\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt+3\underbrace{\int_0^1\frac{\ln x}{1-x}\left(\int_0^x \frac{\ln^3 t}{1-t}dt\right)dx}_{\text{IBP}}-\\ &3\underbrace{\int_0^1\frac{\ln^2 x}{1-x}\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)dx}_{=2\zeta(3)^2}+A\\ &=24\zeta(2)\zeta(4)-30\zeta(6)-6\zeta(3)^2-\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt-2A\\ &=8\zeta(2)\zeta(4)-10\zeta(6)-2\zeta(3)^2-\frac{1}{3}\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt\\ J&=-\frac{2}{3}\zeta(6)+\frac{1}{3}\zeta(2)\zeta(4)-\frac{1}{3}\zeta(3)^2+\frac{1}{9}\underbrace{\int_0^1\frac{\ln(1-t)\ln^4 t}{1-t}dt}_{=B} \end{align} \begin{align} B&\overset{\text{IBP}}=\left[\left(\int_0^t\frac{\ln^4 x}{1-x}dx-\int_0^1\frac{\ln^4 x}{1-x}dx\right)\ln(1-t)\right]_0^1+\\&\int_0^1\int_0^1\frac{1}{1-t}\left(\frac{t\ln^4(tx)}{1-tx}-\frac{\ln^4 x}{1-x}\right)dtdx\\ &=\int_0^1\int_0^1\left(\frac{\ln^4(tx)}{(1-t)(1-x)}-\frac{\ln^4(tx)}{(1-x)(1-tx)}-\frac{\ln^4 x}{(1-t)(1-x)}\right)dtdx\\ &=\int_0^1\int_0^1\frac{4\ln t\ln^3 x+6\ln^2 t\ln^2 x+4\ln^3 t\ln x}{(1-t)(1-x)}dtdx+\\ &\int_0^1\int_0^1\left(\frac{\ln^4 t}{(1-t)(1-x)}-\frac{\ln^4(tx)}{(1-x)(1-tx)}\right)dtdx\\ &\overset{\text{Fubini}}=24\zeta(2)\zeta(4)+24\zeta(3)^2+24\zeta(2)\zeta(4)-\int_0^1\frac{1}{x}\int_0^x\frac{\ln^4 t}{1-t}dt+\int_0^1\frac{1}{1-x}\int_x^1\frac{\ln^4 t}{1-t}dt\\ &\overset{\text{IBP}}=48\zeta(2)\zeta(4)+24\zeta(3)^2-120\zeta(6)-B\\ &=\boxed{24\zeta(2)\zeta(4)+12\zeta(3)^2-60\zeta(6)} \end{align} Therefore, \begin{align}\boxed{J=-\frac{22}{3}\zeta(6)+3\zeta(2)\zeta(4)+\zeta(3)^2=\zeta(3)^2-\frac{5\pi^6}{2268}}\end{align}