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*}
The blue sum, with the denominator rearranged, comes immediately from the result given in Section 4.59, page $313$, from the book (Almost) Impossible Integrals, Sums, and Series.
$$\zeta(4)$$
$$=\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{ \left(H_{2 n}\right)^2}{ (2 n+1)^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^3}$$
$$-\frac{8}{5}\sum _{n=1}^{ \infty } \frac{\left(H_{2 n}\right){}^2}{ n^2}-\frac{32}{5}\underbrace{\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n+1)^2}}_{\text{The series you need}}-\frac{64}{5}\log(2)\sum _{n=1}^{ \infty } \frac{H_{2 n}}{(2 n+1)^2}-\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{n^2}.$$
In fact, in the book the author nicely exploits the fact that for the linear Euler sum of the type $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^m}$, with $m=3$, we arrive at $5/4 \zeta(4)$ which allows us to express the $\zeta(4)$ value in terms of a sum of seven series. You might not need this precise representation, but almost all the steps presented in the solution. It is precisely the same strategy as for the weight $5$ case which is given in On the calculation of two essential harmonic series with a weight 5 structure, involving harmonic numbers of the type $H_{2n}$. In this case we play with weight $4$ series. Observe that all the other series above are known or easily reducible to known series.
A note: In this question Two very advanced harmonic series of weight $5$, if you take a look at the second and third series you may see how they look like when having $2n-1$ and $2n+1$ in denominator (the latter version looks way better in terms of a closed-form). Well, like our case, except that there we are on the realm of weight $5$ series.
What about the red part? We want a clever rearrangement of the initial series, that is
$$\sum _{n=1}^{\infty } \frac{2 H_{2 n}-H_n-2 \log (2)}{2 n (2 n-1)^2}$$
$$=2\sum _{n=1}^{\infty } \frac{H_{2 n-1}+1/(2n)}{(2 n-1)^2}-\sum _{n=1}^{\infty } \frac{H_n}{(2 n-1)^2}-\sum _{n=1}^{\infty } \frac{H_n}{2 n (2 n-1)}-2 \log (2)\sum _{n=1}^{\infty } \frac{1}{(2 n-1)^2}$$
$$+2 \sum _{n=1}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{2 n (2 n-1)}.$$
Both the first and the second series are done by using the results from this paper A new powerful strategy of calculating a classof alternating Euler sums by Cornel Ioan Valean, particularly the main theorem and lemma $4$. Then, the third and the fourth sums are trivial.
Finally, there is a nice thing to observe about the fifth sum, that is if we reindex it and start from $n=0$, we can simply use the series from the second step in this answer Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}$, which is finalized elementarily.
End of story.
Best Answer
I'll just show an idea that avoids those type of sum, but skip the calculations. You might also have better ideas to solve them.
For start we will denote $a=\ln(1-x)$ and $b=\ln(1+x)$ and use the following identity: $$a^2=\frac12 (a+b)^2+\frac12(a-b)^2-b^2$$ $$\Rightarrow I=\frac12 \underbrace{\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx}_{I_1}+\frac12\underbrace{ \int_0^1 \frac{\ln x\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx}_{I_2}-\underbrace{\int_0^1 \frac{\ln x\ln^2(1+x)}{1+x}dx}_{I_3}$$
For the first integral we will write the denominator as: $$\frac{1}{1+x}=\frac{1}{1-x^2}-\frac{x}{1-x^2}$$ $$\Rightarrow I_1=\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1-x^2}dx-{\int_0^1 \frac{x\ln x\ln^2(1-x^2)}{1-x^2}dx}$$ $$\overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}\frac{dx}{\sqrt x}-\frac14\int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}dx$$ Those two integral can be found using the second identity from here.
Let's also take $I_2$ and substitute $\frac{1-x}{1+x}\to x$. $$\Rightarrow I_2=\underbrace{\int_0^1 \frac{\ln(1-x)\ln^2 x}{1+x}dx}_{P}-\underbrace{\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx}_{Q}$$ $$P-Q=I_2;\quad P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1+x}dx$$ And again with the same trick done for $I_1$, we have: $$P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1-x^2}dx-\int_0^1 \frac{x\ln(1-x^2)\ln^2 x}{1-x^2}dx$$ $$\overset{x^2\to x}=\frac18\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}\frac{dx}{\sqrt x}-\frac18 \int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx$$ Henceforth we can extract our second integral, $I_2$ as: $$I_2=P-Q=(P+Q)-2Q$$ Note that $P+Q$ can again be found using the second identity from here.
Finally, we only need to find $Q$.
$$Q=\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx=\sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{n}\ln^2 x=2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{(n+1)^3}$$ So $Q$ is actually an Euler sum in disguise, but you nicely found it here.
Also, $I_3$ is pretty easy, one just needs to use the same approach as done for $I_1$ in your following post. $$I_3=\int_0^1 \frac{\ln x \ln^2(1+x)}{1+x}dx\overset{IBP}=-\frac12\int_0^1 \frac{\ln^3(1+x)}{x}dx$$