TITLE: Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$
QUESTION [38 upvotes]: We have the following result ($\text{Li}_{n}$ being the polylogarithm):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}_4(\frac{1}{2}){}^2-\frac{19}{15} \pi ^4 \text{Li}_4(\frac{1}{2})+\\ 12 \pi ^2 \text{Li}_6(\frac{1}{2})+8 \text{Li}_4(\frac{1}{2}) \log ^4(2)-2 \pi ^2 \text{Li}_4(\frac{1}{2}) \log ^2(2)+12 \pi ^2 \text{Li}_5(\frac{1}{2}) \log (2)+\frac{87 \pi ^2 \zeta (3)^2}{16}+\\ \frac{447 \zeta (3) \zeta (5)}{16}+\frac{7}{5} \zeta (3) \log ^5(2)-\frac{7}{12} \pi ^2 \zeta (3) \log ^3(2)-\frac{133}{120} \pi ^4 \zeta (3) \log (2)-\frac{\pi ^8}{9600}+\frac{\log ^8(2)}{6}- \\ \frac{1}{6} \pi ^2 \log ^6(2)-\frac{1}{90} \pi ^4 \log ^4(2)+\frac{19}{360} \pi ^6 \log ^2(2) }$$
This is extremely amazing: almost all other similar integrals are not expressible via ordinary polylogarithm.
The solution is however non-trivial. There are two methods: first is to find enough linear relations between similar integrals, once the rank is high enough, solving the system gives $(*)$; second method is to convert the integral into multiple zeta values, then use known linear relations between them. None of these methods can explain the result's simplicity.

Question: Is there a simpler method to prove (*), or a conceptual explanation of its elegance?

Any thought is welcomed. Thank you very much.

I wrote a Mathematica package, it can calculate the integral in subject and many similar ones. The following command calculates $(*)$:
MZIntegrate[Log[1-x]^2*Log[x]^2*Log[1+x]^3/x, {x,0,1}]

It can also solve some other integrals.
The package can be obtained here. I hope it can benefit those interested in related integral/series.

Remarks on the question:

It's known that $\zeta(\bar{3},1,\bar{3},1)$ is
very reminiscent to the RHS of $(*)$. But both the simplicity of
$\zeta(\bar{3},1,\bar{3},1)$ and its connection to the integral are
elusive to me.
(Added by Iridescent) This contains nearly all known general formulas
of these log integrals. However it does not help much on solving OP's
problem.

REPLY [3 votes]: Some values of $\text{Li}_k(z)$ are presented in the table $(1).$
\begin{vmatrix}
\hspace{-5mu}^{\overline{\hspace{52pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-10mu} 
&\hspace{-10mu}^{\overline{\hspace{186pt}}}\hspace{-10mu} 
&\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-8mu} \\[-4pt]
\text{Li}_k(z) & z = -1 & z = \dfrac12 & z = 1 \\[-0pt]
\hspace{-5mu}^{\overline{\hspace{52pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-10mu} 
&\hspace{-10mu}^{\overline{\hspace{186pt}}}\hspace{-10mu} 
&\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-8mu} \\[-2pt]
k=1 & -\log(2) & \log(2) & \infty \\[4pt]
k=2 & -\dfrac{\pi^2}{12} & \dfrac{\pi^2}{12} - \dfrac12 \log^2(2) & \dfrac{\pi^2}{6} \\[4pt]
k=3 & -\dfrac34 \zeta(3) & -\dfrac1{12} \pi^2 \log(2) + \dfrac16 \log^3(2) + \dfrac{21}{24} \zeta(3)) & \zeta(3) \\[4pt]
k=4 & -\dfrac{7 \pi^4}{720} & \text{Li}_4\left(\dfrac12\right) & \dfrac{\pi^4}{90} \\[4pt]
k=5 & -\dfrac{15}{16} \zeta(5) & \text{Li}_5\left(\dfrac12\right) & \zeta(5)\\[-2pt]
\hspace{-7mu}\_\_\_\_\_\_\_\_\_\_\_\hspace{-9mu} &\hspace{-9mu}\_\_\_\_\_\_\_\_\_\_\_\_\_\hspace{-9mu} &\hspace{-9mu} \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\hspace{-9mu} & \hspace{-11mu}\_\_\_\_\_\_\_\_\_\_\_\_\_\hspace{-5mu} \tag1
\end{vmatrix}
Also, are known the next antiderivatives below.
$$\int\dfrac{\log (1-x)\log^2 (x)}x\,\text dx = 
-2\,\text{Li}_4(x)+2\text{Li}_3(x)\log(x)-\text{Li}_2(x)\log^2(x)
+ \text{const},\tag2$$

$$\begin{align}
&\int\dfrac{\log^3(1+x)}{1-x}\,\text dx = -6\text{ Li}_4\dfrac {1+x}2
+6\text{ Li}_3\dfrac{1+x}2\log(1+x)\\[4pt]
&-3\text{Li}_2\dfrac{1+x}2\log^2(1+x) - \log\dfrac{1-x}2\log^3(1+x)+\text{const},
\end{align}\tag3$$

$$\begin{align}
&\int\dfrac{\log^2 (1+x) \log(1-x)}{1+x}\,\text dx = -2\text{ Li}_4\dfrac{1+x} 2
+2\text{ Li}_3 \dfrac {1+x} 2 \log(1+x)\\[4pt]
&-\text{Li}_2 \dfrac{1+x} 2\log^2(1+x)+\dfrac13\log(2)\log^3(1+x)+\text{const}.
\end{align}\tag4$$

\begin{align}
&\int_0^1 \log(1-x) \log^3(1+x)\,\frac{\log (1-x) \log^2 (x)}{x}\,\text dx
\\[5mm]
&\overset{IBP(2)}{=\!=\!=\!=}\,
\log (1-x) \log^3(1+x) \left(-2\,\text{Li}_4(x)+2\text{Li}_3(x)\log(x)-\text{Li}_2(x)\log^2(x)\right)\bigg|_0^1\\[4pt] 
&-\int_0^1 \left(-2\,\text{Li}_4(x)+2\text{Li}_3(x)\log(x)
-\text{Li}_2(x)\log^2(x)\right) \frac{\log^3(1+x)}{1-x}\,\text dx\\[4pt]
&-3\int_0^1 \left(-2\,\text{Li}_4(x)+2\text{Li}_3(1+x)\log(x)
-\text{Li}_2(x)\log^2(x)\right) \frac{\log (1- x) \log^2 (1+x)}{1+x} \,\text dx\\[4pt]
& \overset{(3),(4)}{=\!=\!=\!=}\, \int_0^1 \left(-2\,\text{Li}_4(x) + 2\text{Li}_3(x)\log(x) - \text{Li}_2(x)\log^2(x)\right)\text{ d}\Bigl(\log(1-x)\log^3(1+x)\Bigr),
\end{align}
without suitable continuation.