Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$

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.

Here are some ideas towards explaining the form of the right hand side. I'm a bit stuck and my main approach hasn't worked out. This may just be rephrasing things in terms of other log-integrals, but hopefully this is a useful way of looking at the problem.

Taking the integral $$ I = \int_0^1 \log^2(1-x) \log^2(x) \log^3(1+x) \frac{dx}{x} $$ we can also rewrite this as $$ I = \int_0^\infty \log^2(1-e^{-x}) \log^2(e^{-x}) \log^3(1+e^{-x}) \; dx $$ which is suited for interpretation as a Mellin transform. Specifically, the power of $x$, is controlled by the power on $\log(x)$ in the original integral format as $$ I = \int_0^\infty x^2 \log^2(1-e^{-x})\log^3(1+e^{-x}) \; dx $$ according to Mathematica we have in general a result for the Mellin transform of the other components $$ \mathcal{M}[\log^n(1\pm e^{-x})](s) = (-1)^n n! \Gamma(s) S_{s,n}(\mp 1) $$ invoking the Neilsen Generalisation of the polylogarithm, $S_{s,n}$. This does recreate the series expansion for $\log(1+e^{-x})$ but the series for $\log(1-e^{-x})$ has a $\log(x)$ term, which might be causing a problem.

We could toy with the idea of a formal series via the Ramanujan Master Theorem, using these Mellin transforms $$ \log^n(1\pm e^{-x}) = \sum_{k=0}^\infty \frac{(-1)^{k+n} n!}{k!} S_{-k,n}(\mp 1)x^k $$ and then the Cauchy product $$ \log^a(1 + e^{-x})\log^b(1 - e^{-x}) = \left( \sum_{k=0}^\infty \frac{(-1)^{k+a} a!}{k!} S_{-k,a}(-1)x^k \right)\left( \sum_{k=0}^\infty \frac{(-1)^{k+b} b!}{k!} S_{-k,b}(1)x^k \right) $$ $$ \log^a(1 + e^{-x})\log^b(1 - e^{-x}) = \sum_{k=0}^\infty \left(\sum_{l=0}^k \frac{(-1)^{a+b+k} a! b!}{l!(k-l)!} S_{-l,a}(-1) S_{l-k,b}(1)\right) x^k $$ alternatively $$ \log^a(1 + e^{-x})\log^b(1 - e^{-x}) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \left(\sum_{l=0}^k (-1)^{a+b} a! b! \binom{k}{l} S_{-l,a}(-1) S_{l-k,b}(1)\right) x^k $$ plausibly leading to (via RMT) $$ \mathcal{M}\left[ \log^a(1 + e^{-x})\log^b(1 - e^{-x})\right](s) = \Gamma(s) \sum_{l=0}^{-s} (-1)^{a+b} a! b! \binom{-s}{l} S_{-l,a}(-1) S_{l-k,b}(1) $$ then we would conceptually have (with some dodgy negative parts) an answer for the integral as a sum over (four?) pairs of generalized Polylogs, specifically in the case that $s=3$.

This motivates an expression in terms of pairs of $S_{n,k}(z)$, we can guess a term and quickly find $$ -8\cdot3 \cdot 19 S_{2,2}(1)S_{1,3}(-1) = -\frac{19}{15} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{133}{120} \pi ^4 \zeta (3) \log (2)+\frac{19 \pi ^8}{1350}+\frac{19}{360} \pi ^6 \log ^2(2)-\frac{19}{360} \pi ^4 \log ^4(2) $$ this covers a few of the terms in your expression R.H.S. It is likely that other terms contribute to $\pi^8$ for example. I can't get an explicit value for $S_{2,3}(-1)$ to explore this further, but I would assume this holds a $\mathrm{Li}_5(1/2)$ term among others, and the other factor is $S_{1,2}(1) = \zeta(3)$. Perhaps your linear combinations method can be rephrased in terms of the generalized polylogarithm?

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.