Is there a closed form for $\int_0^1\frac{\ln^4(1-x)\operatorname{Li}_4(x)}{x}dx\ ?$

How to compute

$$I=\int_0^1\frac{\ln^4(1-x)\operatorname{Li}_4(x)}{x}dx\ ?$$

where $\operatorname{Li}_r$ is the polylogarithm function.

First I tried integration by parts but got complicated, so I tried subbing $$\operatorname{Li}_4(x)=-\frac16\int_0^1\frac{x\ln^3u}{1-xu}du$$ which leads to

$$I=-\frac16\int_0^1\ln^3u\left(\frac{x\ln^4(1-x)}{1-ux} dx\right)du$$

and I am stuck here. I used Mathematica for $I$ and the inside integral but it failed to give any. I am not sure if $I$ has a closed form though.

A similar question here was solved using multiple zeta value and wondering if this method works here.

By expanding $\text{Li}_4$ and computing Beta derivatives the integral equals to $$S(1,1,1,1;5)+6S(1,1,2;5)+8S(1,3;5)+3S(2,2;5)+6S(4,5)$$ Where $S$ denotes multi-Euler sum (for instance $S(1,1,1,1;5)=\sum_{n=1}^{\infty} \frac{H_n^4}{n^5}$ and so on). All of these sums are solvable due to MZV decomposition, therefore

• $ \int_0^1 \frac{\text{Li}_4(x) \log ^4(1-x)}{x} \, dx=-12 \zeta (3)^3+\frac{17 \pi ^6 \zeta (3)}{378}+\frac{6 \pi ^4 \zeta (5)}{5}+20 \pi ^2 \zeta (7)-\frac{1949 \zeta (9)}{6}$

For the generalization given by @FDP, one have:

• $ \int_0^1 \frac{\text{Li}_5(x) \log ^5(1-x)}{x} \, dx=180 \zeta (3) \zeta(6,2)+180\zeta(8,2,1)+10 \pi ^2 \zeta (3)^3-660 \zeta (5) \zeta (3)^2+\frac{373 \pi ^8 \zeta (3)}{15120}-\frac{143 \pi ^6 \zeta (5)}{189}-\frac{28 \pi ^4 \zeta (7)}{3}-275 \pi ^2 \zeta (9)+\frac{153475 \zeta (11)}{32}$