real analysis - Challenging integral: Evaluate $int_0^1frac{ln^3(1-x)operatorname{Li}_3(x)}{x}dx$

How to evaluate $$I=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_3(x)}{x}dx\ ?$$

I came across this integral $I$ while I was trying to compute two advanced sums of weight 7. The problem with my approach is that when I tried to evaluate $I_5$

(shown below), the main integral $I$ appeared there which cancels out from both sides, so any idea how to evaluate $I_5$ or $I$?

Thanks.

Here is my trial:

Using the two generalized integral expressions of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.

$$\int_0^1\frac{x\ln^n(u)}{1-xu}du=(-1)^n n!\operatorname{Li}_{n+1} (x)\Longrightarrow \operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2(u)}{1-xu}du\tag{1}$$

$$\small{u\int_0^1\frac{\ln^n(x)}{1-u+ux}dx=(-1)^{n-1}n!\operatorname{Li}_{n+1}\left(\frac{u}{u-1}\right)\Longrightarrow\int_0^1\frac{\ln^3x}{1-u+ux}dx=\frac6u\operatorname{Li}_{3}\left(\frac{u}{u-1}\right)}\tag{2}$$

We have

$$\begin{align} I&=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_3(x)}{x}dx\overset{\text{use} (1)}{=}\frac12\int_0^1\frac{\ln^3(1-x)}{x}\left(\int_0^1\frac{x\ln^2u}{1-xu}du\right)dx\\ &=\frac12\int_0^1\ln^2u\left(\frac{\ln^3(1-x)}{1-xu}dx\right)\ du\overset{1-x\ \mapsto\ x}{=}\frac12\int_0^1\ln^2u\left(\int_0^1\frac{\ln^3x}{1-u+ux}dx\right)\ du\\ &\overset{\text{use}\ (2)}{=}3\int_0^1\frac{\ln^2u}{u}\operatorname{Li}_4\left(\frac{u}{u-1}\right)du\overset{IBP}{=}-\int_0^1\frac{\ln^3u}{u(1-u)}\operatorname{Li}_3\left(\frac{u}{u-1}\right)du \end{align}$$

Now we need the trilogarithmic identity:

$$\operatorname{Li}_3\left(\frac{x-1}{x}\right)=\zeta(2)\ln x-\frac12\ln^2x\ln(1-x)+\frac16\ln^3x-\operatorname{Li}_3(1-x)-\operatorname{Li}_3(x)+\zeta(3)$$

set $1-x=u$ to get

$$\small{\operatorname{Li}_3\left(\frac{u}{u-1}\right)=\zeta(2)\ln(1-u)-\frac12\ln^2(1-u)\ln u+\frac16\ln^3(1-u)-\operatorname{Li}_3(u)-\operatorname{Li}_3(1-u)+\zeta(3)}$$

Going back to our integral

$$\begin{align} I&=\small{-\int_0^1\frac{\ln^3u}{u(1-u)}\left(\zeta(2)\ln(1-u)-\frac12\ln^2(1-u)\ln x+\frac16\ln^3(1-u)-\operatorname{Li}_3(u)-\operatorname{Li}_3(1-u)+\zeta(3)\right)du}\\ &=-\zeta(2)\underbrace{\int_0^1\frac{\ln^3u\ln(1-u)}{u(1-u)}du}_{\Large I_1}+\frac12\underbrace{\int_0^1\frac{\ln^4u\ln^2(1-u)}{u(1-u)}du}_{\Large I_2}-\frac16\underbrace{\int_0^1\frac{\ln^3u\ln^3(1-u)}{u(1-u)}du}_{\Large I_3}\\ &\quad+\underbrace{\int_0^1\frac{\ln^3u\operatorname{Li}_3(u)}{u(1-u)}\ du}_{\Large I_4}+\underbrace{\int_0^1\frac{\ln^3u}{u(1-u)}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)du}_{\Large I_5} \end{align}$$

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$$\begin{align} I_1=\int_0^1\frac{\ln^3u\ln(1-u)}{u(1-u)}du=-\sum_{n=1}^\infty H_n\int_0^1 u^{n-1}\ln^3udu=6\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align}$$
.

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$$\begin{align} I_2&=\int_0^1\frac{\ln^4u\ln^2(1-u)}{u(1-u)}du=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^1 u^{n-1}\ln^4udu\\ &=24\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^5}=24\sum_{n=1}^\infty\frac{H_n^2}{n^5}-24\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5} \end{align}$$

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$$\begin{align} I_3&=\int_0^1\frac{\ln^3u\ln^3(1-u)}{u(1-u)}du=\int_0^1\frac{\ln^3u\ln^3(1-u)}{u}du+\underbrace{\int_0^1\frac{\ln^3u\ln^3(1-u)}{1-u}du}_{1-x\ \mapsto\ x}\\ &=2\int_0^1\frac{\ln^3u\ln^3(1-u)}{u}\ du\overset{IBP}{=}\frac32\int_0^1\frac{\ln^4u\ln^2(1-u)}{1-u}du\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^1 u^n\ln^4udu, \quad \text{reindex}\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\int_0^1 u^{n-1}\ln^4u du\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\left(\frac{24}{n^5}\right)\\ &=36\sum_{n=1}^\infty\frac{H_n^2}{n^5}-36\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}-72\sum_{n=1}^\infty\frac{H_n}{n^6}+72\zeta(7) \end{align}$$

.

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$$\begin{align} I_4&=\int_0^1\frac{\ln^3u\operatorname{Li}_3(u)}{u(1-u)}du=\sum_{n=1}^\infty H_n^{(3)}\int_0^1 u^{n-1}\ln^3u du=-6\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^4} \end{align}$$

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$$\begin{align} I_5&=\int_0^1\frac{\ln^3u}{u(1-u)}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)du\\ &=\underbrace{\int_0^1\frac{\ln^3u}{u}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)du}_{IBP}+\underbrace{\int_0^1\frac{\ln^3u}{1-u}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)\ du}_{1-u\ \mapsto\ u}\\ &=\frac14\int_0^1\frac{\ln^4u\operatorname{Li}_2(1-u)}{1-u}du+\underbrace{\int_0^1\frac{\ln^3(1-u)\operatorname{Li}_3(u)}{u}du}_{\large \text{our main integral}}-\zeta(3)\int_0^1\frac{\ln^3u}{1-u}du\\ &=\frac14\int_0^1\frac{\ln^4u\operatorname{Li}_2(1-u)}{1-u}du+I+6\zeta(3)\zeta(4) \end{align}$$

In my solution here I came across the remaining integral and here is the result:

$$\frac14\int_0^1\frac{\ln^4u\operatorname{Li}_2(1-u)}{1-u}du=6\zeta(2)\zeta(5)+36\zeta(7)-30\sum_{n=1}^\infty\frac{H_n}{n^6}-6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}$$

Then

$$I_5=I+6\zeta(3)\zeta(4)+6\zeta(2)\zeta(5)+36\zeta(7)-30\sum_{n=1}^\infty\frac{H_n}{n^6}-6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}$$
.

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Note: We can not use the two sums $\sum_{n=1}^\infty\frac{H_n^3}{n^4}$ and $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}} {n^4}$ in our solution because the integral $I$ is the key to evaluate these two sums.