Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$

A related problem. You can have the following closed form

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, dx= -\sum _{k=1}^{\infty }{\frac {\psi' \left( k+1 \right) }{{k}^{3}}}\sim -0.7115661976, $$

where $\psi(x)$ is the digamma function.

Another possible solution:

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, dx = \zeta(5) -\sum _{k=1}^{\infty }{\frac {\psi' \left( k \right) }{{k}^{3}}}\sim -0.7115661976. $$

Added: For the first one just use the power series expansions of $\operatorname{Li}_3(x)$ and $\frac{1}{1-x}$ and you end up with

$$\sum_{k=1}^{\infty}\frac{1}{k^3}\sum_{n=0}^{\infty} \int_{0}^{1} x^{k+n}\ln(x)dx=-\sum_{k=1}^{\infty}\frac{1}{k^3}\sum_{n=0}^{\infty}\frac{1}{(n+k+1)^2} $$

$$ = -\sum_{k=1}^{\infty}\frac{\psi'(k+1)}{k^3}. $$

If you manipulate the last sum, you will be able to relate it to the Euler sums as

$$ -\sum_{k=1}^{\infty}\frac{\psi'(k+1)}{k^3}= \zeta(5)-\sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n^2}.$$

Note: Notice that, we are getting identities for $\zeta(5)$.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x: \ {\large ?}}$.

In this post it's shown, in general grounds, that: $$ \int_{0}^{1}{{\rm Li}_{q}\pars{x}\ln^{r - 1}\pars{x} \over 1 - x}\,\dd x =\pars{-1}^{r - 1}\pars{r - 1}!\bracks{\zeta\pars{r}\zeta\pars{q}% -\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{r}} \over n^{q}}} $$ where $\ds{\zeta\pars{z}}$ is the Riemann Zeta Function and $\ds{H_{n}^{\rm \pars{r}} \equiv \sum_{k = 1}^{n}{1 \over k^{\rm r}}}$ is a Generalized Harmonic Number

such that $\ds{\pars{~\mbox{with}\ q = 3\ \mbox{and}\ r = 2~}}$: $$ \int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x =\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{2}} \over n^{3}} - \zeta\pars{2}\zeta\pars{3}\,,\qquad H_{n}^{\rm\pars{2}} = \sum_{k = 1}^{n}{1 \over k^{2}} $$

In a comment of this answer, the OP $\pars{\tt @Zaid\ Alfayeai}$ pointed out that $\ds{\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{2}} \over n^{3}} = 3\zeta\pars{2}\zeta\pars{3} - {9 \over 2}\,\zeta\pars{5}}$ such that

$$\color{#66f}{\large% \int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x =2\zeta\pars{2}\zeta\pars{3} - {9 \over 2}\,\zeta\pars{5}} \approx {\tt -0.7115} $$

\begin{align} \int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}{\rm d}x &=\sum^\infty_{n=1}H_n^{(3)}\int^1_0x^n\log{x} \ {\rm d}x\\ &=-\sum^\infty_{n=1}\frac{H_n^{(3)}}{(n+1)^2}\\ &=\sum^\infty_{n=1}\frac{1}{(n+1)^5}-\sum^\infty_{n=1}\frac{H_{n+1}^{(3)}}{(n+1)^2}\\ &=\zeta(5)-\underbrace{\sum^\infty_{n=1}\frac{H_{n}^{(3)}}{n^2}}_{S} \end{align} Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(2)}(-z)}{z^2}$. We know that \begin{align}\pi\cot{\pi z}&=\frac{1}{z-n}-2\sum^\infty_{k=1}\zeta(2k)(z-n)^{2k-1}\\&\approx\frac{1}{z-n}-2\zeta(2)(z-n)\end{align} (see here for a proof) and \begin{align}\Psi^{(2)}(-z)&=\frac{2}{(z-n)^3}+\sum^\infty_{k=2}(-1)^{k}k(k-1)\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^{k-2}\\&\approx\frac{2}{(z-n)^3}+2\left(H_n^{(3)}-\zeta(3)\right)\end{align} At the positive integers, \begin{align} {\rm Res}(f,n) &=\operatorname*{Res}_{z=n}\left[\frac{2}{z^2(z-n)^4}-\frac{4\zeta(2)}{z^2(z-n)^2}+\frac{2\left(H_n^{(3)}-\zeta(3)\right)}{z^2(z-n)}\right]\\ &=-\frac{8}{n^5}+\frac{8\zeta(2)}{n^3}+\frac{2H_n^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2} \end{align} At the negative integers, \begin{align} {\rm Res}(f,-n) &=\frac{\Psi^{(2)}(n)}{n^2}\\ &=\frac{2H_{n}^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2}-\frac{2}{n^5} \end{align} At $z=0$, \begin{align} {\rm Res}(f,0) &=[z^1]\left(\frac{1}{z}-2\zeta(2)z\right)\left(-2\zeta(3)-12\zeta(5)z^2\right)\\ &=-12\zeta(5)+4\zeta(2)\zeta(3) \end{align} Hence \begin{align}4S&=8\zeta(5)-8\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(5)+12\zeta(5)-4\zeta(2)\zeta(3)\\&=22\zeta(5)-8\zeta(2)\zeta(3)\end{align} which implies $$\color{blue}{\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}}=\zeta(5)-\frac{22\zeta(5)-8\zeta(2)\zeta(3)}{4}=\color{blue}{2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)}$$