<s>[INST] Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$ How would you solve the following 
$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$$
I might be able to relate the integral to Euler sums .
 [/INST] \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
$$\color{blue}{\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}}=\zeta(5)-\frac{22\zeta(5)-8\zeta(2)\zeta(3)}{4}=\color{blue}{2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)}$$
</s>