Integral $\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left({x} \right)\,dx$ Is there a closed form for this integral?
$\displaystyle \int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left({x} \right)\,dx\\$
All I have been able to find, so far, is a numeric approximation of $-1.13348$
 A: Yes there is:
$$\mathcal I=\zeta(2)\zeta(3)-3\zeta(5).$$
See e.g. this answer.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\Li{3}{x}\,\dd x} =
\int_{0}^{1}{\ln\pars{1 - x} \over x}\,
\sum_{n = 1}^{\infty}{x^{n} \over n^{3}}\,\dd x
\\[5mm] = &\
\sum_{n = 1}^{\infty}{1 \over n^{3}}\int_{0}^{1}\ln\pars{1 - x}x^{n - 1}\,\dd x
\label{1}\tag{1}
\end{align}

However,
\begin{align}
&\fbox{$\ds{\ \int_{0}^{1}\ln\pars{1 - x}x^{n - 1}\,\dd x\ }$} =
\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}\pars{1 - x}^{\mu}x^{n - 1}\,\dd x
\\[5mm] = &\
\lim_{\mu \to 0}\partiald{}{\mu}
\bracks{\Gamma\pars{\mu + 1}\Gamma\pars{n} \over \Gamma\pars{\mu + n + 1}} =\
\fbox{$\ds{\ -\,{H_{n} \over n}\ }$}
\end{align}

Then $\ds{\pars{~\mbox{see expression}\ \eqref{1}~}}$,
$$
\color{#f00}{\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\Li{3}{x}\,\dd x} =
-\sum_{n = 1}^{\infty}{H_{n} \over n^{4}} =
\color{#f00}{{1 \over 6}\,\pi^{2}\zeta\pars{3} - 3\zeta\pars{5}}
$$

$\ds{\sum_{n = 1}^{\infty}{H_{n} \over n^{4}} =
{3\zeta\pars{5} - {1 \over 6}\,\pi^{2}}\,\zeta\pars{3} - }$ is given as formula $\pars{20}$
  in the MathWorld Harmonic Number page and it has been evaluated in a
  1995 David Borwein and Jonathan Borwein paper.

A: \begin{align}
I&=\int_0^1\frac{\ln(1-x)}{x}\operatorname{Li_3}(x)\ dx=\sum_{n=1}^\infty\frac1{n^3}\int_0^1x^{n-1}\ln(1-x)\ dx=-\sum_{n=1}^\infty\frac{H_n}{n^4}=\zeta(2)\zeta(3)-3\zeta(5)
\end{align}
A: I recommend using following recursion:
$$\operatorname{Li}_{0}(z) =\frac{z}{1-z}; \quad \operatorname{Li}_{n+1}(z) = \int_0^z \frac{\operatorname{Li}_n(t)}{t}$$
