sequences and series - Alternating harmonic sum $sum_{kgeq 1}frac{(-1)^k}{k^3}H_k$

How to analytically prove

$$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3) $$

As O.L answer

where
$$H_k = \sum_{n\geq 1}^{k}\frac{1}{n}.$$

Addition

So far I developed the following

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)$$

where $\text{Li}_3(x)$ is the trilogarithm .

For the derivation see http://www.mathhelpboards.com/f10/interesting-logarithm-integral-5301/

Update

A frined on another site gave the following answer

Answer

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\begin{align}
\sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} & =
\sum_{k = 1}^{\infty}\pars{-1}^{k}H_{k}\
\overbrace{\bracks{{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}x^{k - 1}\,\dd x}}

^{\ds{1 \over k^{3}}}
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}
\bracks{\sum_{k = 1}^{\infty}H_{k}\pars{-x}^{k}}\,{\dd x \over x}
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}
\bracks{-\,{\ln\pars{1 + x} \over 1 + x}}\,{\dd x \over x} =
-\,{1 \over 2}\int_{0}^{1}
{\ln^{2}\pars{x}\ln\pars{1 + x} \over \pars{1 + x}x}\,\dd x
\\[5mm] & =

{1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over 1 + x}\,\dd x -
{1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x}\,\dd x
\\[1cm] & =
{1 \over 6}\int_{0}^{1}{3\ln^{2}\pars{x}\ln\pars{1 + x} -
3\ln\pars{x}\ln^{2}\pars{1 + x} \over 1 + x}\,\dd x
\\[5mm] & +
{1 \over 2}\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 + x} \over 1 + x}\,\dd x +
{1 \over 2}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x
\\[1cm] & =
{1 \over 6}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x -

{1 \over 6}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over 1 + x}\,\dd x
\\[5mm] &-
{1 \over 6}\int_{0}^{1}\ln^{3}\pars{x \over 1 + x}\,{\dd x \over 1 + x} -
{1 \over 6}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x
\\[5mm] & -
\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\ln\pars{-x}\,\dd x
\\[1cm] & =
-\,{1 \over 6}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x -
{1 \over 24}\,\ln^{4}\pars{2} -
{1 \over 6}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x

\\[5mm] &
+{1 \over 6}\int_{1}^{2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x\ +\
\underbrace{\quad\int_{0}^{-1}\mrm{Li}_{4}'\pars{x}\,\dd x\quad}
_{\ds{= \,\mrm{Li}_{4}\pars{-1} = -\,{7 \over 720}\,\pi^{4}}}\label{1}\tag{1}
\end{align}

The remaining integrals are evaluated by successive integration by parts. Namely,

\begin{align}
\int{\ln^{3}\pars{\pm x} \over 1 - x}\,\dd x & =
-\ln\pars{1 - x}\ln^{3}\pars{\pm x} -
3\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x
\\[5mm] & =
-\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} +
6\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x
\\[1cm] & =
-\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} +
6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x}

\\[5mm] & - 6\int\mrm{Li}_{4}'\pars{x}\,\dd x
\\[1cm] & =
-\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} +
6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x}
\\[5mm] & - 6\,\mrm{Li}_{4}\pars{x}\label{2}\tag{2}
\end{align}

---

With \eqref{1} and \eqref{2}:
$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{
\sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} =

-\,{11 \over 360}\,\pi^{4} - {1 \over 12}\ln^{2}\pars{2}\pi^{2} +
{1 \over 12}\,\ln^{4}\pars{2} + 2\,\mrm{Li}_{4}\pars{1 \over 2} +
{7 \over 4}\,\ln\pars{2}\zeta\pars{3}}}
$$