Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$ 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
 A: Let us first recall that harmonic numbers have generating function
\begin{align}
\sum_{k=1}^{\infty}H_kx^k=-\frac{\ln(1-x)}{1-x},
\end{align}
and therefore
\begin{align}
S=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}H_k&=\frac{1}{2}\sum_{k=1}^{\infty}(-1)^kH_k\int_0^{\infty}e^{-kx}x^2dx=\\
&=-\frac{1}{2}\int_0^{\infty}\frac{\ln(1+e^{-x})}{1+e^{-x}}x^2dx.
\end{align}
Mathematica knows how to evaluate the last integral in terms of zeta values and polylogarithms. Its answer is
$$S=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3).$$
It is unlikely that it can be simplified further: Wolfram Alpha proposes alternative expressions for $\mathrm{Li}_{2}\left(\frac12\right)$ and $\mathrm{Li}_{3}\left(\frac12\right)$ in terms of elementary functions and zeta values, but does not suggest anything simpler for $\mathrm{Li}_{4}\left(\frac12\right)$.
A: Related problems: (I), (II), (III). Your sum is a special case of the following general case which I derived an integral representation for it

$$ A(p,q) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q} = \frac{\left( -1 \right) ^{q}}{\Gamma(q)}\int _{0}^{1}\!{\frac { \left( \ln\left( u
 \right)\right)^{q-1}{\mathrm{Li}_{p}(-u)} }{ u\left( 1+u \right) }}{du}. $$

where $ \mathrm{Li}_{p}(z) $ is the polylogarithm function. So, letting $p=1$ and $q=3$ in the above formula gives an integral representation for your sum

$$ A(1,3) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(1)}_k}{k^3} = \frac{\left( -1 \right) ^{3}}{\Gamma(3)}\int _{0}^{1}\!{\frac { \left( \ln\left( u\right) \right)^{3-1}{\mathrm{Li}_{1}(-u)} }{ u (1+u) }}{du}.$$

$$ \implies A(1,3) =\frac{1}{2}\int _{0}^{1}\!{\frac { \left( \ln  \left( u \right)  \right) ^{2}
\ln  \left( 1+u \right) }{u\left(1+u\right)}}{du} \sim 0.8592471579. $$
See here for related techniques.
Note:

1) $$ \mathrm{Li}_{1}(-u)=-\ln(1+u).  $$ 

A: A magical way proposed by Cornel Ioan Valean

Let's consider a powerful form of the Beta function which is presented in the book, (Almost) Impossible Integrals, Sums, and Series, $\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} \textrm{d}x = \operatorname{B}(a,b)$, used for elegant calculations in the Section $3.7$, pages $72$-$73$.
Here is the magic ...
$$\underbrace{\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{3}}{\partial a^2 \partial b}\operatorname{B}(a,b)}_{\displaystyle -5/2\zeta(4)}=3\underbrace{\int_0^1\frac{\log(x)\log^2(1+x)}{x}\textrm{d}x}_{\displaystyle -7/4 \zeta(4)+2 \sum _{n=1}^{\infty} (-1)^{n-1}H_n/n^3}-\underbrace{\int_0^1\frac{\log^2(x)\log(1+x)}{x}\textrm{d}x}_{\displaystyle 7/4\zeta(4)}$$
$$-2\underbrace{\int_0^1 \frac{\log^3(1+x)}{x}\textrm{d}x}_{\displaystyle  6\zeta(4)+3/2\log^2(2)\zeta(2)-21/4\log(2)\zeta(3)\\\displaystyle -\log^4(2)/4-6\operatorname{Li}_4(1/2)},$$
whence we conclude that
  $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n}{n^3}$$
$$=\frac{11}{4}\zeta(4)-\frac{7}{4}\log(2)\zeta(3)+\frac{1}{2}\log^2(2)\zeta(2)-\frac{1}{12}\log^4(2)-2\operatorname{Li_4}\left(\frac12\right).$$
Q.E.D.

A first note: Observe the last two integrals in the right-hand side and known and trivial.
A second note: The Beta function limit can be approached in more ways. An elegant way is achieved by Cornel's Master Theorem of Series from the article A master theorem of series and an evaluation of a cubic harmonic series, which is also given in the book, (Almost) Impossible Integrals, Sums, and Series. For a different approach, note that the limit can be brought to the form, $\displaystyle \int_0^1 \frac{\log (1-x) \log ^2(x)}{(1-x) x} \textrm{d}x$, where we notice easily that behind the scene there is a classical Euler sums, a well-known one!
A third note: a similar strategy, with some more machinery, has been used in this answer https://math.stackexchange.com/q/3531956.
The work will be turned into an article soon.
