Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that 

$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$

This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.

Any way, I tried hard on this one but no success yet. here is what I did:
Using the identity  
$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$
$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$
We can write 
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$
where
\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}
Therefore 
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$
The sum can be done using the following identity
$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
Differentiate both sides with respect to $a$ then set $a=1/2$ we get 
$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$

and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks


I asked Cornel for a solution to the nice key result from Ahmad Bow's solution. Here is a solution in large steps.
We need two known results, that is $\displaystyle \int_0^1\frac{x^n}{1+x}\textrm{d}x=H_{n/2}-H_n+\log(2)$ and $\displaystyle \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$. If replacing $p$ by $i \sqrt{p}$ in the last series, make rearrangements and reindexing the series, we obtain$\displaystyle \sum _{n=0}^{\infty } (-1)^n y^n \cos ((2 n+1) x)=\frac{(1+y) \cos (x)}{1+2y \cos (2 x)+y^2}$. 
Then, we have 
\begin{equation*}
\sum_{n=0}^\infty (-1)^n\left(H_{n/2}-H_n+\ln2\right)\cos((2n+1)x)=\int_0^1\frac{1}{1+y}\sum_{n=0}^\infty (-1)^n y^n\cos((2n+1)x)\textrm{d}y
\end{equation*}
\begin{equation*}
=\cos (x)\int_0^1 \frac{1}{1+2y \cos (2 x)+y^2}\textrm{d}y=\cos (x)\int_{\cos(2x)}^{1+\cos(2x)} \frac{1}{t^2+\sin^2(2x)}\textrm{d}t=\frac{x}{2\sin(x)}.
\end{equation*}
End of story.
