The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $ How to Prove that  

$$  \sum_{n=0}^{\infty}\frac{(-1)^nH^{(2)}_{n}}{(2n+1)^2} \;\;=\;\;\frac{7 \pi \;  \zeta(3)}{4}-\frac{\zeta(2)G}{2}+\frac{45\zeta(4)}{8}-\frac{\Psi^{(3)}\big(\frac{1}{4}\big)}{128}$$ 

where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the $n$th generalized harmonic number of order $m$, $\zeta$ is the Riemann zeta function and $G$ is Catalan constant?
This problem proposed by Ahmad Albow 
 A: Solution by Cornel Valean.

\begin{align}
S&=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^2}\\
&=\sum_{n=1}^\infty(-1)^{n-1}H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\
&=\int_0^1\ln x\sum_{n=1}^\infty(-x^2)^nH_n^{(2)}\ dx\\
&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{x\mapsto 1/x}\\
&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln x\operatorname{Li}_2(-1/x^2)}{1+x^2}\ dx\\
2S&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln x[\color{red}{\operatorname{Li}_2(-x^2)+\operatorname{Li}_2(-1/x^2)}]}{1+x^2}\ dx\\
S&=\frac12\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln x[\color{red}{-2\ln^2x-\zeta(2)}]}{1+x^2}\ dx\\
&=\frac12\underbrace{\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{I}-\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}-\frac12\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}\ dx}_{-G}\\
&=\frac12I+6\beta(4)+\frac12G\zeta(2)\tag1
\end{align}

\begin{align}
I&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln x}{1+x^2}\left(\int_0^1\frac{x^2\ln y}{1+yx^2}\ dy\right)\ dx\\
&=\int_0^1\ln y\left(\int_0^\infty\frac{x^2\ln x}{(1+x^2)(1+yx^2)}\ dx\right)\ dy\\
&=\int_0^1\frac{\ln y}{1-y}\left(\int_0^\infty\frac{\ln x}{1+yx^2}\ dx-\underbrace{\int_0^\infty\frac{\ln x}{1+x^2}\ dx}_{0}\right)\ dy\\
&=\int_0^1\frac{\ln y}{1-y}\left(-\frac{\pi}{4}.\frac{\ln y}{\sqrt{y}}\right)\ dy,\quad \sqrt{y}=x\\
&=-2\pi\int_0^1\frac{\ln^2x}{1-x^2}\ dx=-2\pi\left(-\frac74\zeta(3)\right)=\boxed{-\frac72\pi\zeta(3)}\tag2
\end{align}
Plug (2) and (1) we get

$$S=-\frac74\pi\zeta(3)+6\beta(4)+\frac12G\zeta(2)$$

where $\beta(4)=\frac{\psi_3(1/4)}{768}-\frac{15}{16}\zeta(4)$

Addendum:
Another approach is by applying integration by parts to the integral $S$ then we change the limits from $(0,1)$ to $(0,\infty)$ as we did in our solution above.
A: A fancy way of getting the desired result
If we make use of the following result, 

$$\sum _{n=1}^{\infty } (-1)^{n-1} \frac{ H_n^{(2)}}{(2 n+1)^2}+8\sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_{2 n+1}}{(2 n+1)^3}-\sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_n}{(2 n+1)^3}$$
$$=8+\frac{1}{12}\pi ^2 G-\frac{1}{16}\log (2)\pi ^3+\frac{7 }{96}\pi ^4 -\frac{7}{768} \psi ^{(3)}\left(\frac{1}{4}\right),$$

which may also be extracted immediately from the following result proposed by Cornel I. Valean in Romanian Mathematical Magazine, 

$$6 \int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x) }{x} \textrm{d}x+2\int_0^1 \frac{\arctan(x)\operatorname{Li}_2(-x)}{x} \textrm{d}x-2 \int_0^1 \frac{\arctan(x)\operatorname{Li}_2\left(-x^2\right)}{x} \textrm{d}x$$
$$=6 \zeta(2)G+\frac{45 }{8}\zeta (4)-\frac{1}{128}\psi ^{(3)}\left(\frac{1}{4}\right),$$

that allows a simple, elegant proof without requiring any of the series above (built on simple tweaked ideas from (Almost) Impossible Integrals, Sums, and Series), we are done. I assumed knowledge of the values of the other two series which are well-known now (for example, see Compute $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}$ and $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}$ and Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$).
