What is the closed form of $\sum_{n\geq 1}(-1)^{n-1}\psi'(n)^2$? This problem was proposed by Cornel Ioan Valean.

What is the closed form of
  $$ S=\sum_{n\geq 1}(-1)^{n+1}\psi'(n)^2 $$
  ?

I recall that $\psi'(z)=\frac{d^2}{dz^2}\log\Gamma(z)=\sum_{m\geq 0}\frac{1}{(m+z)^2}$ for any $z>0$.
My attempt was to perform the following manipulation
$$ \sum_{n\geq 1}(-1)^{n+1}\psi'(n)^2 = \sum_{\substack{m,n\geq 1 \\ \min(m,n)\text{ odd}}}\frac{1}{m^2 n^2} \tag{A}$$
in order to turn the original series into
$$\begin{eqnarray*} 2\sum_{n\geq 0}\frac{\zeta(2)-H_{2n+1}^{(2)}}{(2n+1)^2}+\sum_{n\geq 0}\frac{1}{(2n+1)^4}&=&\frac{5\pi^4}{96}-2\sum_{n\geq 0}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}\\&=&\frac{19\pi^4}{1440}+\frac{1}{2}\color{blue}{\sum_{n\geq 1}\frac{H_{2n}^{(2)}}{n^2}}\end{eqnarray*} \tag{B}$$
not so bad after all. My issue is that now I am not able to find a decent closed form for the last series. If we apply summation by parts we have:
$$\begin{eqnarray*} \color{blue}{\sum_{n\geq 1}\frac{H_{2n}^{(2)}}{n^2}}&=&\frac{\pi^4}{36}-\sum_{m\geq 1}\frac{H_m^{(2)}}{(2m+2)^2}-\sum_{m\geq 1}\frac{H_m^{(2)}}{(2m+1)^2}\\&=&\frac{37 \pi^4}{1440}-\color{green}{\sum_{m\geq 1}\frac{H_{m}^{(2)}}{(2m+1)^2}}\end{eqnarray*}\tag{C} $$
but the green series does not seem to be really "better" than the blue one.
Maybe it is relevant to point out that
$$ \color{green}{\sum_{m\geq 1}\frac{H_{m}^{(2)}}{(2m+1)^2}} = -\int_{0}^{1}\frac{\text{Li}_2(z^2)}{1-z^2}\log(z)\,dz.\tag{D}$$
Do you see a way to tackle the last series, or the original one through a different approach? Numerically, $S\approx 2.3949463369266426$.
 A: Assume that $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{2}}z^{n}=f\left(z\right),\,\left|z\right|\leq1.\tag{1}$$ Then if we take $z=\exp\left(\pi i\right)$ we get $$f\left(\exp\left(\pi i\right)\right)=\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{2}}\exp\left(n\pi i\right)=\sum_{n\geq1}\frac{H_{2n}^{\left(2\right)}}{\left(2n\right)^{2}}-\sum_{n\geq1}\frac{H_{2n-1}^{\left(2\right)}}{\left(2n-1\right)^{2}}$$ and taking $z=-\exp\left(\pi i\right)$ we have $$f\left(-\exp\left(\pi i\right)\right)=\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}\left(-1\right)^{n}}{n^{2}}\exp\left(n\pi i\right)=\sum_{n\geq1}\frac{H_{2n}^{\left(2\right)}}{\left(2n\right)^{2}}+\sum_{n\geq1}\frac{H_{2n-1}^{\left(2\right)}}{\left(2n-1\right)^{2}}$$ hence $$\frac{1}{2}\sum_{n\geq1}\frac{H_{2n}^{\left(2\right)}}{n^{2}}=f\left(\exp\left(\pi i\right)\right)+f\left(-\exp\left(\pi i\right)\right).$$ So we have to find the closed form of $(1)$. From the generating function $$\sum_{n\geq1}H_{n}^{\left(2\right)}z^{n}=\frac{\textrm{Li}_{2}\left(z\right)}{1-z}$$ we have, twice integrating and dividing $z$, that $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{2}}z^{n}=3\textrm{Li}_{4}\left(z\right)+\frac{1}{2}\textrm{Li}_{3}^{2}\left(z\right)-2\sum_{n\geq1}\frac{H_{n}}{n^{3}}z^{n}\tag{2}$$ and the closed form of the series in $(2)$ can be found here, answers of Tunk-Fey and Robert Israel, so we have essentially done. I'm too lazy to do all the calculations.
