Infinite Poly-summation

$\begin{array}{c}S & = \displaystyle \sum_{k=1}^\infty \frac{1}{k^2} \psi^{(1)}\big(\tfrac{k+1}{2}\big) \; = \; \sum_{k=1}^\infty \frac{1}{k^2} \int_0^\infty \frac{t}{1-e^{-t}} e^{-\frac12(k+1)t}\,dt \\ & = \displaystyle \int_0^\infty \frac{te^{-\frac12t}}{1 - e^{-t}} \mathrm{Li}_2\big(e^{-\frac12t}\big)\,dt \; = \; -4\int_0^1 \frac{\ln u\, \mathrm{Li}_2(u)}{1-u^2}\,du \end{array}$ Since $\frac{\mathrm{Li}_2(u)}{1-u} \; = \; \sum_{n=1}^\infty H^{(2)}_n u^n \hspace{2cm} |u| < 1$ we deduce that $\int_0^1 \frac{\mathrm{Li}_2(u)\,\ln u}{1-u}\,du \; = \; -\sum_{n=1}^\infty \frac{H^{(2)}_n}{(n+1)^2} \; = \; -\zeta(2,2:1,1) \; = \; -\tfrac{1}{120}\pi^4$ On the other hand $\frac{\mathrm{Li}_2(u)}{1+u} \; = \; \left(\sum_{j \ge 0}u^j\right)\left(\sum_{k \ge 1} \frac{u^k}{k^2}\right) \; = \; \sum_{N=1}^\infty (-1)^N\left(\sum_{k=1}^N \frac{(-1)^k}{k^2}\right)u^N$ and hence $\int_0^1 \frac{\mathrm{Li}_2(u)\,\ln u}{1+u}\,du \; = \; -\sum_{N \ge 1} \frac{(-1)^N}{(N+1)^2}\left(\sum_{k=1}^N \frac{(-1)^k}{k^2}\right) \; = \; -\zeta(2,2:-1,-1) \; = \;-\tfrac{1}{480}\pi^4$ and hence $S \; = \; -2\left( \int_0^1 \frac{\mathrm{Li}_2(u)\,\ln u}{1-u}\,du + \int_0^1 \frac{\mathrm{Li}_2(u)\,\ln u}{1+u}\,du \right) \; = \; \tfrac{1}{48}\pi^4$ making the answer $4 + 48 = \boxed{52}$ .