Is it possible to evaluate, in closed-form, the following sum:
$$\sum_{n=1}^{\infty} \frac{(-1)^n \, \psi^{(1)} (\sqrt{2}n)}{n}$$ or perhaps more generally $$f(x):=\sum_{n=1}^{\infty} \frac{(-1)^n \, \psi^{(1)} (n x)}{n}$$ where $\psi^{(1)}(x)$ is the trigamma function.
Context:
This series comes from an attempt of mine at this question whereby I managed to show $$\int_{-\infty}^{\infty} \frac{\csc \left(\frac{\pi+2iz}{2\sqrt{2}}\right)\operatorname{sech}^2 (z)}{\pi+2iz}\,dz=\frac{(16\sqrt{2}-3)\zeta(3)}{8\pi^2}-\frac{2}{\pi^2}\sum_{n=1}^{\infty} \frac{(-1)^n \, \psi^{(1)} (\sqrt{2}n)}{n}$$
Here is my attempt at evaluating the sum:
By using $$\psi^{(1)} (z) = \int_{0}^{1} \frac{x^{z-1} \ln (x)}{x-1} \,dx$$ we can determine that $$f(\sqrt{2})=\sum_{n=1}^{\infty} \frac{(-1)^n \, \psi^{(1)} (\sqrt{2}n)}{n}=\int_{0}^{1} \frac{\ln (x) \ln\left(1+x^{\sqrt{2}}\right)}{x(1-x)} \, dx\\=-\frac{3\zeta (3)}{8}+\int_{0}^{1} \frac{\ln (x) \ln\left(1+x^{\sqrt{2}}\right)}{x-1} \, dx$$ which appears in similar nature to other logarithmic integrals with irrational exponents that are evaluable via the methods of algebraic number theory such as: $$\int_{0}^{1} \frac{\log (1+x^{2+\sqrt{3}})}{1+x} \, dx=\frac{\pi^2}{12} (1-\sqrt{3})+\log (2) \log(1+\sqrt{3})$$ $$\int_{0}^{1} \frac{\arctan(x^{3+2\sqrt{2}})}{1+x^2}\,dx=\frac{1}{16}\log (2) \log(3+2\sqrt{2})$$ i.e Herglotz-type integrals.
Instead introducing $$\psi^{(1)} (z) = \sum_{n=0}^{\infty} \frac{1}{(z+n)^2}$$ we find $$f(\sqrt{2})=-\frac{3\zeta(3)}{8} -\frac{\pi^2}{6} \ln (2)+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \left(\psi^{(0)}\left(\frac{4+\sqrt{2}n}{4}\right)-\psi^{(0)}\left(\frac{2+\sqrt{2}n}{4}\right)\right)-\frac{\sqrt{2}}{8} \sum_{n=1}^{\infty}\frac{1}{n}\left(\psi^{(1)}\left(\frac{4+\sqrt{2}n}{4}\right)-\psi^{(1)}\left(\frac{2+\sqrt{2}n}{4}\right)\right)$$
I credit @TheSimpliFire for the following representation: $$f(\sqrt{2}) = \left(2\sqrt{2}-\frac{3}{8}\right)\zeta (3) - \frac{\pi^2}{6} \log (2)-\sqrt{2} \sum_{n=1}^{\infty} \int_{1}^{\infty} \frac{n \log x+1}{n^2 (x^{n+1}+x^{n+1-\sqrt{2}})}\,dx$$
When $x \in \mathbb{N}$ it seems that $f(x)$ can be computed with some strenuous effort by turning it into a similar integral as said above, namely $$f(x) = \int_{0}^{1} \frac{\ln (t) \ln (1+t^{x})}{t(1-t)}\,dt$$ for which there exist multiple such logarithmic integrals here on MSE that @pisco's MZIntegrate Mathematica package can generally readily compute such as:
$$f(1) = \frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$$
$$f(2) = \frac{29}{16} \zeta (3) - \frac{3\pi^2}{16} \log (2)-\frac{\pi G}{2}$$ where $G$ is Catalan's constant.
