I came across the following result:
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n}H_{4n-3} = \frac{\pi}{4} + \frac{\pi}{\sqrt{2}}-\frac{5\pi^2}{32}+\frac{\ln^2(2)}{8}-\frac{3\ln(2)}{2}+\frac{\ln^2(1+\sqrt{2})}{2}+\frac{\ln(3-2\sqrt{2})}{\sqrt{2}}$$
$H_{4n-3}$ refers to Harmonic Numbers
However Mathematica is not able to compute it as it gives an enormous result full of PolyLogs and Roots of functions.
Could someone please provide a proof for the above result?
The opposite series is $$ \sum_{n\geq 0}\frac{(-1)^n}{n+1}H_{4n+1} = \int_{0}^{1}\sum_{n\geq 0}\frac{(-1)^n}{n+1}\cdot\frac{x^{4n+1}-1}{x-1}\,dx =\int_{0}^{1}\frac{x^3\log(2)-\log(1+x^4)}{x^3(1-x)}\,dx$$ and Mathematica is able to deal with the integral, even if the simplification of dilogarithms has to be carried out by hand. By using integration by parts the RHS boils down to elementary integrals and $$ I_1 = \int_{0}^{1}\frac{x^3}{1+x^4}\log(x)\,dx,\qquad I_2=\int_{0}^{1}\frac{x^3}{1+x^4}\log(1-x)\,dx. $$ Of course $$ I_1 = \sum_{n\geq 0}\int_{0}^{1} (-1)^n x^{4n+3}\log(x)\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(4n+4)^2}=\frac{\pi^2}{192}$$ so the whole problem is more or less equivalent to the computation of $I_2$, which can be performed by enforcing the substitution $x=1-z$, computing a partial fraction decomposition and exploiting $$ \int_{0}^{1}\frac{\log(x)}{x-a}\,dx =\text{Li}_2\left(\frac{1}{a}\right).$$ This is strongly reminiscent of the trick I used at page $3$ here.