How to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$ Edit
In this post I computed the following integral
$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)$$
Now I am trying to compute
$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2}G}$$
What I did is
$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\int_{0}^{1}\frac{\log(1-x)\log[(1-x^2)(1+x^2)]}{x}dx$$
$$=\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$$
$$=\frac{11}{8}\zeta(3)+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$$
$$=\frac{11}{8}\zeta(3)+I$$

$$I=\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx=\sum_{k=1}^{\infty}\frac{(-1)^{n-1}}{k}\int_{0}^{1}x^{2k-1}\log(1-x)dx$$
Integrating by parts
$$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{\frac{\log(1-x)(x^{2n}-1)}{2n}\Big|_{0}^{1}+\frac{1}{2n}\int_{0}^{1}\frac{x^{2n}-1}{1-x}dx \bigg\}$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}\int_{0}^{1}\frac{1-x^{2n}}{1-x}dx \bigg\}$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}H_{2n} \bigg\}$$
$$\boxed{I=\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}}$$
So now, the task is to evaluate this Sum

 A: First note that
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}=4\Re\sum_{n=1}^\infty\frac{i^nH_{n}}{n^2}.$$
By Cauchy product we have
$$-\ln(1-x)\operatorname{Li}_2(x)=2\sum_{n=1}^\infty \frac{H_n}{n^2}x^n+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}x^n-3\operatorname{Li}_3(x).$$
Set $x=i$ and consider the real parts,
$$-\Re\{\ln(1-i)\operatorname{Li}_2(i)\}=2\Re\sum_{n=1}^\infty \frac{i^nH_n}{n^2}+\Re\sum_{n=1}^\infty \frac{i^nH_n^{(2)}}{n}-3\Re\operatorname{Li}_3(i).$$
So we just need to calculate $\Re\sum_{n=1}^\infty \frac{i^nH_n^{(2)}}{n}$which is equivalent to $\frac12\sum_{n=1}^\infty \frac{(-1)^nH_{2n}^{(2)}}{n}$ whose integral representation $\int_0^1\frac{\ln(x)\ln(1+x^2)}{1-x}dx$ is calculated here. To show this conversion, expand $\ln(1+x^2)$ in Taylor series then integrate using $\int_0^1\frac{x^{2n}\ln(x)}{1-x}dx=\zeta(2)-H_{2n}^{(2)}.$
A: One idea is to start with the generating function
$$\sum _{n=1}^{\infty } (-1)^n H_{2 n} \,x^{2 n-1}=-\frac{\log \left(x^2+1\right)+2 x \tan ^{-1}(x)}{2x \left(x^2+1\right)}\tag{1}$$
and integrate twice (not forgetting to multiplying the result by $4$). Thus
$$2I=\sum _{n=1}^{\infty } (-1)^n \frac{H_{2n}}{n^2} =-4 \int_0^1 \frac{1}{x} \int \frac{\log \left(x^2+1\right)+2 x \tan ^{-1}(x)}{x \left(2 x^2+2\right)} \, dx\, dx$$
which becomes
$$2I=-2 \int_0^1 \frac{1}{x}\left( \frac{1}{2}\text{Li}_2\left(-x^2\right)+\frac{1}{4} \log ^2\left(x^2+1\right)-\tan ^{-1}(x)^2\right)\, dx\tag{2}$$
alternatively we could of  started with
$$\sum _{n=1}^{\infty } (-1)^n H_{2 n} x^{n-1}=-\frac{\log (x+1)+2 \sqrt{x} \tan ^{-1}\left(\sqrt{x}\right)}{x (2 x+2)}\tag{3}$$
to give
$$2I=-\int_0^1 \frac{1}{x} \left( \frac{\text{Li}_2(-x)}{2}+\frac{1}{4} \log ^2(x+1)-\tan ^{-1}\left(\sqrt{x}\right)^2 \right)\, dx\tag{4}$$
So you pick either set of three functions to integrate, or mix and match, since only a factor of 2 is between them.
One of the integrals can be easily integrated:
$$-\frac{1}{2} \int \frac{\text{Li}_2(-x)}{x} \, dx=-\frac{\text{Li}_3(-x)}{2}$$
with
$$-\frac{1}{2} \int_0^1 \frac{\text{Li}_2(-x)}{x} \, dx=\frac{3 \zeta (3)}{8}\tag{5}$$
The next integral also evaluates to $\zeta(3)$
$$-\frac{1}{4} \int_0^1 \frac{\log ^2(x+1)}{x} \, dx=-\frac{\zeta (3)}{16}\tag{6}$$
So interestingly the problem really boils down to integrating: $$2 \int_0^1 \frac{\left(\tan ^{-1}(x)\right)^2}{x} \, dx \;\;\;\;\; \text{or} \;\;\;\;\; \int_0^1 \frac{\left(\tan ^{-1}\left(\sqrt{x}\right)\right)^2}{x} \, dx$$
With an example equivalent integral being
$$4 \int_0^{\frac{\pi }{4}} y^2 \csc (2 y) \, dy=\pi \, G-\frac{7 \,\zeta (3)}{4}\tag{7}$$
$2I$ is the sum of results $(5)$, $(6)$ and $(7)$
$$2I=\frac{3 \,\zeta (3)}{8}-\frac{\zeta (3)}{16}+ \pi \, G -\frac{7\, \zeta (3)}{4}=\pi \, G-\frac{23\, \zeta (3)}{16}$$
