<s>[INST] Integral $\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx$ I am trying to solve by a different approach the fourth sum from here,  namely:
$$S= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(4n+m)} =\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx= \frac{67}{32} \zeta(3) -\frac{\pi}{2}G$$

One way to solve it is similarly to my answer from there:
$$S=\int_0^1 \frac{\ln(1-x)\ln(1-x^2)}{x}dx+\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx$$
From here we know that:
$$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$
Thus by setting $m=0,n,p,q=1$ in the first integral we get that:
$$S=\frac{11}{8}\zeta(3)+\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx=\frac{11}{8}\zeta(3)+I$$
$$I=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 x^{2n-1} \ln(1-x)dx=\frac12\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{n^2}=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G$$
And the result for $S$ follows. The last sum appears to be known, see $(659)$ from here, or alternatively since $I=2\Re\left( S(i)\right)$ just use the following identity:
$$S(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$
However I am trying to find a different method since the result is quite nice and I believe there's a nicer way to solve the integral without using such sums. 
Thus I would appreciate to get some help with the following problem:

Prove without using Euler's sum or polylogs that
  $$\int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}dx=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G$$

I also tried to consider the following integral:
$$J=\int_0^1 \frac{\ln(1+x)\ln(1+x^2)}{x}dx$$
$$\Rightarrow I+J=\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{x}dx\overset{x^2=t}=\frac12 \int_0^1\frac{\ln(1-t)\ln(1+t)}{t}dt=-\frac{5}{16}\zeta(3)$$
So now I am after the following integral:
$$I-J=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln(1+x^2)}{x}dx=\frac74 \zeta(3)-\pi G$$
 [/INST] First let's calculate the integral $\int_0^1\frac{\ln(y)\ln(1+y^2)}{1-y}dy$:
$$\int_0^1\frac{x\text{Li}_2(x)}{1+x^2}dx=\int_0^1\frac{\text{Li}_2(x)}{x}dx-\int_0^1\frac{\text{Li}_2(x)}{x(1+x^2)}dx$$
$$=\zeta(3)-\int_0^1\frac{1}{1+x^2}\left(\int_0^1\frac{-\ln(y)}{1-xy}dy\right)dx$$
$$=\zeta(3)-\int_0^1\ln(y)\left(\int_0^1\frac{-dx}{(1+x^2)(1-yx)}\right)dy$$
$$=\zeta(3)-\int_0^1\ln(y)\left(\frac{y\ln(1-y)}{1+y^2}-\frac{\ln(2)y}{2(1+y^2)}-\frac{\pi}{4(1+y^2)}\right)dy$$
$$=\zeta(3)-\int_0^1\frac{y\ln(y)\ln(1-y)}{1+y^2}dy-\frac1{16}\ln(2)\zeta(2)-\frac{\pi}{4}G$$
$$\small{\overset{IBP}{=}\zeta(3)-\frac12\int_0^1\frac{\ln(y)\ln(1+y^2)}{1-y}dy+\frac12\int_0^1\frac{\ln(1-y)\ln(1+y^2)}{y}dy-\frac1{16}\ln(2)\zeta(2)-\frac{\pi}{4}G. \quad(*)}$$
On the other hand, by integration by parts we have
$$\int_0^1\frac{x\text{Li}_2(x)}{1+x^2}dx=\frac12\ln(2)\zeta(2)+\frac12\int_0^1\frac{\ln(1-x)\ln(1+x^2)}{x}dx.\quad (**)$$
Combining $(*)$ and $(**)$ gives
$$\int_0^1\frac{\ln(y)\ln(1+y^2)}{1-y}dy=2\zeta(3)-\frac98\ln(2)\zeta(2)-\frac{\pi}{2}G.$$
This integral can be converted to $\sum_{n=1}^\infty (-1)^n \frac{H_{2n}^{(2)}}{n}$ by expanding $\ln(1+x^2)$  in series, which can be related to $\sum_{n=1}^\infty (-1)^n \frac{H_{2n}}{n^2}$ through the Cauchy product of $-\ln(1-x)\text{Li}_2(x)$, which is related to your integral as you showed in your question body. this way, we avoided using the generating function of $\sum_{n=1}^\infty\frac{H_n}{n^2}x^n$
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