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:
$$I-J=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln(1+x^2)}{x}dx=\frac74 \zeta(3)-\pi G$$
 A: Here is my attempt to compute $I-J$ using complex analysis method (I hope you don't mind.) Let $f(z) $ be an analytic function defined by $\displaystyle f(z) = \frac{2\log(1+z)\log(1+iz)}z$ on the unit disk. By Cauchy's integral theorem, we have that
\begin{align*}
\int_{[0,i]} f(z)dz - \int_{[0,1]} f(z) dz =&  \int_{[1,i]} f(z)dz.
\end{align*} Then the LHS is ($[0,i]$ is parametrized by $z = ix, x\in [0,1]$)
\begin{align*}
\int_{[0,i]} f(z)dz - \int_{[0,1]} f(z) dz =&\int_0^1 \frac{2\log(1+ix)\log(1-x)}{x} dx -\int_0^1 \frac{2\log(1+x)\log(1+ix)}x dx  \\
 =& \color{red}{\int_0^1 \frac{2\log\left(\frac{1-x}{1+x}\right)\log(1+ix)}x dx}.
\end{align*} On the other hand, the RHS is ($[1,i]$ is parametrized by $z = e^{i\theta}, \theta \in [0,\frac\pi 2]$)
\begin{align*}
\int_{[1,i]} f(z)dz =&2i \int_0^{\frac\pi 2} \log(1+e^{i\theta})\log(1+ie^{i\theta}) d\theta \\
=&\color{blue}{2i\int_0^{\frac \pi 2}\Big[\log\left(2\cos(\theta/2)\right) + i\theta/2\Big]\Big[\log(2\cos\left(\theta/2 +\pi /4\right)+i(\theta/2+\pi/4)\Big]d\theta }
\end{align*} where we have used $\log(1+e^{i\theta}) = \log(2\cos (\theta/2)) + i\theta/2$ for $|\theta|<\pi$.
Note that for all real $x$, it holds that $2\Re[\log(1+ix)] =  \ln(1+x^2)$. So by equating the real parts of $\color{red} {\text{red}}$ and $\color{blue} {\text{blue}}$ integrals, we get
\begin{align*}
I-J =& \Re\left[\int_0^1 \frac{2\log\left(\frac{1-x}{1+x}\right)\log(1+ix)}x dx\right]\\
=&-\int_0^{\frac \pi 2}  \left(\theta+ \frac \pi 2\right)\log(2\cos (\theta/2)) -\int_0^{\frac \pi 2} \theta \log(2\cos(\theta/2 + \pi /4)) d\theta\\
=& -\int_0^{\frac \pi 2}  \left(\theta+ \frac \pi 2\right)\log(2\cos (\theta/2)) -\int_0^{\frac \pi 2} \left(\frac \pi 2 -\theta\right) \log(2\sin(\theta/2)) d\theta \\
=& \int_0^{\frac \pi 2} \theta \log (\tan (\theta/2))d\theta -\frac \pi 2\left(\int_0^{\frac \pi 2} \log(2\cos(\theta/2)) d\theta+\int_0^{\frac \pi 2} \log(2\sin(\theta/2)) d\theta\right)\\
=& \int_0^{\frac \pi 2} \theta \log (\tan (\theta/2))d\theta 
\end{align*} because
\begin{align*}
\int_0^{\frac \pi 2} \log(2\cos(\theta/2)) d\theta+\int_0^{\frac \pi 2} \log(2\sin(\theta/2)) d\theta =& \int_0^{\frac \pi 2} \log(2\cos(\theta/2)) d\theta+\int_{\frac \pi 2}^\pi \log(2\cos(\theta/2)) d\theta\\
=& \int_0^{\pi } \log(2\cos(\theta/2)) d\theta \\=& 0.
\end{align*}
Finally, using the Fourier series of $\displaystyle \log\left(\tan\left(\theta/2\right)\right) =  \sum_{k=1}^\infty \frac{(-1)^k-1}{k}\cos(k\theta)$, we get
\begin{align*}
I - J =&\sum_{k=1}^\infty \frac{(-1)^k-1}{k}\int_0^{\frac \pi 2}\theta\cos(k\theta)d\theta\\
=&\sum_{k=1}^\infty \frac{(-1)^k-1}{k}\left(\frac{\pi\sin(k\pi /2)}{2k}+\frac{\cos(k\pi /2) - 1}{k^2}\right)\\
\overset{k=2j+1}=&\sum_{j=0}^\infty \left[\frac{\pi (-1)^{j+1}}{(2j+1)^2} +\frac 2{(2j+1)^3}\right]\\
=& -\pi \text{G} + \frac 7 4\zeta(3).
\end{align*}
A: Applying the integration by parts and rearranging everything, we obtain 
$$
J=\int_0^1 \frac{\log (1+x) \log \left(1+x^2\right)}{x} \textrm{d}x=-2\int_0^1\frac{x \log (x) \log (1+x)}{1+x^2} \textrm{d}x$$
$$+\int_0^1 \left(\frac{(1-x) \log (x) \log \left(1-x^2\right)}{1-x^2}-\frac{(1-x) \left(1+x^2\right) \log (x) \log \left(1-x^4\right)}{1-x^4}\right)\textrm{d}x.$$
The integrals $\displaystyle U=\int_0^1\frac{x \log (x) \log (1-x)}{1+x^2} \textrm{d}x$ and $\displaystyle V=\int_0^1\frac{x \log (x) \log (1+x)}{1+x^2} \textrm{d}x$ are easily calculated in the book, (Almost) Impossible Integrals, Sums, and Series, (see pages $8$-$9$) by calculating $U-V$ and $U+V$. No need to use Euler sums or Polylogs, but you might need to accept Beta function.
Note that by expanding 
$$\int_0^1 \left(\frac{(1-x)\log (x) \log \left(1-x^2\right)}{1-x^2}-\frac{(1-x) \left(1+x^2\right) \log (x) \log \left(1-x^4\right)}{1-x^4}\right)\textrm{d}x$$
you only have Beta functions.
Similar approach for $I$.
That's all.
ADDENDUM: Since I mentioned the use of the integrals  $\displaystyle U=\int_0^1\frac{x \log (x) \log (1-x)}{1+x^2} \textrm{d}x$ and $\displaystyle V=\int_0^1\frac{x \log (x) \log (1+x)}{1+x^2} \textrm{d}x$, the sum $U+V$ reduces to the calculation of the integral $\displaystyle \int_0^1 \frac{\log(x)\log(1-x)}{1+x}\textrm{d}x$ evaluated separately in Section $2$, page $4$, in the new preprint  A note presenting the generalization of a special logarithmic integral by Cornel Ioan Valean, with no use of Beta function, Polylogarithm, or Euler sums.
More generally (if you're possibly interested), we have 
$$\int_0^1 \frac{\log ^{2n-1}(x) \log(1-x)}{1+x} \textrm{d}x$$
$$=\frac{1}{2}(2n)!\zeta (2n+1)-2\log(2)(1 -2^{-2n})(2n-1)!\zeta (2n)$$
$$-2^{-1-2n} (2n+1-2^{1+2n})(2n-1)!\zeta(2n+1)$$
$$-(2n-1)!\sum_{k=1}^{n-1}\zeta (2k)\zeta (2n-2k+1)+2^{-2n}(2n-1)!\sum_{k=1}^{n-1}2^{2k}\zeta (2k)\zeta (2n-2k+1),$$
where $\zeta$ represents the Riemann zeta function.
This last result could be new in the literature.
The case $U-V$ is easy to see it can be calculated again with no use of Beta function, Polylogarithm, or Euler sums.
A: Alternatively, the integrals $\displaystyle \int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}\textrm{d}x$ and $\displaystyle \int_0^1 \frac{\ln(1-x)\ln(1+x^2)}{x}\textrm{d}x$ may be seen as disguised forms of the integrals $\displaystyle \int_0^1 \frac{x\operatorname{Li}_2(x)}{1+x^2}\textrm{d}x$ and $\displaystyle \int_0^1 \frac{x\operatorname{Li}_2(-x)}{1+x^2}\textrm{d}x$ (to see this integrate by parts), which both appear in the book (Almost) Impossible Integrals, Sums, and Series, pages $123$-$126$ and that are evaluated by real methods exclusively.
A: 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 $\sum_{n=1}^\infty\frac{H_n}{n^2}x^n$
