Evaluating $\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}$ 
How to evaluate $$I=\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}?$$

It equals $\frac5{64}\pi^3-\frac92G\ln2+\frac14\pi\ln^22$ according to Mathematica, where $G$ denotes Catalan's constant.
Attempt
$$I=\frac d{ds}\int_0^1\ln(x^2+x^3)\frac{dx}{(1+x^2)^{1-s}}$$
or, $$I=\int_0^{\pi/4}2\ln\sec t\ln(\tan^2t(1+\tan t))dt$$
$$=2\int_0^{\pi/4}\left(\ln2+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}n\right)\left(-2\sum_{n=1}^\infty\frac{\cos(4n-2)x}{2n-1}+\ln(1+\tan x)\right)dx$$
$$=-4G\ln2+\frac14\pi\ln^22+2\sum_{n=1}^\infty\frac{(-1)^n}n\int_0^{\pi/4}\cos(2nx)\ln(\tan^2 x+\tan^3x)dx$$
 A: $$I=\int_0^1\frac{\ln(1+x^2)(2\ln(x)+\ln(1+x))}{1+x^2}dx$$
$$=2\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx+\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{1+x^2}dx$$
$$=2I_1+I_2$$
$I_1$ is calculated here:
$$\boxed{I_1=-2\,\Im\operatorname{Li_3}(1+i)+\frac{3\pi^3}{32}+\frac{\pi}8\ln^2(2)-\ln(2)G}$$
For $I_2$, let $x\to (1-x)/(1+x)$
$$I_2=\int_0^1\frac{\ln\left(\frac{2}{1+x}\right)\ln\left(\frac{2(1+x^2)}{(1+x)^2}\right)}{1+x^2}dx$$
$$=\ln(2)\underbrace{\int_0^1\frac{\ln\left(\frac{2(1+x^2)}{(1+x)^2}\right)}{1+x^2}dx}_{x\to (1-x)/(1+x)}-\ln(2)\int_0^1\frac{\ln(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx-I_2$$
$$=\ln(2)\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx-\ln(2)\int_0^1\frac{\ln(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx-I_2$$
$$=\ln(2)\left(\frac{\pi}{2}\ln(2)-G\right)-\ln(2)\left(\frac{\pi}{8}\ln(2)\right)$$
$$+2\left(4\,\mathfrak{J}\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^2(2)-2\ln(2)G\right)-I_2$$
$$\Longrightarrow \boxed{I_2=4\,\mathfrak{J}\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac52\ln(2)G}$$
$$\Longrightarrow I=2I_1+I_2=\frac{5\pi^3}{64}+\frac{\pi}{4}\ln^2(2)-\frac92\ln(2)G.$$
