Evaluating $\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$ 
How can we find the value of $$\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$$ using elementary methods?

With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.
Thoughts of this integral
Since I have asked this question and Pisco gave a brilliant answer, I tried to convert $$I_1=\int_0^1\arctan x\ln(1+x)\frac{dx}x\text{ and }I_2=\int_0^1\arctan x\ln(1+x)\frac{dx}{1+x}$$ into the form of integral Pisco gave.
Integrating by parts to the second integral converts $I_2$ into $\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx$.
But for $I_1$? Integrating by parts gives a dilog function and I tried substitution $x=\frac{1-t}{1+t}$ and got $$\frac{\ln\frac{2}{t+1} \arctan\frac{1-t}{1+t}}{1-t^2}$$ which is not what I want.

$$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}\ dx$$
Apply IBP for the second integral, we get, 
$$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-\frac{3\pi}{8}\ln^22+\frac32\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx$$
The first integral was calculated here :

$$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2-3\Im\operatorname{Li}_3(1+i)$$

And the second integral was calculated here:

$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2G\ln2$$

Substituting these two results, we get $\quad\displaystyle \boxed{I={\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}}$
