TITLE: Integral $\int_0^1\frac{\ln(1+x)}{1+x^2}\left(\frac{3\arctan x}{x}+2\ln x\right)dx$
QUESTION [6 upvotes]: I was playing around with PARI GP to generate a challenging integral.
The method:
The function lindep is intended to detect integer dependence.
The constants used:
$\text{G}\ln 2,\pi^3,\pi\ln^2 2,\ln^3 2,\pi^2\ln 2,\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx$.
All these numbers are of "degree"  $3$.
To get a challenging integral you multiply, $\ln(1+x),\ln(1+x^2),\dfrac{1}{1+x^2},\arctan x,\dfrac{1}{1+x},\dfrac{1}{x}$
The idea is to sum up two integrals to cancel out the awful term $\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx$
After ten minutes, I have found:
\begin{align}A&=\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}\,dx\\
&=-3\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx+...\\
B&=\int_0^1\frac{\ln(1+x)\arctan x}{x(1+x^2)}\,dx\\
&=2\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx+...\\
\end{align}
Now, if you take $2A+3B$ you cancel out the term  $\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx$.
The result is:
\begin{align}\int_0^1\frac{\ln(1+x)}{1+x^2}\left(\frac{3\arctan x}{x}+2\ln x\right)dx=\frac{5}{128}\pi^3-\frac{7}{4}\text{G}\ln 2+\frac{3}{16}\pi \ln^2 2\end{align}
The question is: How to prove this?
Addendum:
My first candidate for $B$ was: $\int_0^1\frac{\ln(1+x)\arctan x}{1+x^2}\,dx$ but it's not expressible as an integer linear combination of the chosen constants. Probably not related to constants of "degree" $3$.
To decrease the "degree" i have multiply by $\dfrac{1}{x}$ and voilà !

REPLY [2 votes]: $$2A+3B=2\int_0^1 \frac{ \ln(1+x)\ln x}{1+x^2}dx+3 \int_0^1 \frac{ \arctan x\ln(1+x)}{x(1+x^2)}dx$$
$$=\color{blue}{2\underbrace{\int_0^1 \frac{ \ln(1+x)\ln x}{1+x^2}dx}_{=A}}+\color{green}{3\underbrace{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}_{=C}}\color{red}{-3\underbrace{\int_0^1 \frac{x\arctan x \ln(1+x)}{1+x^2}dx}_{=D}}$$

$$A\overset{IBP}=\underbrace{\arctan x\ln x\ln(1+x)\bigg|_0^1}_{=0}-\int_0^1 \frac{\arctan x\ln(1+x)}{x}dx-{\int_0^1 \frac{\arctan x \ln x}{1+x}dx}$$
$$=-C+\frac{\pi^3}{64}-\frac{G\ln 2}{2}$$
See here for above the integral.

For $D$ let $x=\frac{1-t}{1+t}$ to get:
$$D=\int_0^1 \frac{x\arctan x\ln(1+x)}{1+x^2}dx=\int_0^1 \left(\frac{\pi}{4}-\arctan t\right)\ln\left(\frac{2}{1+t}\right)\left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt$$
Sum up the two integral from above to get:
$$2D=\frac{\pi}{4}\int_0^1 \ln\left(\frac{2}{1+t}\right)\left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt$$
$$-\ln 2\int_0^1 \arctan t \left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt$$
$$\require{cancel}+\int_0^1 \arctan t \ln(1+t)\left(\frac{1}{1+t}-\cancel{\frac{t}{1+t^2}}\right)dt$$
Here, I showed combined with your approach that:
$$\boxed{\int_0^1\frac{\arctan x\ln(1+x)}{1+x}dx=\frac23 \underbrace{\int_0^1\frac{\arctan x\ln(1+x)}{x}dx}_{=C}-\frac{\pi^3}{128}+\frac{3\pi}{32}\ln^2 2}$$
And the first two integrals are pretty trivial.
$$\frac{\pi}{4}\int_0^1 \ln\left(\frac{2}{1+t}\right)\left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt=\frac{\pi}{4}\int_0^1 \ln\left(1+\frac{1-t}{1+t}\right)\frac{1-t}{1+t} \frac{dt}{1+t^2}$$
$$=\frac{\pi}{4}\int_0^1 \frac{x\ln(1+x)}{1+x^2}dx=\frac{\pi^3}{384}+\frac{\pi}{32}\ln^2 2$$
See here. Similarly for the next integral:
$$\ln 2\int_0^1 \arctan t \left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt=\ln 2 \int_0^1 \frac{x\left(\frac{\pi}{4}-\arctan x\right)}{1+x^2}dx=\frac{\pi}{4}\ln^2 2-\frac{G}{2}\ln 2$$
$$\Rightarrow D=\boxed{-\frac{\pi^3}{384}-\frac{\pi}{16}\ln^2 2+\frac{G\ln 2}{4}+\frac13 C}$$

Now we can breathe, because:
$$2A+3B=\color{blue}{-\cancel{2C}+\frac{\pi^3}{32}-G\ln 2} \color{green}{+\cancel{3C}}\color{red}{+\frac{\pi^3}{128}+\frac{3\pi}{16}\ln^2 2-\frac{3G\ln 2}{4}-\cancel{C}}$$
$$\Rightarrow \boxed{2\int_0^1 \frac{ \ln(1+x)\ln x}{1+x^2}dx+3 \int_0^1 \frac{ \arctan x\ln(1+x)}{x(1+x^2)}dx=\frac{5\pi^3}{128}+\frac{3\pi}{16}\ln^2 2-\frac{7G\ln 2}{4}}$$