Evaluating $\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$

Here is an elementary approach, although it turned into a crossover with FDP's answer.

First note that from here we have: $$\color{blue}{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}=\frac{3}{2}\int_0^1 \frac{\arctan x\ln(1+x^2)}{x}dx$$ $$\overset{IBP}=\frac32 \underbrace{\ln x\arctan x\ln(1+x^2)\bigg|_0^1}_{=0}-\frac32 \left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right) $$ Back to the original integral, we have: $$I=\color{blue}{2\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}-\color{red}{3\int_0^1 \frac{\arctan x \ln(1+x)}{1+x}dx} $$ $$=\color{blue}{-3\left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right)}-\color{red}{3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}dx}$$ $$\Rightarrow I=-3(B+2A+J)\quad \quad (1)$$ Where I kept the notation like in FDP's answer. Namely: $$\begin{align*} \displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\ \displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\ \displaystyle J&=\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx \end{align*}$$ Also another two identities follows from that post, see $(8)$ and $(9)$: $$J=\dfrac{5}{3}G\ln 2-\dfrac{\pi^3}{128}+\dfrac{3\pi\left(\ln 2\right)^2}{32}+B+\dfrac{2}{3}\left(\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}\right)-\dfrac{2}{3}\cdot\frac{\pi^3}{32} $$ $$\Rightarrow \color{purple}{J=2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B} \tag 2$$ $$\color{magenta}{A=\dfrac{1}{64}\pi^3-B-G\ln 2} \tag 3$$ Now plugging $(2)$ and $(3)$ in $(1)$ yields: $$I=-3\left(B+2\left(\color{magenta}{\dfrac{1}{64}\pi^3-B-G\ln 2}\right)+ \color{purple}{2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B}\right)$$ $$\Rightarrow I=-3\left(-\frac{\pi^3}{128}+\frac{3\pi}{32}\ln^2 2\right)=\boxed{\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}$$ Credits to FDP for his amazing answer there!