Can I get some help calculating the following integral
$$I = \int_0^1 x(\arctan x)\ln(\operatorname{arctanh}x)dx$$
By using $\operatorname{arctanh}x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ we have
$$I=\left(\frac{1}{2}-\frac{\pi}{4}\right)\ln2+\int_{0}^{1}x\left(\arctan x\right)\ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx$$
substituting $x\to\frac{1+t}{1-t}$ we get
$$\begin{align} I&=\left(\frac{1}{2}-\frac{\pi}{4}\right)\ln2+2\int_{1}^{\infty}\frac{t-1}{\left(t+1\right)^{3}}\arctan\left(\frac{t-1}{t+1}\right)\ln\left(\ln\left(t\right)\right)dt\\ &=\left(\frac{1}{2}-\frac{\pi}{4}\right)\ln2+2\int_{1}^{\infty}\frac{t-1}{\left(t+1\right)^{3}}\arctan\left(t\right)\ln\left(\ln\left(t\right)\right)dt-\frac{\pi}{2}\int_{1}^{\infty}\frac{t-1}{\left(t+1\right)^{3}}\ln\left(\ln\left(t\right)\right)dt \end{align}$$
This form bears similarity to Malmsten's integrals detailed by Blagouchine.
The potential key would be finding the anti-derivative of $$\frac{t-1}{\left(t+1\right)^{3}}\ln\left(\ln\left(t\right)\right)$$ then performing integration by parts.
Small update $$\int\frac{t-1}{\left(t+1\right)^{3}}\ln\left(\ln\left(t\right)\right)=\frac{\operatorname{li}(t)-t\ln(\ln(t))}{(t+1)^2}+2\int\frac{\operatorname{li}(t)}{(t+1)^3}dt$$
Edit
I found that
$$2I = \left(2-\frac{\pi}{2}\right)\ln2-\ln\pi+\gamma+\pi\ln\left(\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\sqrt{2\pi}\right)+\int_{0}^{1}\frac{\arctan t}{\operatorname{arctanh}t}dt$$
where the last integral is the infamous egg. So finding a closed form may prove very difficult.
This comes from the fact that $$\frac{d}{dx}(x^2-1)\arctan(x)\ln(\operatorname{arctanh}(x)) = 2x\arctan(x)\ln(\operatorname{arctanh}(x))+\left(\frac{x^2-1}{x^2+1}\right)\ln(\operatorname{arctanh}(x))-\frac{\arctan(x)}{\operatorname{arctanh}(x)}$$
then evaluating $$\int_0^1 \left(\frac{x^2-1}{x^2+1}\right)\ln(\operatorname{arctanh}(x)) dx$$