Calculating $\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du$ What real tools would you employ for calculating the integral below? Some useful ideas I mean.
$$\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi  \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du$$
 A: Continuing from Leucippus' answer, through $\text{arctanh}(z)=\frac{1}{2}\log\frac{1+z}{1-z}$, we just need to compute:
$$ I_1 = \int_{0}^{1}\frac{dt}{(1+t)\sqrt{1+t^2}}=\frac{1}{\sqrt{2}}\text{arcsinh}(1)=\frac{1}{\sqrt{2}}\log(1+\sqrt{2}),$$
$$ I_2 = \int_{0}^{1}\frac{\log(\sqrt{2}+\sqrt{1+t^2})}{(1+t)\sqrt{1+t^2}}\,dt,\qquad I_3 = \int_{0}^{1}\frac{\log(\sqrt{2}-\sqrt{1+t^2})}{(1+t)\sqrt{1+t^2}}\,dt$$
and since 
$$ \int\frac{dt}{(1+t)\sqrt{1+t^2}} = \frac{1}{\sqrt{2}}\,\log\left(1-t+\sqrt{2}\sqrt{1+t^2}\right) $$
integration by parts looks like a good way to find $I_2$ and $I_3$ in terms of dilogarithms.
I have the strong feeling that the Catalan-Harmonic identity is involved, or some special value of the Rogers L-function.

If we substitute $t=\tan\theta$ in Leucippus' $(1)$ we get:
$$ I = 2\sqrt{\pi}\int_{0}^{\pi/4}\text{arctanh}\left(\frac{1}{\sqrt{2}\cos\theta}\right)\frac{d\theta}{\sin\theta+\cos\theta}$$
so we just need to compute:
$$ J_k = \int_{0}^{\pi/4}\frac{d\theta}{\cos^{2k+1}(\theta)(\sin\theta+\cos\theta)}=\int_{0}^{1}(1+t^2)^{k+1}\frac{dt}{1+t}$$
then exploit the Taylor series of $\text{arctanh}$.
