How to calculate this hard integral $\int_0^{\infty} \frac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx$? How to prove that $\displaystyle \int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx= \frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2\left(1+\sqrt{2}\right)$ ?
It's very difficult and I have no idea.

Let $x = \tan u \implies dx = \sec^2u\ du$
$$\begin{align}\int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx =& \int_0^{\pi/2} \dfrac{\arctan(\tan u)\sqrt{\sec u-1}\sec u}{\tan u }\,du &\\=& \int_0^{\pi/2} \dfrac{u\sqrt{\sec u-1}}{\sin u }\,du&\\=& \int_0^{\pi/2} \dfrac{u\sqrt{1-\cos u}}{\sin u \sqrt{\cos u}}\,du&\\=& \sqrt{2}\int_0^{\pi/2} \dfrac{u\sin u/2}{\sin u \sqrt{\cos u}}\,du &\\=& \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} \dfrac{u}{\cos (u/2) \sqrt{\cos u}}\,du&\\=& \dfrac{1}{\sqrt{2}}\left(\left[2u \sin^{-1}\left(\tan \dfrac u2\right)\right]_0^{\pi/2} - 2\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du\right)&\\=& \dfrac{\pi^2}{2\sqrt{2}} - \dfrac{2}{\sqrt{2}}\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du \end{align}  $$
Let $z = u/2 \implies 2dz = du$
$$\begin{align}\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du&=& 2\int_0^{\pi/4}\sin^{-1}(\tan z )\, dz\end{align}$$
Let $t = \tan z$
$$\begin{align} 2\int_0^{\pi/4}\sin^{-1}(\tan z )\, dz=& 2\int_0^1\dfrac{\sin^{-1} t}{t^2 + 1} dt&\\=&  2[\tan^{-1} x \sin^{-1} x]_0^1 -
 2\int_0^1 \dfrac{\tan^{-1}(t)}{\sqrt{1-t^2}}\, dt&\\=&  2\dfrac{\pi^2}{8} -
 2\int_0^1 \dfrac{\tan^{-1}(t)}{\sqrt{1-t^2}}\, dt\end{align}$$
The integral $$\int_0^1 \dfrac{\tan^{-1}(t)}{\sqrt{1-t^2}}\, dt =  -\frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} \tag 1$$
So,
$$\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du = \ln^2(1+ \sqrt{2})$$ 
The final answer is then 
$$\dfrac{\pi^2}{2\sqrt{2}} - \sqrt{2}\ln^2(1+\sqrt{2})$$.
For the proof of $(1)$ see here.
