Compute integral $\int_0^1\int_0^1\int_0^1 \sqrt{x^2+y^2+z^2} \,\mathrm{d}x\mathrm{d}y\mathrm{d}z$ How to Compute 
$$\int_0^1\int_0^1\int_0^1 \sqrt{x^2+y^2+z^2} \,\mathrm{d}x\mathrm{d}y\mathrm{d}z $$
The Mathematica 9.01 give a result is $$-\frac{\pi}{24}+\frac{1}{4}(\sqrt{3}+\log(7+4\sqrt{3})).$$ I want to know how to get it.

Consider the following figure:

Due to symmetry one sixth of the rays emanating from $(0,0,0)$ leave the cube through the red triangle $T$. We therefore use the parametrization
$$\psi:\quad \left.\eqalign{x&=t\> u\cr y&=t\>v \cr z&=t\cr}\right\}\qquad(0\leq v\leq u, \ 0\leq u\leq 1, \ 0\leq t\leq 1)$$
with Jacobian $J_\psi=t^2$. It follows that the integral $(=:Q)$ in question is given by
$$Q=6\int_0^1\int_0^1\int_0^u t\sqrt{1+u^2+v^2}\>t^2 dv\ du\ dt={3\over2}\int_0^1\int_0^u \sqrt{1+u^2+v^2}\  dv\ du\ .$$
Herewith the geometrical difficulties are out of the way and it remains to compute the nested integral using standard calculus techniques. A first step could be introducing polar coordinates
$$u=r\cos\phi, \quad v=r\sin\phi\qquad\left(0\leq r\leq{1\over\cos\phi},\quad 0\leq\phi\leq{\pi\over4}\right)\ ,$$
which leads to
$$Q={3\over2}\int_0^{\pi/4}\int_0^{1/\cos\phi}r\>\sqrt{1+r^2}\ dr\ d\phi\ .$$
The inner integral is easy, and we are left with
$$Q={1\over2}\int_0^{\pi/4}\left((1+\cos^{-2}\phi)^{3/2}-1\right)\ d\phi\ ,$$
which has the quoted value $\ldots$
