TITLE: A triple integral dancing in the unit cube
QUESTION [18 upvotes]: Straight integration seems pretty tedious and difficult, and I guess the symmetry may open some new ways of which I'm not aware. What would your idea be?
Calculate
$$\int_0^1 \int_0^1 \int_0^1 \frac{x^2}{\sqrt{x^2+1} \left(x^2-y^2\right) \left(x^2-z^2\right)}+\frac{y^2}{\sqrt{y^2+1} \left(y^2-x^2\right) \left(y^2-z^2\right)}+\frac{z^2}{\sqrt{z^2+1} \left(z^2-x^2\right) \left(z^2-y^2\right)} \, dx \ dy \ dz.$$
A 300 points bounty moment: After 2 years and 10 months since the problem was posed no full solution was provided yet. Is it possible to find a slick solution (like a bolt of lightning)? Good luck!

REPLY [7 votes]: The value of your triple integral is 

$$\color{royalblue}{-\frac{3}{16}\text{Li}_3\left(\frac{1}{2}\left(1-\sqrt{2}\right)\right)-\frac{3\text{Li}_3\left(2-\sqrt{2}\right)}{16}+\frac{9}{16}\text{Li}_3\left(2\left(-1+\sqrt{2}\right)\right)-\frac{3}{16}\text{Li}_3\left(\frac{1}{2+\sqrt{2}}\right)+\frac{3}{16}\text{Li}_3\left(\frac{1}{4}\left(2+\sqrt{2}\right)\right)+\frac{3}{16}\text{Li}_3\left(\frac{1}{2\left(2+\sqrt{2}\right)}\right)-\frac{3}{64}\text{Li}_3\left(3-2\sqrt{2}\right)+\frac{3}{64}\text{Li}_3\left(17-12\sqrt{2}\right)-\frac{3}{16}\text{Li}_2\left(\frac{1}{2}\left(1-\sqrt{2}\right)\right)\ln2-\frac{9}{32}\text{Li}_2\left(-1+\sqrt{2}\right)\ln2-\frac{9}{16}\text{Li}_2\left(2\left(-1+\sqrt{2}\right)\right)\ln2+\frac{3}{8}\text{Li}_2\left(\frac{1}{4}\left(2+\sqrt{2}\right)\right)\ln2+\frac{3}{8}\text{Li}_2\left(\frac{1}{\sqrt{2}}\right)\ln\left(1+\sqrt{2}\right)+\frac{3}{16}\text{Li}_2\left(-1+\sqrt{2}\right)\ln\left(1+\sqrt{2}\right)-\frac{3}{16}\text{Li}_2\left(\frac{1}{2+\sqrt{2}}\right)\ln\left(2\left(2+\sqrt{2}\right)\right)-\frac{3}{16}\text{Li}_2\left(\frac{1}{4}\left(2+\sqrt{2}\right)\right)\ln\left(2+\sqrt{2}\right)+\frac{3}{16}\text{Li}_2\left(\frac{1}{2\left(2+\sqrt{2}\right)}\right)\ln\left(2\left(2+\sqrt{2}\right)\right)+\frac{3}{8}\text{Li}_2\left(-3+2\sqrt{2}\right)\ln\left(1+\sqrt{2}\right)-\frac{3}{16}\text{Li}_2\left(3-2\sqrt{2}\right)\ln\left(1+\sqrt{2}\right)-\frac{21\zeta(3)}{512}+\frac{73\ln^32}{128}-\frac{7}{16}\ln^3\left(1+\sqrt{2}\right)-\frac{1}{16}\ln^3\left(2+\sqrt{2}\right)+\frac{39}{32}\ln\left(1+\sqrt{2}\right)\ln^22-\frac{45}{64}\ln\left(2+\sqrt{2}\right)\ln^22-\frac{21}{32}\ln^2\left(1+\sqrt{2}\right)\ln2+\frac{15}{32}\ln^2\left(2+\sqrt{2}\right)\ln2+\frac{3}{16}\ln\left(1+\sqrt{2}\right)\ln^2\left(2+\sqrt{2}\right)+\frac{3}{16}\ln^2\left(1+\sqrt{2}\right)\ln\left(2+\sqrt{2}\right)+\frac{47}{256}\pi^2\ln2-\frac{9}{16}\ln\left(1+\sqrt{2}\right)\ln\left(2+\sqrt{2}\right)\ln2+\frac{3}{16}\pi^2\ln\left(1+\sqrt{2}\right)-\frac{37}{128}\pi^2\ln\left(2+\sqrt{2}\right)}$$

Here is the Mathematica code of the result, I hope someone can simplify it further.
47/256\[Pi]^2Log[2]+(73Log[2]^3)/128+3/16\[Pi]^2Log[1+Sqrt[2]]+39/32Log[2]^2Log[1+Sqrt[2]]-21/32Log[2]Log[1+Sqrt[2]]^2-7/16Log[1+Sqrt[2]]^3-37/128\[Pi]^2Log[2+Sqrt[2]]-45/64Log[2]^2Log[2+Sqrt[2]]-9/16Log[2]Log[1+Sqrt[2]]Log[2+Sqrt[2]]+3/16Log[1+Sqrt[2]]^2Log[2+Sqrt[2]]+15/32Log[2]Log[2+Sqrt[2]]^2+3/16Log[1+Sqrt[2]]Log[2+Sqrt[2]]^2-1/16Log[2+Sqrt[2]]^3+3/8Log[1+Sqrt[2]]PolyLog[2,1/Sqrt[2]]-3/16Log[1+Sqrt[2]]PolyLog[2,3-2Sqrt[2]]-3/16Log[2]PolyLog[2,1/2(1-Sqrt[2])]-9/32Log[2]PolyLog[2,-1+Sqrt[2]]+3/16Log[1+Sqrt[2]]PolyLog[2,-1+Sqrt[2]]-9/16Log[2]PolyLog[2,2(-1+Sqrt[2])]+3/16Log[2(2+Sqrt[2])]PolyLog[2,1/(2(2+Sqrt[2]))]-3/16Log[2(2+Sqrt[2])]PolyLog[2,1/(2+Sqrt[2])]+3/8Log[2]PolyLog[2,1/4(2+Sqrt[2])]-3/16Log[2+Sqrt[2]]PolyLog[2,1/4(2+Sqrt[2])]+3/8Log[1+Sqrt[2]]PolyLog[2,-3+2Sqrt[2]]+3/64PolyLog[3,17-12Sqrt[2]]-3/64PolyLog[3,3-2Sqrt[2]]-3/16PolyLog[3,1/2(1-Sqrt[2])]-3/16PolyLog[3,2-Sqrt[2]]+9/16PolyLog[3,2(-1+Sqrt[2])]+3/16PolyLog[3,1/(2(2+Sqrt[2]))]-3/16PolyLog[3,1/(2+Sqrt[2])]+3/16PolyLog[3,1/4(2+Sqrt[2])]-(21Zeta[3])/512


I will continue from an integral representation given by Jack D'Aurizio 
$$\int_0^1 {\frac{{{{\arctan }^3}x}}{{x\sqrt {1 - {x^2}} }}dx}  = \int_0^{\pi/4} {\frac{{{x^3}}}{{\sin x\sqrt {\cos 2x} }}dx}  = \frac{1}{8}\int_{ - \pi /2}^{\pi /2} {\frac{{{x^3}\cos \frac{x}{2}}}{{\sin x\sqrt {\cos x} }}dx} $$

Next assume $a,b,c \in \mathbb{R}$, not necessarily integer with $a-b-c>0$. 
$$\begin{aligned}
&\int_{ - \pi /2}^{\pi /2} {{e^{iax}}{{( - 2i\sin x)}^b}{{(2\cos x)}^c}dx}  =  - i\int_{ - \pi /2}^{\pi /2} {{e^{i(a - 1)x}}{{({e^{ - ix}} - {e^{ix}})}^b}{{({e^{ix}} + {e^{ - ix}})}^c}d({e^{ix}})}  \\ &=  - i\int_C {{z^{a - 1}}{{\left( {\frac{1}{z} - z} \right)}^b}{{\left( {z + \frac{1}{z}} \right)}^c}dz}  =  - i\int_{ - i}^i {{z^{a - 1}}{{\left( {\frac{1}{z} - z} \right)}^b}{{\left( {z + \frac{1}{z}} \right)}^c}dz} 
\\&= {e^{ - \frac{\pi }{2}i(a - b - c - 1)}}\int_0^1 {{u^{a - 1}}{{\left( {u + \frac{1}{u}} \right)}^b}{{\left( {\frac{1}{u} - u} \right)}^c}du}  + {e^{\frac{\pi }{2}i(a - b - c - 1)}}\int_0^1 {{u^{a - 1}}{{\left( {u + \frac{1}{u}} \right)}^b}{{\left( {\frac{1}{u} - u} \right)}^c}du} \\
&=\sin \left( {\frac{\pi }{2}(a - b - c)} \right)\int_0^1 {{x^{(a - b - c)/2 - 1}}{{(1 - x)}^c}{{(1 + x)}^b}dx} \end{aligned}$$ where $C$ is the semicircle with ends at $\pm i$ and lying in the right half plane. You may want to convince yourself that all details of the contour integration is justified.
Take $a=1/2,b=-1,c=-1/2$, 
$$\int_{ - \pi /2}^{\pi /2} {{e^{iax}}\frac{1}{{( - 2i\sin x)\sqrt {2\cos x} }}dx}  = \sin \left( {\frac{\pi }{2}(a + \frac{3}{2})} \right)\int_0^1 {{x^{\frac{a}{2} - \frac{1}{4}}}{{(1 - x)}^{ - \frac{1}{2}}}{{(1 + x)}^{ - 1}}dx} $$
note that integral does not converge, but we can remedy it by first differentiating with respect to $a$, then let $b \to 1$. In any case, we have
$$\frac{1}{{\sqrt 2 }}\int_0^{\pi /2} {\frac{{{x^3}\cos \frac{x}{2}}}{{\sin x\sqrt {\cos x} }}dx}  =  - \frac{{3\pi }}{8}\color{red}{\int_0^1 {\frac{{{{\ln }^2}x}}{{\sqrt {1 - x} (1 + x)}}dx}}  + \frac{\pi^3}{8}\underbrace{\int_0^1 {\frac{1}{{\sqrt {1 - x} (1 + x)}}dx}}_J $$
Thanks to the fact that $a-b-c = 2$ in our situation, so we don't need to handle integral involving $\ln^3 x$: the red integral already poses major hindrance.
Note that $J = \sqrt{2} \ln(1+\sqrt{2})$, for the red one:
$$I  = \int_0^1 {\frac{{{{\ln }^2}(1 - x)}}{{\sqrt x (2 - x)}}dx}  = 2\int_0^1 {\frac{{{{\ln }^2}(1 - {x^2})}}{{2 - {x^2}}}dx} $$
so it boils down to the four integrals:
$$\int_0^1 {\frac{{{{\ln }^2}(1 - x)}}{{2 - {x^2}}}dx} \quad \int_0^1 {\frac{{{{\ln }^2}(1 + x)}}{{2 - {x^2}}}dx} \quad \int_0^1 {\frac{{\ln (1 + x)\ln (1 - x)}}{{\sqrt 2  - x}}dx} \quad \int_0^1 {\frac{{\ln (1 + x)\ln (1 - x)}}{{\sqrt 2  + x}}dx} $$
each one is straightforward, albeit tedious evaluation of trilogarithm/dilogarithm. Combining them gives the result.