How can I integrate this function? $\frac{\sqrt{e^{2x+2y+z}}}{(1+e^x+e^y+e^{x+y+z})^2}$ I want to evaluate the following integration
$$
\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \int_{-\infty}^{\infty} dz \frac{\sqrt{e^{2x+2y+z}}}{(1+e^x+e^y+e^{x+y+z})^2}.
$$
According to Mathematica12, the answer is $\pi^2$.
How can I get the answer?

 A: Let $x= \ln X$, $y=\ln Y$ and $z=2\ln Z$. Then
$$ I = \int_0^\infty\int_0^\infty\int_0^\infty\frac{XYZ}{(1+X+Y+XYZ^2)^2} \frac{dX}{X}\frac{dY}{Y}\frac{2\cdot dZ}{Z}.$$
This is a rational function, and thus can be easily integrated. For instance integrate first against X:
$$ I = 2\int_0^\infty\int_0^\infty \frac{1}{1+YZ^2}\frac{1}{1+Y}dY dZ. $$
Then integrate against $Z$:
$$ I = 2\int_0^\infty \frac{\pi}{2\sqrt{Y}}\frac{1}{1+Y}dY . $$
Then, let $u=\sqrt{Y}$ to obtain
$$I= \pi\int_0^\infty \frac{2u }{u}\frac{1}{1+u^2}du=\pi^2.$$
A: This is not an answer.
For the most inner integral, the antiderivative is not too bad. Hoping no mistake
$$I=\int \frac{\sqrt{e^{2(x+y)+z}}}{(1+e^x+e^y+e^{x+y+z})^2}\,dz$$
$$I=\frac{\sqrt{e^{2( x+ y)+z}}}{\left(1+e^x+e^y\right)
   \left(1+e^x+e^y+e^{x+y+z}\right)}+\frac{\sqrt{e^{x+y}}}{\left(1+e^x+e^y\right)^{3/2}}\tan^{-1} \left(\sqrt{\frac{e^{x+y+z}}{1+e^x+e^y} }\right)$$ The problem is for the definite integral.
If you are able to find it, may be the next could be easier.
