I have the following spherical density distribution:
$\rho(x, z) = \frac{1}{\sqrt{x^2 + z^2}\left(1+\sqrt{x^2+z^2}\right)^2}$
which I have broken into a "line of sight" dimension $z$ and a "transverse" dimension $x$. Integrating this profile along the line of sight gives the projected 2d density $\Sigma$:
$\Sigma(x) = 2\int_0^\infty\rho(x,z)dz$
I wish to compute this for any generic upper bound $\zeta$, i.e.
$\Sigma(x; \zeta) = 2\int_0^\zeta\rho(x,z)dz$
(that is, $\zeta=\infty$ corresponds to the case of projecting the entire distribution to the transverse plane, while $\zeta<\infty$ corresponds to a projection which is truncated in the $z$-dimension).
It turns out this has to be solved piecewise; the solution for $x>1$, via Mathematica 11.3, is
$$ \left.\int_0^\zeta\rho(x, z)dz\right\rvert_{x>1} = \frac{\zeta \left(\sqrt{x^2+\zeta^2}-1\right)}{\left(x^2-1\right) \left(x^2+\zeta^2-1\right)}+\frac{\tan ^{-1}\left(\frac{\zeta}{\sqrt{\left(x^2-1\right) \left(x^2+\zeta^2\right)}}\right)-\tan ^{-1}\left(\frac{\zeta}{\sqrt{x^2-1}}\right)}{\left(x^2-1\right)^{3/2}}$$
However, I am unable to obtain the solution for the case $x<1$. I currently only have access to Mathematica 12.0, rather than 11.3 which reproduces the form above, and it is failing on this integral. Performing
Assuming[{x < 1, ζ \[Element] Reals, ζ > 0},
FullSimplify[Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, ζ}]]]
returns a HyperGeometric function, though I suspect that the $x<1$ case should not be much more complicated than $x>1$. Can anyone confirm? Or see any issue?
