Help finding integral of chord length divided by Cauchy PDF

Question

I've tried many different ways to find this integral but can't find it. Is anyone able to help me by finding the integral for this? If it's easier to find the definite integral, I only need to integrate it from $a-r$ to $a+r$. Also, all variables will always be positive, real integers. If it helps, this equation is based on the chord length of a circle divided by the Cauchy-Lorentz distribution function.

$\int \frac{\sqrt{r^2-(a-x)^2}}{(x-x_0)^2+\gamma^2}\,dx$

Answer 1

Using Mathematica (and lots of simplification) I get the following:

$$\frac{\pi \sqrt{-a^2+\sqrt{\left(-(a-x_0)^2+\gamma ^2+r^2\right)^2+4 \gamma ^2 (a-x_0)^2}+2 a x_0+\gamma ^2+r^2-x_0^2}}{\sqrt{2} \gamma }-\pi$$

For $a=3$, $r=5$, $x_0=1$, and $\gamma=2$, I get

$$\frac{1}{4} \left(\sqrt{2 \left(\sqrt{689}+25\right)}-4\right) \pi \simeq 4.80987$$

If I use numerical integration, I get the same value:

parms = {a -> 3, r -> 5, x0 -> 1, γ -> 2};
NIntegrate[Sqrt[r^2 - (a - x)^2]/((x - x0)^2 + γ^2) /. parms,
  {x, a - r, a + r} /. parms]
(* 4.80987 *)