Let me set
$$k^2=\frac{4ar}{(r+a)^2} <1. $$
If we replace $\theta$ with $2\theta$ we reduce this to an integral;
$$ 2 \underbrace{\int_0^\pi \sqrt{1-k^2\cos^2\theta} d\theta}_{=I}. $$
Now set
$$ x=\cos\theta $$
so that $$ dx=-\sqrt{1-x^2} d\theta $$
and
$$ I=\int_{-1}^1\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx= 2\underbrace{\int_0^2 2\underbrace{\int_0^1 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx}_{=:E_2(k)}. $$
The integral $E_2(k)$ is called Jacobi's complete elliptic integral of the second kind. There is no simple formula for it but you can have a look at the beautiful book by H. McKean and V. Moll, Elliptic Curves. Function Theory, Geometry Arithmetic Cambridge University Press,1997.
