How to evaluate $\int_0^{2\pi}\frac{1}{(1+a\cos {\theta})^2}\,d\theta$ without contour integration?

It is not difficult to check that for any $b>1$ we have: $$ J(b) = \int_{0}^{2\pi}\frac{d\theta}{b+\cos\theta} = 4\int_{0}^{\pi/2}\frac{d\varphi}{b+\cos(2\varphi)}\\=4\int_{0}^{+\infty}\frac{dt}{(1+t^2)(b-1+2\cos^2(\arctan t))}=\frac{2\pi}{\sqrt{b^2-1}}$$ hence it follows that: $$ -J'(b) = \int_{0}^{2\pi}\frac{d\theta}{(b+\cos\theta)^2} = \frac{2b\pi}{(b^2-1)^{3/2}} $$ and by taking $a=\frac{1}{b}$ we get:

$$\forall a\in(0,1),\qquad \int_{0}^{2\pi}\frac{d\theta}{(1+a\cos\theta)^2} = \color{red}{\frac{2\pi}{ (1-a^2)^{3/2}}}.$$