Evaluate ∫02π1(1+acos⁡θ)2dθ, 0≤a<1

It is not difficult to check that for any ${\displaystyle b>1}$ we have:
${\displaystyle J\left(b\right)={\int }_0^{2\pi }\frac{d\theta }{b+\cos \theta }=4{\int }_0^{\frac{\pi }{2}}\frac{d\varphi }{b+\cos \left(2\varphi \right)}}$
${\displaystyle =4{\int }_0^{+\mathrm{\infty }}\frac{dt}{(1+t^2)(b-1+2{\cos }^2\left(\arctan t\right))}=\frac{2\pi }{\sqrt{b^2-1}}}$
hence it follows that:
${\displaystyle -J^{\prime }\left(b\right)={\int }_0^{2\pi }\frac{d\theta }{{(b+\cos \theta )}^2}=\frac{2b\pi }{{(b^2-1)}^{\frac{3}{2}}}}$
and by taking ${\displaystyle a=\frac{1}{b}}$ we get:
${\displaystyle \mathrm{\forall }a\in (0,1),{\int }_0^{2\pi }\frac{d\theta }{{(1+a\cos \theta )}^2}=\frac{2}{{(1-a^2)}^{\frac{3}{2}}}}$

We can actually evaluate this integral indefinitely.
This is the most basic thing I could come up with,
Consider
${\displaystyle f\left(x\right)=\frac{\sin x}{1+a\cos x}}$
${\displaystyle \Rightarrow f^{\prime }\left(x\right)=\frac{1+\cos x}{{(1+a\cos x)}^2}}$
Now integrate both sides,
${\displaystyle \Rightarrow f\left(x\right)=\int \frac{(\cos x+a)dx}{{(1+a\cos x)}^2}}$
Now integrate both sides,
${\displaystyle \Rightarrow f\left(x\right)=\int \frac{(\cos x+a)dx}{{(1+a\cos x)}^2}}$
${\displaystyle \Rightarrow f\left(x\right)=\frac{1}{a}\int \frac{(a\cos x+1)dx}{{(1+a\cos x)}^2}+\frac{a^2-1}{a}\int \frac{dx}{{(1+a\cos x)}^2}}$
${\displaystyle \Rightarrow \frac{\sin x}{1+a\cos x}=\frac{1}{a}\int \frac{dx}{1+a\cos x}+\frac{a^2-1}{a}\cdot I}$
Now, 'I' is the integral we wanted to evaluate and the only problem left is
${\displaystyle \int \frac{dx}{1+a\cos x}}$
which is easily calculated by putting ${\displaystyle \cos x=\frac{1-\frac{{\tan }^{2}x}{2}}{1+\frac{{\tan }^{2}x}{2}}}$
${\displaystyle \int \frac{dx}{1+a\cos x}=\frac{2}{1-a}\sqrt{\frac{1-a}{1+a}}\arctan \left(t\sqrt{\frac{1-a}{1+a}}\right)+C}$
Where ${\displaystyle t=\frac{\tan x}{2}}$

$$\begin{gathered} {\int }_0^{2\pi }\frac{d\theta }{[1+a\cos (\theta ){]}^2}={\int }_{-\pi }^{\pi }\frac{d\theta }{[1-a\cos (\theta ){]}^2}=-2\frac{d}{db}{\int }_0^{\pi }\frac{d\theta }{b-a\cos (\theta )}{|}_{b=1} \\ =-2\frac{d}{db}[{\int }_0^{\pi /2}\frac{d\theta }{b-a\cos (\theta )}+{\int }_0^{\pi /2}\frac{d\theta }{b+a\cos (\theta )}{]}_{b=1} \\ =-4\frac{d}{db}[b{\int }_0^{\pi /2}\frac{d\theta }{b^2-a^2{\cos }^2(\theta )}{]}_{b=1} \\ =-4\frac{d}{db}[b{\int }_0^{\pi /2}\frac{{\sec }^2(\theta )d\theta }{b^2{\tan }^2(\theta )+b^2-a^2}{]}_{b=1} \\ =-4\frac{d}{db}(b{\int }_0^{\mathrm{\infty }}\frac{dt}{b^2{t}^2+b^2-a^2}{)}_{b=1} \\ =-4\frac{d}{db}(\frac{1}{\sqrt{b^2-a^2}}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+1}{)}_{b=1} \\ =-2\pi \frac{d}{db}(\frac{1}{\sqrt{b^2-a^2}}){|}_{b=1}=\frac{2\pi b}{(b^2-a^2{)}^{3/2}}{|}_{b=1} \\ =\frac{2\pi }{(1-a^2{)}^{3/2}} \end{gathered}$$