Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$  ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$  .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved.  So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.

For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$

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UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand).  I have also added an answer based on Quanto's but with the intermediate steps written out.

Here I have taken the compact answer by /u/Quanto (to whom all credit is due) and simply filled in some intermediate steps.  His trick was to (twice) differentiate inside the integral  ( a special case of the Leibniz integral rule ).
Using the known definite integral:- $\int_0^{2\pi}\frac{1}{E+F\cos x}\mathrm{d}x= \frac{2\pi}{\sqrt{E^2-F^2}}$  with $E=1$, $F=-a$ and $x=\theta$, we can write:-
$$
\int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}}
$$
We can multiply both sides by $a$ (so long as $a$ is not a function of $\theta$) to give:-
$$
\int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2a\pi}{\sqrt{1-a^2}}
$$
now differentiate both sides (within the integral on the LHS) with respect to $a$:-
$$\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta
=\frac{2\pi}{(1-a^2)^{3/2}}
$$
again, differentiate both sides (within the integral on the LHS) with respect to $a$:-
$$\int_{0}^{2\pi} \frac{2\cos \theta}{(1-a \cos \theta)^3} d\theta
=\frac{6a\pi}{(1-a^2)^{5/2}}
$$
Hence:-
$$\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^3} d\theta
=\frac{3a\pi}{(1-a^2)^{5/2}}.
$$
