Reduction Formula for $I_n=\int \frac{dx}{(a+b \cos x)^n}$ Reduction Formula for $$I_n=\int \frac{dx}{(a+b \cos x)^n}$$
I considered $$I_{n-1}=\int \frac{(a+b \cos x)dx}{(a+b \cos x)^n}=aI_n+b\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$
Let $$J_n=\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$ using parts for $J_n$ we get
$$J_n=\frac{\sin x}{(a+b \cos x)^n}-nb \int \frac{\sin^2 x\:dx}{{(a+b \cos x)^{n+1}}}$$
Can we proceed here?

This is not a full answer the the original question but instead evidence that any such answer will not lead to the underlying problem at hand:  calculating the given integral.
Mathematica yields:
$$\frac{\csc (x) \sqrt{\frac{b (\cos (x)+1)}{b-a}} \sqrt{\frac{b-b \cos (x)}{a+b}} (a+b
   \cos (x))^{1-n} F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{a+b \cos
   (x)}{a-b},\frac{a+b \cos (x)}{a+b}\right)}{b (n-1)}$$
which strongly suggests you won't find a simple such reduction.  So in principle there may be a way to proceed as you ask, but it seems virtually certain that such an approach will lead to a dead end.
