Reduction Formula for $I_n=\int \frac{dx}{(a+b \cos x)^n}$

Question

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?

Answer 1

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.

Answer 2

If I've got my coefficients right, $$\sin^2 x =1-\cos^2 x=-\frac{1}{b^2}(a+b\cos x)^2+\frac{2a}{b^2}(a+b\cos x)+1-\frac{a^2}{b^2}.$$ Thus $$\frac{\sin x (a+b\cos x)^{-n}-J_n}{nb}=\int\frac{1-\cos^2 x}{(a+b\cos x)^{n+1}}dx=(1-\frac{a^2}{b^2})I_{n+1}+\frac{2a}{b^2}I_n-\frac{1}{b^2}I_{n-1}.$$You'll want to double-check what comes next, but I get:$$(1-n)I_{n-1}=(1-2n)aI_n+\frac{b\sin x}{ (a+b\cos x)^n}-nb^2(1-\frac{a^2}{b^2})I_{n+1}.$$