Solution of Integral $\int_{0}^{a}\frac{1}{(b\cos^{2}\theta + n)^{n}}d\theta $

Question

I'm looking for the solution of the following integral.

$$\int_{0}^{a}\frac{1}{(b\cos^{2}\theta + n)^{n}}d\theta $$

where $a\leqslant \frac{\pi}{2}$ and $n$ is an integer

Thank you very much for the help.

Answer 1

With the help of a CAS:

$\int_0^a \frac{1}{\left(b \cos ^2(x)+n\right)^n} \, dx=\frac{1}{4} i \left(\frac{2^n F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{b+2 n+b \cos (2 a)}{2 n},\frac{b+2 n+b \cos (2 a)}{2 (b+n)}\right) (b+2 n+b \cos (2 a))^{1-n}}{(-1+n) \sqrt{n (b+n)}}-2 \sqrt{n} (b+n)^{\frac{1}{2}-n} \sqrt{\pi } \Gamma (-n) \, _2\tilde{F}_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{b+n}{n}\right)\right)$

Mathematica code:

 HoldForm[Integrate[1/(b Cos[x]^2 + n)^n, {x, 0, a}] == 
 1/4 I ((2^
   n AppellF1[1 - n, 1/2, 1/2, 2 - n, (b + 2 n + b Cos[2 a])/(
    2 n), (b + 2 n + b Cos[2 a])/(
    2 (b + n))] (b + 2 n + b Cos[2 a])^(1 - n))/((-1 + n) Sqrt[
   n (b + n)]) - 
  2 Sqrt[n] (b + n)^(1/2 - n) Sqrt[\[Pi]]
    Gamma[-n] Hypergeometric2F1Regularized[1/2, 1 - n, 3/2 - n, (
    b + n)/n])] // TraditionalForm

Maple code:

 int(1/(b*cos(x)^2+n)^n, x = 0 .. a) = ((1/4)*I)*(2^n*AppellF1(1-n, 1/2, 
 1/2, 2-n, (b+2*n+b*cos(2*a))/(2*b+2*n), (1/2)*(b+2*n+b*cos(2*a))/n)*
 (b+2*n+b*cos(2*a))^(1-n)/((-1+n)*(n*(b+n))^(1/2))-2*n^(1/2)*(b+n)^(1/2-
 n)*Pi^(1/2)*GAMMA(-n)*hypergeom([1/2, 1-n], [3/2-n], (b+n)/n)/GAMMA(3/2-n))