func = a^2 *(1/(1 + b Cos[x]))^2/c
  int = Integrate[func, x]
  (*(a^2 (-((2 ArcTanh[((-1 + b) Tan[x/2])/Sqrt[-1 + b^2]])/(-1 + b^2)^(
      3/2)) + (b Sin[x])/((-1 + b^2) (1 + b Cos[x]))))/c*)
  D[int, x] // FullSimplify 
  (*a^2/(c (1 + b Cos[x])^2)*)
  func == D[int, x] // FullSimplify
  (*True*)

The derivative of antiderivative gives the original function .

for y[0]= 0 then we have:

 Limit[int, x -> 0](*Then C = 0 *)
 y[x] == int // Expand
 (*y[x] == -((2 ArcTanh[Tan[x/2]/Sqrt[3]])/(3 Sqrt[3])) + (2 Sin[x])/(
   3 (1 + 2 Cos[x]))*)

Solve[y[x] == -((2 ArcTanh[Tan[x/2]/Sqrt[3]])/(3 Sqrt[3])) + (2 Sin[
   x])/(3 (1 + 2 Cos[x])), x](*Solve can' t solve because is a Transcendental equation .*)

We can solve only numerically automatic and view the solution into on the Graph:

  a = 1; b = 2; c = 1; 
   ContourPlot[y == int, {y, -1, 1}, {x, -Pi, Pi}, 
   Frame -> False, Axes -> True, AxesLabel -> Automatic]

More info for Transcendental equation see here.

y(x)=(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c , y(0)=0

So if we have y , how to find x ? Must compute it everytime ?

What about the maybe simplification ?

Also the derivative of antiderivative doesn't give the original function (1/(1 + a Cos[x]))^2 , WolframAlpha and Mathematica gives
(a cos(x))/((a^2-1) (a cos(x)+1))+(a^2 sin^2(x))/((a^2-1) (a cos(x)+1)^2)-((a-1) sec^2(x/2) coth(x))/(a^2-1)^2+(2 (a-1) tan(x/2) csch^2(x))/(a^2-1)^2 .