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I'm trying to solve the following integral analytically :

Integrate[
 1/(a + b Cos[x - y])*1/(c + d Cos[x - z])*1/(
  e + f Cos[z - y]), {x, 0, 2 Pi} , {y, 0, 2 Pi}, {z, 0, 2 Pi}]

and I want to Solve for arbitrary Real values of a,b,c,d,e,f. the integrate works when I give some values to a,b,c,d,e,f but when I don't ,the process is interminable and doesn't seem to give an answer ! I should also note that the arbitrary values are chosen in a way that the integral is not singular , for example a>b. I appreciate your time .

EDIT

I just found out numerically that the integral can be substituted with this one :

2*Pi*Integrate[
 1/(a + b Cos[x])*1/(c + d Cos[y])*1/(
  e + f Cos[x-y]), {x, 0, 2 Pi} , {y, 0, 2 Pi}]

So now problem reduces to a two dimensional angle integration rather than three.

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    $\begingroup$ While it won't resolve your issue, you should start by replacing all instances of pi with Pi $\endgroup$
    – Bob Hanlon
    Nov 17, 2018 at 20:04
  • 1
    $\begingroup$ Any reason to expect this integral can be expressed in terms of elementary functions? Especially for arbitrary complex $a,b,c,d,e,f$? $\endgroup$ Nov 17, 2018 at 20:51
  • $\begingroup$ Thanks bob @BobHanlon $\endgroup$ Nov 18, 2018 at 4:55
  • $\begingroup$ Sorry I should have mentioned that a,b,c,d,e values are Real . I know that when I sum the angles rather than subtracting them , the result can be turned into three separate integrals over the angles , and then can be solved easily. but I'm not sure about this one .@AccidentalFourierTransform $\endgroup$ Nov 18, 2018 at 5:13
  • $\begingroup$ One may assume $a=1,c=1,e=1,|b|<1,|d|<1,|f|<1$ without loss of generality. $\endgroup$
    – user64494
    Nov 18, 2018 at 9:26

1 Answer 1

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No chance at present.The command

Integrate[ 1/(1 + b Cos[x - y])*1/(1 + d Cos[x - z])*1/(1 + f Cos[z - y]), {x,0, 2 Pi}, 

{y, 0, 2 Pi},{z, 0, 2*Pi},Assumptions -> b < 1 &&b >-1 && d < 1 &&d>-1 &&f < 1 && f > -1] 

fails. Even

Integrate[1/(1 + b Cos[x - y])*1/(1 + d Cos[x - z])*1/(1 - 1/4* Cos[z - y]), {x, 0, 2 Pi}, 
{y, 0, 2 Pi}, {z, 0, 2*Pi},Assumptions -> b < 1 && b > -1 && d < 1 && d > -1]

(1/(15 Sqrt[1-d^2]))16 [Pi] Integrate[(60 d^2+Sqrt[15] Sqrt[1-d^2]+d (15+4 Sqrt[15] Sqrt[1-d^2]) Cos[x-y])/((1+b Cos[x-y]) (2+31 d^2+d^2 Cos[2 x-2 y]+16 d Cos[x-y])),{x,0,2 [Pi]},{y,0,2 [Pi]},Assumptions->b<1&&b>-1&&d<1&&d>-1]

The following works.

Integrate[ 1/(1 + 1/2* Cos[x - y])*1/(1 + 1/3*Cos[x - z])*1/(1 - 
 1/4* Cos[z - y]), {x, 0, 2* Pi}, {y, 0, 2 *Pi}, {z, 0, 2*Pi}]

$$\frac{8}{5} \left(2 \sqrt{5}+\sqrt{6}+\sqrt{30}-6\right) \pi ^3 $$

The result is confirmed numerically.

Addition. The modified question has an affirmative answer:

Integrate[ 1/(1 + b Cos[x])*1/(1 + d Cos[y])*1/(1 + f Cos[x - y]), {x, 0, 2 Pi}, 
{y, 0, 2 Pi}, Assumptions -> b > -1 && b < 1 && d > -1 && d < 1 && f > -1 && f < 1]

ConditionalExpression[(4 (Sqrt[1-b^2] d^2 f^2 (-Sqrt[1-d^2]+Sqrt[1-d^2] f^2-Sqrt[1-f^2]+d^2 Sqrt[1-f^2])-b^2 (Sqrt[1-d^2] f^2+d^2 Sqrt[1-f^2]) Sqrt[-(-1+d^2) (-1+f^2) (d^2+f^2-2 (1+Sqrt[(-1+d^2) (-1+f^2)]))]+b d f (-f^2 Sqrt[(-1+b^2) (-1+f^2)]+Sqrt[1-b^2] d^2 (-Sqrt[1-d^2]+f^2 (Sqrt[1-d^2]+Sqrt[1-f^2]))-(Sqrt[1-d^2]+Sqrt[1-f^2]) Sqrt[-(-1+d^2) (-1+f^2) (d^2+f^2-2 (1+Sqrt[(-1+d^2) (-1+f^2)]))])) [Pi]^2)/(Sqrt[(-1+d^2) (-1+f^2)] (-2 b d f-d^2 f^2+b^2 (-f^2+d^2 (-1+f^2))) Sqrt[(-1+b^2) (-1+d^2) (-1+f^2) (d^2+f^2-2 (1+Sqrt[(-1+d^2) (-1+f^2)]))]),b!=0]

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  • $\begingroup$ Thank you so much for the answer , unfortunately the coefficients are functions of some other variable . maybe I should consider the possible ways of solving it by hand. $\endgroup$ Nov 18, 2018 at 13:01
  • $\begingroup$ Hi , Please take a look at my edit . tnx . @user64494 $\endgroup$ Nov 19, 2018 at 8:09

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