Why can't Mathematica compute this integral $ \int_{0}^{\pi/2} \frac{1-a \cos^2x}{1+b\sin^2x} \mathrm e^{-\frac{a}{4}\cos2x}\mathrm d x $?

Question

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I have faced some difficulties to do the following integral

$$ \int_{0}^{\pi/2} \frac{1-a \cos^2x}{1+b\sin^2x}\mathrm e^{-\frac{a}{4}\cos2x}\mathrm d x, $$

where $a$ and $b$ are constants under $x-$variable. I'm sure this can be solved analytically. However, Mathematica doesn't give explicit answer. My code:

 Integrate[((1 - a Cos[x]^2) Exp[-a/4 Cos[2 x]])/(1 + b Sin[x]^2), {x,0, Pi/2},
           Assumptions -> a > 0 && b > 0]

I have found a closed expression in terms of Hypergeometric series. It's Something like that:

Limit[AppellF1[3/2, 1, 1/k, 2, x, k y], k -> 0],

but this not work for me. If you kindly give me some hint or text such that on going through which I can do it by myself. Thanking you.

Edit: Solution in terms of $F_1$:

Here my solution in terms of series. The original integral is that

$$ \int_{0}^{2\pi}\mathrm d\phi\int_{0}^{\pi}\mathrm d\theta~\sin\theta\frac{(1-u^2\sin^2\theta\cos^2\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{u^2}{2}\sin^2\theta\cos^2\phi}, \tag{1} $$

where $x$, $y$ and $u$ are positive constants. I tried to do the $\phi$ integral first. Letting $I_\phi$ this $\phi$-integral, then:

$$ I_\phi=4\int_{0}^{\pi/2}\mathrm d\phi\frac{1-a\cos^2\phi}{A\cos^2\phi+B\sin^2\phi} \mathrm e^{-\frac{a}{2}\cos^2\phi}. $$

The change of variable $\cos^2\phi=t$ give us

$$ I_\phi= \frac{2}{B}\int_{0}^{1}\frac{\mathrm dt}{\sqrt{t}\sqrt{1-t}}\frac{1-at}{1-\nu t}\mathrm e^{-\frac{a}{2}t}, $$

where $B=x^2\sin^2\theta+x^2y^2\cos^2\theta$, $a=u^2\sin^2\theta$, and $\nu=\frac{x^2-y^2}{x^2+x^2y^2\cot^2\theta}$. According to this we have then

$$ I_\phi=\frac{2}{B}\left[B\left(\frac{1}{2}\right)F_1\left(\frac{1}{2},1,-;1;\nu,-\frac{a}{2}\right)-aB\left(\frac{1}{2}\right)F_1\left(\frac{3}{2},1,-;2;\nu,-\frac{a}{2}\right)\right] $$

That is my solution until now.

Edit 2

If you prefer, the more general case of the equation (1) is:

$$ I= \int_{0}^{2\pi}\mathrm d\phi\int_{0}^{\pi}\mathrm d\theta~\sin\theta\int_{0}^{\infty}\mathrm dr~r^2\frac{\cos(u r \sin\theta \cos\phi)3x^2y^2\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{r^2}{2}} \tag{2} $$

EDIT 3

Here is a solution of the equation $(2)$, wich was found by another person.

$$ I_G=\frac{12 \pi x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2 \exp\left(-\frac{u^2}{2}\frac{x^2k^2}{(1-x^2)(1-k^2)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}. $$

Notice that,numerically,

$$ I_G=\sqrt{\frac{2}{\pi}} I$$

as you can see in this code performed in Mathematica:

IG[x_, y_, u_] := 
 Sqrt[Pi/2] NIntegrate[(12  Pi x y)/(1 - x^2)^(3/2)
(v^2 Exp[-(u^2 x^2 v^2)/(2 (1 - x^2) (1 - v^2))])/(Sqrt[1 - v^2] Sqrt[1 - v^2 (1 - y^2)/(1 - x^2)]), {v, 0, Sqrt[1 - x^2]}]
IG[.3, .4, 1]
** 4.53251 **

I[x_, y_, u_] := 
 NIntegrate[(r^2 Sin[a] Cos[
      u r Sin[a] Cos[b]] 3 x^2 y^2 Cos[a]^2 Exp[-r^2/
       2])/((y^2 Cos[b]^2 + x^2 Sin[b]^2) Sin[a]^2 + 
     x^2 y^2 Cos[a]^2), {r, 0, Infinity}, {a, 0, Pi}, {b, 0, 2 Pi}]
I[.3, .4, 1]
** 4.53251 **

Therefore, it is possible to find an elementary closed form for $I_\phi$. What do you say?

Edit: that is just a detail

My question comes down to: what steps were taken to get from $I$ to $I_G$.

I let Mathematica run overnight, but it was unable to compute the integral $I$. However, it also didn’t “give up”. What do you say about that? Maybe another CAS will work?

Answer 1

@user3321: if you report the solution in your possession it is more likely that you can help in finding a trick to use in MMA.

For now, the only thing that comes to mind is to get an idea of the solutions when the positive parameters a and b change:

points = Table[{a, b, NIntegrate[((1 - a Cos[x]^2) Exp[-a/4 Cos[2 x]])/(1 + b Sin[x]^2), 
               {x, 0, Pi/2}]}, {a, 0, 20, .5}, {b, 0, 20, .5}];
ListPointPlot3D[points, PlotRange -> All]