Difficult Gaussian Integral Involving Two Trig Functions in the Exponent: Any Help? Here's the integral:

$$\int_d^e \exp\left(-a\left((b+c)\cos(x)-\sqrt{b^2 - (b+c)^2 \sin^2(x)}\right)^2 \right) \, dx$$

I've tried using Mathematica: it fails.
Can anyone help evaluate it?
Perhaps some kind of series expansion of the exponential might help...
Have any of you dealt with these kinds of Gaussians before?
Offering Respect,
Alex
 A: Assume $b\neq0$ , $c\neq0$ and $b+c\neq0$ to maintain the key meaning of the question.
$\int_d^e\exp\left(-a\left((b+c)\cos x-\sqrt{b^2-(b+c)^2\sin^2x}\right)^2 \right)~dx$
$=\int_d^e\sum\limits_{n=0}^\infty\dfrac{(-1)^na^n\left((b+c)\cos x-\sqrt{b^2-(b+c)^2\sin^2x}\right)^{2n}}{n!}dx$
$=\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{(-1)^nC_{2m}^{2n}a^n(b+c)^{2n-2m}\cos^{2n-2m}x\left(\sqrt{b^2-(b+c)^2\sin^2x}\right)^{2m}}{n!}dx-\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^nC_{2m-1}^{2n}a^n(b+c)^{2n-2m+1}\cos^{2n-2m+1}x\left(\sqrt{b^2-(b+c)^2\sin^2x}\right)^{2m-1}}{n!}dx$
$=\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{(-1)^n(2n)!a^n(b+c)^{2n-2m}\cos^{2n-2m}x\left(b^2-(b+c)^2+(b+c)^2\cos^2x\right)^m}{n!(2m)!(2n-2m)!}dx-\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^n(2n)!a^n(b+c)^{2n-2m+1}\cos^{2n-2m+1}x\left(b^2-(b+c)^2\sin^2x\right)^{m-\frac{1}{2}}}{n!(2m-1)!(2n-2m+1)!}dx$
Note that both $\int\cos^{2n-2m}x\left(b^2-(b+c)^2+(b+c)^2\cos^2x\right)^m~dx$ and $\int\cos^{2n-2m+1}x\left(b^2-(b+c)^2\sin^2x\right)^{m-\frac{1}{2}}~dx$ , where $m$ and $n$ are any non-negative integers have close-form.
