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

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.
