Evaluating a double integral involving exponential of trigonometric functions I am having trouble evaluating the following double integral:
$$\int\limits_0^\pi\int\limits_0^{2\pi}\exp\left[a\sin\theta\cos\psi+b\sin\theta\sin\psi+c\cos\theta\right]\sin\theta d\theta\, d\psi$$
I will be satisfied with an answer involving special functions (e.g. modified Bessel function).
I tried expanding the product terms using the product-to-sum identities, as well applied other trigonometric identities, to no avail.  Can anyone help?
I suspect that the solution will involve a modified Bessel function $I_n(x)$ with argument in the form $\sqrt{a^2+b^2+c^2}$.  The reason I think so is because the above integral is related to a likelihood function used in evaluation of a non-coherent detection of signals that are corrupted by additive white Gaussian noise (for example, see Proakis "Digital Communications" 4th Edition pp. 304-306). However, these signals are in the two-dimensional domain (they are complex numbers) and the corresponding integral is as follows: $\int_{0}^{2\pi}\exp[a\sin\theta+b\cos\theta]d\theta=2\pi I_0(\sqrt{a^2+b^2})$.  The double integral that I am trying to evaluate is the extension of this problem to a three-dimensional space, with the corresponding change from polar to spherical coordinate system.
 A: $$\begin{align}a \sin{\theta} \cos{\psi} + b \sin{\theta} \sin{\psi}  &=  (a \cos{\psi} + b \sin{\psi}) \sin{\theta}\\ &= \sqrt{a^2+b^2} \cos{(\psi- \psi_0)} \sin{\theta} \end{align}$$
where
$$\tan{\psi_0} = \frac{b}{a}$$
Now, when we integrate the periodic function over $\psi \in [0,2 \pi)$, the result does not depend on $\psi_0$ and we get
$$\int_0^{2 \pi} d\psi \: e^{\sqrt{a^2+b^2} \sin{\theta} \cos{(\psi- \psi_0)}} = \int_0^{2 \pi} d\psi \: e^{\sqrt{a^2+b^2} \sin{\theta} \cos{\psi}}$$
This turns out to be
$$2 \pi I_0(\sqrt{a^2+b^2} \sin{\theta})$$
Now the integral is
$$\int_0^{\pi} d\theta \sin{\theta} I_0(\sqrt{a^2+b^2} \sin{\theta}) e^{c \cos{\theta}}$$
A: First, I would like to thank Ron Gordon for the partial answer above, and, very importantly, for pointing out the error in the original formulation question in the comments.  Here I show a closed-form expression for the last integral in his answer:
$$\int_0^{\pi} d\theta \sin{\theta} I_0(\sqrt{a^2+b^2} \sin{\theta}) e^{c \cos{\theta}}$$
Let $t=-\cos\theta$.  Then $dt=d\theta\sin\theta$, $\theta=\arccos(-t)\Rightarrow\sin\theta=\sqrt{1-t^2}$, and the limits of integration $\{-1,1\}$.  Substitution all of this yields the following integral:
$$\int_{-1}^{1}I_0(\sqrt{a^2+b^2}\sqrt{1-t^2})e^{-ct}dt$$
Gradshteyn&Ryzhik, 7th edition has the following integral as formula 6.616.5: $\int_{-1}^{1}e^{-ax}I_0(b\sqrt{1-x^2})dx=2(a^2+b^2)^{-1/2}\sinh\sqrt{a^2+b^2}$.
Thus, the integral in my question evaluates to:
$$\frac{2\sinh\sqrt{a^2+b^2+c^2}}{\sqrt{a^2+b^2+c^2}}$$
