In my research, I am attempting to solve the following integral: $$\int\limits_0^{2\pi}\exp(A\cos x+B\sin x +ilx)dx.$$ As of now, my method to solve it is as follows:
Use trigonometric identity $A\cos x+B\sin x=\sqrt{A^2+B^2}\cos[x-\arctan(B/A)]$ to express the $\sin$ and $\cos$ as a single $\cos$,
Make a change of variable $y=x-\arctan(B/A)$,
Shift my limits of integration as needed (integrating over period of periodic function),
Use definition of Bessel function: $$J_{n}(z)=\frac{e^{in\pi/2}}{2\pi} \int\limits_{0}^{2\pi} e^{\large i(n\tau - z \cos{\tau})}d\tau.$$
These steps lead me to the solution:
$$2\pi J_l\left(-\sqrt{A^2+B^2}\right)e^{\large il\left[\arctan\left(\frac BA\right)-\frac\pi2\right]}.$$
The issue is that I now need to integrate this, and A and B contain variables too complicated to integrate simply, considering the term $e^{il\arctan(B/A)}$. Is there any way to simplify this last expression, or a simpler solution to my integral? Thanks!
Edit: Mathematica doesn't seem to know how to solve my integral.
Attempt to simplify: we know that $\cos(\arctan z)=\dfrac{1}{\sqrt{1+z^2}}$, $\sin(\arctan z)=\dfrac{z}{\sqrt{1+z^2}}$, however, I am not aware of a formula for $\cos(l\arctan z)$, etc.
Edit 2:It doesn't seem like there is a nice way to simplify $e^{il\arctan(B/A)}$. Are there any other special functions that can be used to solve this integral?