Elliptic integration with exponential numerator. I was wondering if someone could help me with the evaluation of an integral: 
\begin{equation}
\int_{x_1}^{x_2} \frac{e^{ax}}{\sqrt{1-b\cos(x)}}dx
\end{equation}
I'm familiar with elliptic integrals of the first and second kinds, and I can obtain solutions for 
\begin{equation}
\int_{x_1}^{x_2} \frac{1}{\sqrt{1-b\cos(x)}}dx
\end{equation}
or 
\begin{equation}
\int_{x_1}^{x_2} \sqrt{1-b\cos(x)} dx
\end{equation}
but I am not sure how to deal with the exponential in the numerator. Any help will be much appreciated. 
 A: Not a closed form, but some attempt to get a series solution.
$$\int_0^p \frac{e^{ax}}{\sqrt{1-b\cos(x)}}dx=\sum_{k=0}^\infty \frac{(2k)!}{k!^2} \frac{b^k}{4^k} \int_0^p \cos^k x~e^{ax}~dx=\sum_{k=0}^\infty \frac{(2k)!}{k!^2} \frac{b^k}{4^k} I_k(a)$$
The integrals inside the series are still complicated, but using integration by parts, it's quite simple to find a recurrence relation for them:

$$I_k(a)=\frac{k(k-1)I_{k-2}(a) +(a \cos p+k \sin p) \cos^{k-1} p~e^{ap}-a}{k^2+a^2}$$

We know that:
$$I_0(a)=\frac{e^{ap}-1}{a}$$
And we can formally take $I_{-1}$ to be any finite number because of the $(k-1)$ factor.
Then we find all the $I_k$ for $k=1,2,3,\dots$ from these initial values and the recurrence  in a straightforward way.

In fact, from numerical experiments the series converges quite fast. Here's a code example in Mathematica, where only $20$ terms of the series give $8$ correct digits for the integral:
a=1;
p=1;
b=1/2;
Nm=10;
J0=0;
I0=(Exp[a p]-1)/a;
Ik=Table[I0,{k,1,Nm+2}];
Jk=Table[J0,{k,1,Nm+2}];
Do[Jk[[n+1]]=N[((2 n-1) (2 n-2) Jk[[n]]+(a Cos[p]+(2 n-1) Sin[p]) Cos[p]^(2 n-2) Exp[a p]-a)/((2 n-1)^2+a^2),20];Ik[[n+1]]=N[((2 n) (2 n-1) Ik[[n]]+(a Cos[p]+(2 n) Sin[p]) Cos[p]^(2 n-1) Exp[a p]-a)/((2 n)^2+a^2),20],{n,1,Nm+1}];
N[Sum[((4k)! ((b/4)^(2k)) )/((2k)!)^2 Ik[[k+1]]+((4k+2)! ((b/4)^(2k+1)) )/((2k+1)!)^2 Jk[[k+2]],{k,0,Nm}],20]
NIntegrate[Exp[a x]/Sqrt[1-b Cos[x]],{x,0,p},WorkingPrecision->20]

The result is:
$$\color{blue}{2.2319784}804859150574\dots$$
While the numerical value of the integral is:
$$\color{green}{2.2319784976660188548}\dots$$

For $a<0$ and $p = +\infty$ the recurrence should still be valid, and it will have a more simple form because the exponential term will go to $0$.
So for $a=-s$ and $p =\infty$, we have:

$$I_k(s)=\frac{k(k-1)I_{k-2}(s) +s}{k^2+s^2}$$

