Evaluate the integral $\int_{-1}^1\exp\left(ax+b\sqrt{1-x^2}\right)\,dx$ I wonder if there is a closed form expression for the following integral with respect to $x$:
\begin{equation}
\int_{-1}^1\exp\left(ax+b\sqrt{1-x^2}\right)\,dx
\end{equation}
where $a,b$ are some constant. Thanks so much! Any help would be greatly appreciated!
 A: $$\int_{-1}^1 \exp \left(a x+b \sqrt{1-x^2}\right) \, dx=\int_{-1}^1 \exp (a x) \exp \left(b \sqrt{1-x^2}\right) \, dx=\int_{-1}^1
   \exp (a x) \sum _{n=0}^{\infty } \frac{\left(b \sqrt{1-x^2}\right)^n}{n!} \, dx=\sum _{n=0}^{\infty } \int_{-1}^1 \frac{\exp (a
   x) \left(b \sqrt{1-x^2}\right)^n}{n!} \, dx=\sum _{n=0}^{\infty } \frac{2^{\frac{1}{2}-\frac{n}{2}} a^{-\frac{1}{2}-\frac{n}{2}}
   b^n \pi  I_{\frac{1+n}{2}}(a)}{\Gamma \left(\frac{1}{2}+\frac{n}{2}\right)}=\sqrt{\frac{2}{a}} \pi  \sum _{n=0}^{\infty }
   \frac{(2 a)^{-\frac{n}{2}} b^n I_{\frac{1+n}{2}}(a)}{\Gamma \left(\frac{1}{2}+\frac{n}{2}\right)}$$
For integral to solve I'm use a CAS.
A: $\int_{-1}^1e^{ax+b\sqrt{1-x^2}}~dx$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}e^{a\sin x+b\sqrt{1-\sin^2x}}~d(\sin x)$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}e^{a\sin x+b\cos x}\cos x~dx$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(a\sin x+b\cos x)^n\cos x}{n!}~dx$
$=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^na^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{n!}~dx$
$=\int_{-\frac{\pi}{2}}^0\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\int_\frac{\pi}{2}^0\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}(-x)\cos^{k+1}(-x)}{k!(n-k)!}~d(-x)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{((-1)^{n-k}+1)a^{n-k}b^k\sin^{n-k}x\cos^{k+1}x}{k!(n-k)!}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{((-1)^{n-k}+1)a^{n-k}b^kB\left(\dfrac{n-k+1}{2},\dfrac{k}{2}+1\right)}{2k!(n-k)!}$ (according to http://mathworld.wolfram.com/BetaFunction.html)
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{((-1)^{n-k}+1)a^{n-k}b^k\Gamma\left(\dfrac{n-k+1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{2k!(n-k)!\Gamma\left(\dfrac{n+3}{2}\right)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{((-1)^{n-k}+1)a^{n-k}b^k\Gamma\left(\dfrac{n-k+1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{2k!(n-k)!\Gamma\left(\dfrac{n+3}{2}\right)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{((-1)^n+1)a^nb^k\Gamma\left(\dfrac{n+1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{2k!n!\Gamma\left(\dfrac{n+k+3}{2}\right)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^k\Gamma\left(n+\dfrac{1}{2}\right)\Gamma\left(\dfrac{k}{2}+1\right)}{(2n)!k!\Gamma\left(n+\dfrac{k+3}{2}\right)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k}\Gamma\left(n+\dfrac{1}{2}\right)\Gamma(k+1)}{(2n)!(2k)!\Gamma\left(n+k+\dfrac{3}{2}\right)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k+1}\Gamma\left(n+\dfrac{1}{2}\right)\Gamma\left(k+\dfrac{3}{2}\right)}{(2n)!(2k+1)!\Gamma(n+k+2)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k}\pi}{4^{n+k}n!\Gamma\left(k+\dfrac{1}{2}\right)\Gamma\left(n+k+\dfrac{3}{2}\right)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k+1}\pi}{2^{2n+2k+1}n!k!(n+k+1)!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n}b^{2k}\pi}{4^{n+k}n!\Gamma\left(k+\dfrac{1}{2}\right)\Gamma\left(n+k+\dfrac{3}{2}\right)}+\dfrac{b\pi}{2}{_0F_1}\left(-;2;\dfrac{a^2+b^2}{4}\right)$ (according to Some properties about the Kampé de Fériet function)
