Solving the integral $\int_{-k}^{k} \frac{e^{-ax-bx^2}}{\sqrt{1-x^2}}dx$ I need to compute the definite integral.

$$\int_{-k}^{k} \frac{e^{-ax-bx^2}}{\sqrt{1-x^2}}dx$$

Although the integral looks neat and fairly simple in form, I'm not able to evaluate it further. I tried evaluating it using integration by parts but without success. I also couldn't find the solution in the book, "Table of Integrals, Series, and Products" by Gradshteyn and Ryzhik.
Can anyone help me in solving this?
Thank you.
 A: You never specified a closed form in your question, so here is an exact form answer using the following: integration source, hypergeometric argument simplification, main simplification, signum or sign function, proof of series expansion, an expansion of exp(y), binomial theorem,incomplete beta function and hypergeometric function.
Using the Maclaurin series expansion for exp(y) and binomial theorem:
$$\mathrm{\int_{-k}^k \frac{e^{-bx^2-ax}}{\sqrt{1-x^2}}dx= \int_{-k}^k \frac{1}{\sqrt{1-x^2}}\sum_{n=0}^\infty\frac{(-bx^2-ax)^n}{n!}dx= \int_{-k}^k \frac{1}{\sqrt{1-x^2}}\sum_{n=0}^\infty\frac{(-1)^n}{n!} \sum_{m=0}^n \binom nm b^ma^{n-m}x^{m+n}dx}$$
Integrating with the hypergeometric function seen above gives after a little simplification. The end is sped up with bound substitution already applied:
$$\mathrm{\int_{-k}^k \frac{e^{-bx^2-ax}}{\sqrt{1-x^2}}dx= \sum_{n=0}^\infty(-1)^n\sum_{m=0}^n \frac{b^ma^{n-m}}{m!(n-m)!} \int_{-k}^k \frac{x^{m+n}}{\sqrt{1-x^2}}dx=\quad\sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}k^{m+n+1}\left[(-1)^m+(-1)^n\right]\,_2F_1\left(\frac12,\frac{m+n+1}{2}, \frac{m+n+3}{2},k^2\right)}{m!(n-m)!(m+n+1)}= \quad \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}k^{m+n+1}\left[(-1)^m+(-1)^n\right]\left(\frac{m+n+1}{2|k|^{m+n+1}}B_{k^2}\left(\frac{m+n+1}{2},\frac12\right)\right)}{m!(n-m)!(m+n+1)}=  \quad\frac12 \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}sgn\left(k^{m+n+1}\right)\left[(-1)^m+(-1)^n\right]B_{k^2}\left(\frac{m+n+1}{2},\frac12\right)}{m!(n-m)!}}$$
Here is graphical proof of this result. Notice the inverse sine look to the graph.
Feel free to simplify this as the consecutive $(-1)^r$ terms look like they can be simplified. Go ahead and try this result if you like. The “proof of expansion link” takes a bit of time to load, so please be patient. Please correct me and give me feedback!
