On a better solution for $\large {\int_{-x}^x\frac{dx}{\sqrt {1-x^2} e^{bx^2+ax}}}$ In

this user’s question

I was able to find out the following. It is for a bigger physics research problem having to do with comparing a Wigner function corresponding to a state of light for the user named @Jayanth Jayakumar. Although, I know almost nothing on this subject, I still find the result very interesting. In case you do not believe the results, here is graphical proof for each step. I would like to support both of us, but I am more interested in finding out a way to significantly simplify the result below:
$$\mathrm{\int_{-x}^x\frac{dx}{\sqrt {1-x^2} e^{bx^2+ax}}=\quad \frac12 \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}sgn\left(x^{m+n+1}\right)\left[(-1)^m+(-1)^n\right]B_{x^2}\left(\frac{m+n+1}{2},\frac12\right)}{m!(n-m)!}=\quad \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}k^{m+n+1}\left[(-1)^m+(-1)^n\right]\,_2F_1\left(\frac12,\frac{m+n+1}{2}, \frac{m+n+3}{2},k^2\right)}{m!(n-m)!(m+n+1)}=\quad \sum_{n=0}^\infty\sum_{m=0}^n\frac{b^{2m}a^{2(n-m)}sgn\left(x^{2(m+n)+1}\right)B_{x^2}\left(\frac{2(m+n)+1}{2},\frac12\right)}{(2m)!(2(n-m))!-}\quad -\sum_{n=0}^\infty\sum_{m=0}^n\frac{b^{2m+1}a^{2(n-m)}sgn\left(x^{2(m+n)+3}\right)B_{x^2}\left(\frac{2(m+n)+3}{2},\frac12\right)}{(2m+1)!(2(n-m))!}, }$$
For the full area under the function, set $x=\pm1$:
$$\mathrm{\int_{-1}^1 \frac{e^{-bx^2-ax}}{\sqrt{1-x^2}}dx=\quad\frac{\sqrt\pi}{2}\sum_{n=0}^\infty\sum_{m=0}^n\frac{b^ma^{n-m}sgn\left(x^{m+n+1}\right)[(-1)^m+(-1)^n]}{m!(n-m)!} \left(\frac{m+n-1}2\right)^{\left(-\frac12\right)},B_1\left(\frac{m+n+1}2,\frac12\right)= B\left(\frac{m+n+1}2,\frac12\right)= \frac{\sqrt\pi\left(\frac{m+n-1}2\right)!}{\left(\frac{m+n}2\right)!}= \sqrt\pi\left(\frac{m+n -1}2\right)^{\left(-\frac12\right)}}$$
This problem uses a double series, the link has a way to turn it into a single sum, Incomplete Beta function properties, hypergeomtric function values, falling factorial $x^{(y)}$ and the Wigner function. An exact solution is needed, so how can one simplify or evaluate this series? I said before, that the special case was a goal, but now the main goal is simplifying the general case or integrating another way. Please correct me and give me feedback!
Here is a single sum solution inspired by @Harry Peter and @James Arathoon. It uses the Lower Incomplete Fox-Wright function and Central Binomial Coefficient:
$$\displaystyle\begin{align}\displaystyle\int_{-t}^t\frac{dx}{\sqrt {1-t^2} e^{bt^2+at}}=\sum _{n=0}^{\infty } \frac{(2 n)! }{2^{2 n} (n!)^2}\sum _{m=0}^{\infty } \frac{a^{2 m} }{(2 m)!} b^{-\big(m+n+\frac{1}{2}\big)} \gamma\left(m+n+\frac{1}{2},bt^2\right)=\\ b^{-\frac12}\sum _{n=0}^{\infty } \binom {2n}n (4b)^{-n}\,_2Ψ^{(\gamma)}_1\left[\,^{\left(1,n+\frac12,bt^2\right),(1,1)}_{\quad \quad (1,2)}\ \frac{a^2}b\right]\end{align}$$
which is super close to this would be closed form of equation $2.27$ in this Korea Science article. If the coefficients of the factorials of $n$ were the same in the denominator and numerator, also for $m$, then we would technically have a closed form.
 A: Since $e^{-a x}=\cosh (a x)-\sinh (a x)$ and $\sinh (a x)$ is an odd function it follows that
$$\int_{-t}^t \frac{e^{-b x^2} \sinh (a x)}{\sqrt{1-x^2}} \, dx=0 $$
and
$$I(t)=\int_{-t}^t \frac{e^{-a x-b x^2}}{\sqrt{1-x^2}} \, dx=\int_{-t}^t \frac{e^{-b x^2} \cosh (a x)}{\sqrt{1-x^2}} \, dx$$
Given this
$$I(t)=\sum _{n=0}^{\infty } \frac{(2 n)! }{2^{2 n} (n!)^2}\sum _{m=0}^{\infty } \frac{a^{2 m} }{(2 m)!}\int_{-t}^t e^{-b x^2} x^{2 (m+n)} \, dx$$
and with Mathematica giving
$$\int_{-t}^t e^{-b x^2} x^{2 (m+n)} \, dx=  b^{-m-n-\frac{1}{2}} \left(\Gamma \left(m+n+\frac{1}{2}\right)-\Gamma \left(m+n+\frac{1}{2},b t^2\right)\right)$$
At the suggestion of @TymaGaidash, this can be simplified using the lower incomplete gamma function, $\gamma(s,x) = \int_0^x t^{s-1} e^{-t} dt$, as
$$\gamma\left(m+n+\frac{1}{2},bt^2\right)=\left(\Gamma \left(m+n+\frac{1}{2}\right)-\Gamma \left(m+n+\frac{1}{2},b t^2\right)\right)$$
Thus
$$\int_{-t}^t e^{-b x^2} x^{2 (m+n)} \, dx=  b^{-m-n-\frac{1}{2}} \gamma\left(m+n+\frac{1}{2},bt^2\right)$$
and finally
$$I(t)=\sum _{n=0}^{\infty } \frac{(2 n)! }{2^{2 n} (n!)^2}\sum _{m=0}^{\infty } \frac{a^{2 m} }{(2 m)!} b^{-(m+n+\frac{1}{2})} \gamma\left(m+n+\frac{1}{2},bt^2\right)$$
Update: Also maybe worth looking at
$$I(t)=\sum _{n=0}^{\infty } \frac{(2 n)!}{2^{2 n} (n!)^2} \int_{-t}^t e^{-b x^2} x^{2 n} \cosh (a x) \, dx$$
for approximations as Mathematica can calculate the integral when $n$ is fixed, for example the largest $n=0$ term for which
$$I(t)\approx \frac{\sqrt{\pi } e^{\frac{a^2}{4 b}} \left(\text{erf}\left(\frac{a+2 b t}{2 \sqrt{b}}\right)-\text{erf}\left(\frac{a-2 b t}{2 \sqrt{b}}\right)\right)}{2 \sqrt{b}}$$
where $\text{erf}(x)$ is the error function
