As per the title, I am looking for a closed-form expression for the integral
$$\frac{1}{B(\alpha,\beta)}\int_{0}^{1}e^{-ax(1 - bx )}x^{\alpha-1}(1-x)^{\beta - 1}dx$$
where $a,\alpha,\beta>0$ and $b \in [0,\frac{1}{2}]$. The factor $\frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta - 1}$ is the beta probability density function where $B$ denotes the beta function. For $b=0$, the above is an integral representation of the confluent hypergeometric function $_{1}F_{1}(\alpha,\alpha+\beta,-a)$.
My first attempts at evaluating this by numerical integration in MATLAB indicate that it will be problematic for $\alpha,\beta < 1$, i.e. when the beta function goes to infinity at the endpoints of the interval. If there is a closed-form expression in terms of special functions, it may be easier to evaluate numerically.
The problematic component of the proposed integral is the term $e^{ab x^2}$. This can be eliminated by expanding this exponential into the corresponding power series for which the integral in question becomes \begin{align} I &= \sum_{n=0}^{\infty} \frac{(ab)^{n}}{n!} \, \int_{0}^{1} \, e^{-ax} \, x^{2n + \mu-1} \, (1-x)^{\nu - 1} \, dx \\ &= \sum_{n=0}^{\infty} \frac{(ab)^{n}}{n!} \, B(2n+\mu, \nu) \, {}_{1}F_{1}(2n+\mu; 2n+\mu+\nu; -a) \\ &= B(\mu, \nu) \, \sum_{n=0}^{\infty} \sum_{r=0}^{\infty} \frac{(\mu)_{2n} (2n+\mu)_{r} }{ (\mu + \nu)_{2n} (2n+\mu + \nu)_{r} } \, \frac{(ab)^{n}}{n!} \, \frac{(-a)^{r}}{r!} \\ &= B(\mu, \nu) \, \sum_{n=0}^{\infty} \sum_{r=0}^{\infty} \frac{(\mu)_{2n+r}} { (\mu + \nu)_{2n+r} } \, \frac{(ab)^{n}}{n!} \, \frac{(-a)^{r}}{r!} \\ &= B(\mu, \nu) \, F_{1:0}^{1:0} \left[ \begin{array}{cc} [(\mu):2,1]: - ; \\ [(\mu + \nu):2,1]: - ; \end{array} \hspace{3mm} ab , \, -a \right] \end{align} where the last function, $F$, is the Srivastava-Daoust function. From this it can be stated that \begin{align} \frac{1}{B(\mu, \nu)} \, \int_{0}^{1} \, e^{-ax(1-bx)} \, x^{\mu-1} \, (1-x)^{\nu - 1} \, dx = F_{1:0}^{1:0} \left[ \begin{array}{cc} [(\mu):2,1]: - ; \\ [(\mu + \nu):2,1]: - ; \end{array} \hspace{3mm} ab , \, -a \right] \end{align}
The Srivastava-Daoust Function is defined by formula (1.1) in the article by Rekha Panda