According to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions the integral
$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt $$
can be expressed as
$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt=\dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_1\left(a,b,k,c;x,\dfrac{y}{k}\right).$$
I know this is true, but gives numeric error. Look that, calculated by Mathematica:
F[a_?NumericQ, b_?NumericQ, c_?NumericQ, x_?NumericQ, y_?NumericQ] :=
NIntegrate[t^{a - 1} (1 - t)^{c - a - 1} (1 - x t)^{-b} Exp[ y t], {t, 0, 1}]
N[F[3/2, 1, 2, .4, .3], 20]
{2.8964403550198865`}
G[a_?NumericQ, b_?NumericQ, c_?NumericQ, x_?NumericQ, y_?NumericQ] :=
Gamma[a] Gamma[c - a]/Gamma[c] Limit[AppellF1[a, b, k, c, x, y/k],k -> Infinity]
N[G[3/2, 1, 2, .4, .3], 20]
{2.2854650559595466`}
Note that $|x|<1,|y|<1$, and $\text{Re}(c)>\text{Re}(a)>0$.
Is this equation true or not?
Edit 1
According to Wolfram Alpha
$$\lim\limits_{k\rightarrow \infty} F_1[3/2,1,k;2;x,y/k] =\frac{2\left(\frac{1}{\sqrt{1-x}}-1\right)}{x}$$
such that $\lim\limits_{k\rightarrow \infty}F_1[3/2,1,k;2;0.4,\text{any/k}]=1.4549$.
So I think
$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt \neq \dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_1\left(a,b,k,c;x,\dfrac{y}{k}\right).$$
What do you say?
Edit 2
It has been solved here.