This is not a complete answer, but I hope it will be helpful.
I assume $\alpha, a, c$ are real and nonzero.
Let
$$J(\alpha,\beta,a,b,c,d) = \int_{-\infty}^\infty e^{-(\alpha t + \beta)^2} \text{erf}(at+b)\; \text{erf}(ct+d)\; dt$$
Now
$$ \dfrac{\partial J}{\partial b} = \dfrac{2}{\sqrt{\pi}}
\int_{-\infty}^\infty e^{-(\alpha t + \beta)^2 - (a t + b)^2} \; \text{erf}(ct+d)\; dt $$
Write $$\eqalign{(\alpha t + \beta)^2 + (a t + b)^2 &= A^2 s^2 + r^2\cr
ct + d &= cs + \delta\cr
A &= \sqrt{\alpha^2 + a^2}\cr s &= t + \dfrac{\alpha \beta + a b}{\alpha^2 + a^2}\cr
r &= \dfrac{\alpha b - a \beta}{\sqrt{a^2+\alpha^2}}\cr
\delta &= d - c \dfrac{\alpha \beta + ab}{\alpha^2 + a^2}}$$
Then
$$\dfrac{\partial J}{\partial b} = \dfrac{2}{\sqrt{\pi}} e^{-r^2}
\int_{-\infty}^\infty e^{-A^2 s^2} \text{erf}(cs + \delta)\; ds
= \dfrac{2}{A} e^{-r^2} \; \text{erf}\left( \dfrac{A\delta}{c^2+A^2}\right)$$
and $J$ is an antiderivative of this with respect to $b$ (note that
$r$ and $\delta$ are affine functions of $b$). Unfortunately the antiderivative
has no closed form AFAIK.