Integral of the product of Normal density and cdf

The horror, the horror... :-)

Recall that $\Phi(x)=P[X\leqslant x]$ for every $x$, where the random variable $X$ is standard normal, and that, for every suitable function $u$, $$ \int_{-\infty}^{+\infty}u(x)\phi(x)\mathrm dx=E[u(Y)], $$ where the random variable $Y$ is standard normal. Using this for $u=\Phi$, one sees that the integral $I$ you are interested in is $$ I=P[X\leqslant B^{-1}(Y-A))]=P[BX\geqslant Y-A]=P[Z\leqslant A], $$ where $Z=Y-BX$, where $X$ and $Y$ are i.i.d. standard normal, and where we used the fact that $B\lt0$ to reverse the inequality sign. Now, the random variable $Z$ is centered gaussian with variance $\sigma^2=1+B^2$, hence $Z=\sigma U$ with $U$ standard normal, and $$ I=P[U\leqslant A/\sigma]=\Phi(A/\sqrt{1+B^2}). $$