Brutal gaussian integral of death $\int_{\mathbb{R}} x \Phi(x) \phi(Bx-b)$

We have $$ \int_{- \infty}^x ds \ \phi(s) = \int_{- \infty}^0 ds \ \phi(s + x) \ .$$ Using Fubini's theorem, we have $$ \int_{-\infty}^\infty dx \ x \Phi(x) \phi(Bx - b) = \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) \ .$$ Computing the inner integral, we have $$ \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \frac{(bB - s)e^{- \frac{(Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$ We have $$ \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \int_{0}^\infty ds \ \frac{(bB + s)e^{- \frac{(- Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$ From here, I suppose one has a representation in terms of the error function. I do not see any other way to simplify this expression.

If we set $B = 1$ and $b = 0$, then the value of this integral agrees with the value that Anne calculated in the comments.