Integral of a shifted Gaussian distribution with an error function In the course of computing a convolution of two functions, I have simplified it to a single variable integral of the form
$$\int_0^\infty xe^{-ax^2+bx}\mathrm{erf}(cx+d) dx$$
where $\mathrm{erf}$ is the error function defined as $$\ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt.$$
I've looked through A Table of Integrals of the Error Functions, especially the formulas on pages 8 and 9. There are a lot of integrals that are similar, but I couldn't find a way to simplify this integral into something that was in that table. I also have tried differentiating under the integral sign using the constants $c$ and $d$ as parameters, but that only seemed to complicate the integration on the parameter after computing the integral with respect to $x$.
Is there any way to find a closed form of this integral?

Not a full answer, more a long comment, but it might get you started. Define $$f_n:=\int_0^\infty x^n e^{-ax^2+bx}erf(cx+d)dx$$so we want $f_1=-\frac{1}{b}\partial_b f_0$. Defining $u:=\text{erf}(cx+d),\,v:=e^{-ax^2+bx}$, integration by parts gives $$(b+\frac{2a}{b}\partial_b)f_0=-2af_1+bf_0=-\text{erf}(d)-\frac{2c}{\sqrt{\pi}}\int_0^\infty e^{-ax^2+bx-(cx+d)^2}dx.$$The right-hand side is easy enough to write in terms of the error function. 
