Integral involving erf and exponential Problem
I would like to compute the integral:
\begin{align}
\int_{0}^{+\infty} \text{erf}(ax+b) \exp(-(cx+d)^2) dx \tag{1}
\end{align}
I have been looking at this popular table of integral of the error functions, and also found here the following expression:
\begin{align}
\int_{-\infty}^{+\infty} e^{-(\alpha x+ \beta)^2} \text{erf}(\gamma x+\delta) dx = \dfrac{\sqrt{\pi}}{\alpha} \text{erf} \left[ \dfrac{\alpha \delta - \beta \gamma}{\sqrt{\alpha^2+ \gamma^2}} \right] \tag{2}
\end{align} 
as well as:
\begin{align}
\int_{0}^{+\infty} e^{-\alpha^2 x^2} \text{erf}(\beta x) dx = \dfrac{\text{arctan}(\beta / \alpha)}{\alpha \sqrt{\pi}} \tag{3}
\end{align}
However the latter (3) is only a particular case of (1), which is what I am looking for. Do you know how to prove (2)? This might help me understand how to compute (1)? Or do you know how to compute (1)?

Partial solution
Define:
$$I\left( \delta  \right)=\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}}}}\text{erf}(\gamma x+\delta )dx$$
now differentiate with respect to $\delta$
$$\begin{align}
  & \frac{dI}{d\delta }=\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}}}}\frac{\partial }{\partial \delta }\left( \text{erf}(\gamma x+\delta ) \right)dx \\ 
 & \quad \quad =\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}}}\left( \frac{2{{e}^{-{{\left( \gamma x+\delta  \right)}^{2}}}}}{\sqrt{\pi }} \right)}dx\  \\ 
 & \quad \quad =\frac{2}{\sqrt{\pi }}\int_{-\infty }^{+\infty }{{{e}^{-{{(\alpha x+\beta )}^{2}}-{{\left( \gamma x+\delta  \right)}^{2}}}}}dx=\frac{2}{\sqrt{\pi }}\left( \frac{\sqrt{\pi }{{e}^{-\frac{{{\left( \alpha \delta -\beta \gamma  \right)}^{2}}}{{{\alpha }^{2}}+{{\gamma }^{2}}}}}}{\sqrt{{{\alpha }^{2}}+{{\gamma }^{2}}}} \right) \\\end{align}$$
Finally, you can find that:
$$I\left( \delta  \right)=\frac{\sqrt{\pi }}{\alpha }\textrm{erf}\left( \frac{\alpha \delta -\beta \gamma }{\sqrt{{{\alpha }^{2}}+{{\gamma }^{2}}}} \right)$$
