I am trying to find the following integral \begin{align} \int_{0}^{\infty} \frac{c y^2}{1+c y^2}\frac{1}{\sqrt{2 \pi}} e^{-\frac{(y+c)^2}{2}} dy \end{align} where $c>0$. I was able to find the following integral
\begin{align} \int_{0}^{\infty} \frac{c y^2}{1+c y^2}\frac{1}{\sqrt{2 \pi}} e^{-\frac{y^2}{2}} dy=\frac{1}{2} - \frac{\sqrt{\pi}\, \mathrm{e}^{\frac{1}{2\, c}}}{\sqrt{8}\, \sqrt{c}} + \frac{\sqrt{\pi}\, \mathrm{e}^{\frac{1}{2\, c}}\, \mathop{\mathrm{erf}}\nolimits\!\left(\frac{\sqrt{2}}{2\, \sqrt{c}}\right)}{\sqrt{8}\, \sqrt{c}} \end{align}
but all attempts to change the exponent to what we want failed.
Consider the integral \begin{align} I = \frac{1}{\sqrt{2 \pi}} \, \int_{0}^{\infty} \frac{c y^2}{1+c y^2} \, e^{-\frac{(y+c)^2}{2}} dy \end{align} for which \begin{align} I &= \frac{1}{\sqrt{2 \pi}} \, e^{- \frac{c^{2}}{2}} \, \sum_{r=0}^{\infty} \frac{(-c)^{r}}{r!} \, J_{r} \end{align} where \begin{align} J_{r} &= \int_{0}^{\infty} \frac{c \, y^{r+2}}{1 + c \, y^{2}} \, e^{- \frac{y^{2}}{2}} \, dy \\ &= 2^{\frac{r-1}{2}} \, \Gamma\left(\frac{r+1}{2}\right) - \int_{0}^{\infty} \frac{y^{r}}{1 + c \, y^{2}} \, e^{- \frac{y^{2}}{2}} \, dy \\ &= 2^{\frac{r-1}{2}} \, \Gamma\left(\frac{r+1}{2}\right) - 2^{\frac{r-1}{2}} \, \int_{0}^{\infty} \frac{x^{\frac{r-1}{2}} \, e^{-x}}{1 + 2 c \, x} \, dx \end{align} where $y = \sqrt{2x}$ was used. Now evaluate the last integral and sum the resulting values.