I am trying to integrate the following function involving the Normal CDF ($\Phi$). I actually need the definite integral $$\int^b_a\Phi\left(\frac{p}{\sqrt{q+rx}}\right)dx$$ for $q+ra,q+rb >0$ but for simplicity wanted to figure out the indefinite integral first. I believe I have a solution mapped out below, but I am not very confident the answer is correct - I would appreciate if you could take a look at what I have and tell me if this is along the correct lines or if I have made a mistake. Thanks.
$I=\int\Phi\left(\frac{p}{\sqrt{q+rx}}\right)dx$
I apply the substitution:
$y=\frac{p}{\sqrt{q+rx}}\\ \frac{dy}{dx}=-\frac{pr}{2\left(q+rx\right)^{\frac{3}{2}}}\\ =-\left(\frac{r}{2p^2}\right)\left(\frac{p}{\sqrt{q+rx}}\right)^3\\ =-\frac{ry^3}{2p^2}\\ dx=-\frac{2p^2}{ry^3}dy\\ I=-\int\frac{2p^2\Phi\left(y\right)}{ry^3}dy$
I then attempt a integration by parts:
$u=\Phi\left(y\right)\Rightarrow\frac{du}{dy}=\phi\left(y\right)\\ \frac{dv}{dy}=\frac{-2p^2}{ry^3}\Rightarrow v=\frac{p^2}{ry^2}\\ I=\frac{p^2\Phi\left(y\right)}{ry^2}-\int\frac{p^2\phi\left(y\right)}{ry^2}dy$
I then attempt another integration by parts:
$u=\phi\left(y\right)\Rightarrow\frac{du}{dy}=-y\phi\left(y\right)\\ \frac{dv}{dy}=\frac{p^2}{ry^2}\Rightarrow v=-\frac{p^2}{ry}\\ I=\frac{p^2}{r}\left[\frac{\Phi\left(y\right)}{y^2}+\frac{\phi\left(y\right)}{y}+\int\phi\left(y\right)dy\right]\\ $
$I=\frac{p^2}{r}\left[\frac{\Phi\left(y\right)}{y^2}+\frac{\phi\left(y\right)}{y}+\Phi\left(y\right)\right]\\ =\frac{p^2}{r}\left[\left(\frac{q+rx+p^2}{p^2}\right)\Phi\left(\frac{p}{\sqrt{q+rx}}\right)+\left(\frac{\sqrt{q+rx}}{p}\right)\phi\left(\frac{p}{\sqrt{q+rx}}\right)\right]$
$$I=\frac{p^2+q+rx}{r}\Phi\left(\frac{p}{\sqrt{q+rx}}\right)+\frac{p\sqrt{q+rx}}{r}\phi\left(\frac{p}{\sqrt{q+rx}}\right)$$