$$ \int_{-\infty}^{\infty}\Phi(a+bx)\phi(x)dx=\Phi\left(\frac{a}{\sqrt{1+b^2}}\right) $$
I have a problem with showing the above result, where $\phi(x)$ and $\Phi(x)$ respectively are the pdf and cdf of the standard normal distribution. I've tried to prove it by calculating the pdf $\phi(a+bx)$ first and combine it with the pdf $\phi(x)$ but the result is complicated and doesn't yield $\phi\left(\dfrac{a}{\sqrt{1+b^2}}\right) $ . I really need your help. Thanks.
Using $N(0,\,1)$ iids $U,\,V$ and defining $W:=U-bV\sim N(0,\,1+b^2)$, we can write the integral as $$\int_{\Bbb R}P(U\le a+bx)dP(V\le x)=P(U\le a+bV)=P(W\le a)=\Phi\bigg(\frac{a}{\sqrt{1+b^2}}\bigg).$$