# Expectation of a transformed normal CDF [duplicate]

If we define a random variable $$X \sim N(0,1)$$ with $$\Phi$$ being the cdf of a standard normal, what would $$E(\Phi(a+bX))$$ be?

I was only able to rewrite $$\Phi(a+bX)$$ as $$P(Z\leq aX+b|X)$$ with $$Z\sim N(0,1)$$.

I could also calculate the cdf of $$\Phi(a+bX)$$ as per this answer but it doesn't help with its expectation.

Any help would be greatly appreciated.

For $$b\ne 0$$,
\begin{align} \mathsf{E}\Phi(a+bX)&=\int_0^{\infty}\mathsf{P}(\Phi(a+bX)>t)\,dt \\ &=\int_0^1\mathsf{P}(X>(\Phi^{-1}(t)-a)/b)\,dt \\ &=1-\int_0^1\Phi((\Phi^{-1}(t)-a)/b)\,dt. \end{align}
When $$a=0$$ and $$b=1$$, the third line reduces to $$1-\int_0^1 t\,dt=\frac{1}{2}.$$