Expected value of normal CDF

Assume that the $\Phi$ function is for some other independent standard normal random variable Y, and simply rewrite the problem like so:

$$E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=P(Y<\frac{a-bX}{c})=P\left( Y+\frac{bX}{c}<\frac{a}{c}\right)$$

Now, since $X$ ~ $N(0,1), \frac{bX}{c}$ ~ $N(0,\frac{b^2}{c^2})$, so $Y+\frac{bX}{c}$ ~ $N(0,1+\frac{b^2}{c^2})$

So,

$$ P\left( Y+\frac{bX}{c}<\frac{a}{c}\right) = P\left( \frac{Y+\frac{bX}{c}}{\sqrt{1+\frac{b^2}{c^2}}}<\frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right) $$

Therefor: $E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)$

Let $c$ in your question be absorbed into $a$ and $b$, so that you are concerned with $\Phi(a-bX)$. Let $X$ and $Y$ be independent standard normals, and consider their joint pdf $\phi(x,y) = \phi(x)\,\phi(y)$ as a surface covering the $X,\!Y$ plane, where $\phi(\cdot)$ is the standard normal pdf.

For any given $x$, $\Phi(a-bx)\,\phi(x)\,\text{d}x$ is the volume under the joint pdf in the thin slice at $x$ with $Y < a-bx$. Then the conditional expected value of $\Phi(a-bX)$, given that $X$ is in some interval, is just the volume under the pdf in the $X$-interval and below the line $Y = a - bX$, divided by all the volume under the pdf in the $X$-interval.

I know of no closed-form expression for the numerator in the general case where the X-interval is of the form $(t,\infty)$.