computing an integral involving standard normal pdf and cdf

recently, i need to compute this kind of integral: $$\int ^\infty _c \Phi(ax+b) \phi(x) dx$$ where a, b and c are all constants and $\Phi(x)$ denotes the CDF of standard normal distribution and $\phi(x)$ denotes the PDF of standard normal

I have looked up similar questions both in "math.stackexchange.com" and here, but i don't find any satisfactory answer. if "c" is negative infinity here, it would be relatively easy. but here it's not.

later i find if i can find a solution to: $$\int ^\infty _c x^2\Phi(ax+b) \phi(x) dx$$ then the former integral could be calculated.

can someone help me? if no closed solution exists, is there any practical approximation to that integral?

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If $a=1, b=0,$ then Maple produces $$1/2\,{\frac { \left( -1/4\,\sqrt {\pi }\sqrt {2} {{\rm erf}\left(1/2\,c\sqrt {2}\right)}-1/8\,\sqrt {\pi }\sqrt {2} \left( {{\rm erf}\left(1/2\,c\sqrt {2}\right)} \right) ^{2}+3/8\, \sqrt {\pi }\sqrt {2} \right) \sqrt {2}}{\sqrt {\pi }}}.$$ –  Mark May 17 '13 at 17:03
The simplified result is $$-1/8\, \left( {{\rm erf}\left(1/2\,c\sqrt {2}\right)} \right) ^{2}-1/4 \,{{\rm erf}\left(1/2\,c\sqrt {2}\right)}+3/8.$$ –  Mark May 17 '13 at 17:09
@Mark: That case is easy to compute by hand. It would be the probability that $X \gt c, X \gt Y$ where $X,Y \sim N(0,1)$. If I calculate correctly, that's $P(X \gt c) - P(Y \gt X \gt c) = P(X \gt c) - 1/2 P(X \gt c)^2$. The general case doesn't simplify like that. –  Douglas Zare May 17 '13 at 18:56
This questions asks for the probability that a standard normal distribution is in a wedge-shaped region, or equivalently for the CDF of a two-dimensional Gaussian distribution with a general covariance matrix. Wikipedia says that there isn't an analytic expression for this, but that approximations are known. One related paper I remember is by Marsaglia, jstatsoft.org/v16/i04/paper, on the distribution of the ratio between two normal distributions, but that would be a double wedge, between two lines instead of two rays. –  Douglas Zare May 17 '13 at 19:16
@Mark: I don't understand what you are confused about. I translated the integral into a statement about random variables. Inside the integral is a density for $X=x, Y \lt x$. Integrating from $x=c$ to $x = \infty$ means $X \gt c, Y \lt X$. –  Douglas Zare May 17 '13 at 20:44

$\phi\left(\frac{b}{\sqrt{1+a^2}}\right)-\Phi_2\left[c,\frac{b}{\sqrt{1+a^2}},\frac{-a}{\sqrt{1+a^2}}\right]$
where $\Phi_2\left(x,y,\rho\right)$ is the bivariate normal cdf with means zero, variances one, and correlation $\rho$.
I found it by differentiating with respect to $b$, then reintegrating. You can get the second integral the same way.