Most of the ways to do what you are requesting is probably due to adding rules to the built-in functions (e.g. Integrate , CDF , PDF , etc.), which may not be a very good option. Here's a slightly softer way using stunt-based Block macros:
ClearAll[withIntegrationRule]; SetAttributes[withIntegrationRule, HoldAll]; withIntegrationRule[code_] := Block[{CDF, PDF, Integrate, NormalDistribution}, Integrate[ CDF[NormalDistribution[0, 1], \[Alpha]_ + \[Beta]_ x_] PDF[ NormalDistribution[\[Mu]_, \[Sigma]_], x_], {x_, -\[Infinity], \[Infinity]}] := CDF[NormalDistribution[0, 1], (\[Alpha] + \[Beta] \[Mu])/ Sqrt[1 + \[Beta]^2 \[Sigma]^2]]; code];
Here is how we can use it:
In[27]:= withIntegrationRule[a=Integrate[CDF[NormalDistribution[0,1],\[Alpha]+\[Beta] x] PDF[NormalDistribution[\[Mu],\[Sigma]],x],{x,-\[Infinity],\[Infinity]}]]; a Out[28]= 1/2 Erfc[-((\[Alpha]+\[Beta] \[Mu])/(Sqrt[2] Sqrt[1+\[Beta]^2 \[Sigma]^2]))]
When our rule does not match, it will still work, automatically switching to the regular evaluation route:
In[36]:= Block[{$Assumptions = \[Alpha]>0&&\[Beta]==0&&\[Mu]>0&&\[Sigma]>0}, withIntegrationRule[b=Integrate[CDF[NormalDistribution[0,1],\[Alpha]+\[Beta] x] PDF[NormalDistribution[\[Mu],\[Sigma]],x],{x,0,\[Infinity]}]]] Out[36]= 1/4 (1+Erf[\[Alpha]/Sqrt[2]]) (1+Erf[\[Mu]/(Sqrt[2] \[Sigma])])
where I set \[Alpha] to 0 under assumptions to make integration possible in closed form.
Another alternative would be to implement your own special purpose integrator.