jmc

Multiplying top and bottom by the conjugate, we have $\frac{1+\sqrt{2}}{2+\sqrt{3}}=\frac{(1+\sqrt{2})(2-\sqrt{3})}{(2+\sqrt{3})((2-\sqrt{3}))}=\frac{2-\sqrt{3}+2\sqrt{2}-\sqrt{6}}{4-3}=2-\sqrt{3}+2\sqrt{2}-\sqrt{6}$. So, we obtain $A=2,B=2,C=3$ and $D=6$ ($C$ and $D$ are interchangeable). So $A+B+C+D=2+2+3+6=\boxed{13}$.