jmc

We expand the expression on the left and attempt to match up the coefficients with those in the expression on the right. $$\begin{array}{rl}(x^2+ax+b)(x^2+cx+d)=x^4+c{x}^3+& d{x}^2\\ a{x}^3+& ac{x}^2+adx\\ +& b{x}^2+bcx+bd\end{array}$$ $$=x^4+x^3-2x^2+17x-5$$ So we have $a+c=1$, $ac+b+d=-2$, $ad+bc=17$, $bd=-5$.

From the final equation, we know that either $b=1,d=-5$ or $b=-1,d=5$. We test each case:

If $b=1,d=-5$, then $ac+b+d=ac-4=-2$, so $ac=2$. We substitute $a=1-c$ from the first equation to get the quadratic $c^2-c+2=0$. This equation does not have any integer solutions, as we can test by finding that the discriminant is less than zero, $(-1{)}^2-4(1)(2)=-7$.

If $b=-1,d=5$, then $ac+b+d=ac+4=-2$, so $ac=-6$. We substitute $a=1-c$ from the first equation to get the quadratic $c^2-c-6=0$, which has solutions $c=-2$ (so $a=3$) or $c=3$ (so $a=-2$). In either case, we get that $a+b+c+d=\boxed{5}$.

The remaining equation, $ad+bc=17$, tells us that the coefficients are $a=3,b=-1,c=-2,d=5.$