jmc

We can factor the numerator and denominator to get $$\frac{x^2+3x}{x^2+4x+3}=\frac{x(x+3)}{(x+3)(x+1)}.$$In this representation we can immediately read that there is a hole at $x=-3$, and a vertical asymptote at $x=-1$. There are no more holes or vertical asymptotes, so $a=1$ and $b=1$. If we cancel out the common factors we have $$\frac{x(x+3)}{(x+3)(x+1)}=\frac{x}{x+1}.$$We can write $\frac{x}{x+1}$ as $1-\frac{1}{x+1}$ which shows that as $x$ becomes very large, the graph tends towards $1$, giving us a horizontal asymptote.

Since the graph cannot have more than one horizontal asymptote, or a horizontal asymptote and a slant asymptote, we have that $c=1$ and $d=0$. Therefore, $a+2b+3c+4d=1+2+3+0=\boxed{6}.$