jmc

Let $a$ be the first term, and let $r$ be the common ratio. Then $a{r}^4-a{r}^3=576,$ so $a{r}^3(r-1)=576.$ Also, $ar-a=9,$ so $a(r-1)=9.$ Then $9r^3=576,$ so $r=4.$ Then $3a=9,$ so $a=3.$ Therefore, $$a_1+a_2+a_3+a_4+a_5=3+3\cdot 4+3\cdot 4^2+3\cdot 4^3+3\cdot 4^4=\frac{3(4^5-1)}{4-1}=\boxed{1023}.$$