4. Given that $a$, $b$, $c$, and $d$ are all prime numbers, and satisfy 10 $<c<d<20$, also $c$ and $a$ differ by a larger prime number, $d^{2}-c^{2}=a^{3} b(a+b)$. Then the value of $a b(c+d)$ is $\qquad$.
4. 180.

Since $a, b, c, d$ are all prime numbers, and $10<c<d<20$, $c, d$ can only be 11, 13, 17, or 19, and $c \neq 19$. Also, $c-a$ is a larger prime number (not even), so $a=2$. By taking $c=11, 13, 17$, $c-a$ are 9, 11, 15, respectively. Only when $c=13, c-a=11$ does it meet the condition. Substituting $c=13, a=2$ into $d^{2}-c^{2}=a^{3} b(a+b)$, we get $d^{2}-13^{2}=8 b(2+b)$.
(1) If $d=17$, substituting gives $17^{2}-13^{2}=8 b(2+b)$, which simplifies to $b^{2}+2 b-15=0$. Solving this, we get $b=3$ or $b=-5$ (discard the latter);
(2) If $d=19$, substituting gives $19^{2}-13^{2}=8 b(2+b)$, which simplifies to $b^{2}+2 b-24=0$. Solving this, we get $b=4$ or $b=-6$, neither of which meet the requirements.
In summary, $a=2, b=3, c=13, d=17$.
Therefore, $a b(c+d)=6 \times(13+17)=180$.