Closed form for the sum $\sum_{a=1}^{b} a^3\cdot (b \bmod a)$

Question

How can we simplify $\sum_{a=1}^{b} a^3\cdot (b \mod a)$? For $a \ge \frac{b+1}{2} $ to $a = b$ it reduces to $$\sum_{a\ge \frac{b+1}{2}}^{b}a^3\cdot (b-a)=b\cdot\sum_{a\ge \frac{b+1}{2}}^{b}a^3-\sum_{a\ge \frac{b+1}{2}}^{b}a^4$$ but getting no idea in $a= 1$ to $a \le \frac{b}{2}$ ?