Let the function $f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0$ be strictly increasing in $\left(-\infty, \alpha_1\right) \cup\left(\alpha_2, \infty\right)$ and strictly decreasing in $\left(\alpha_3, \alpha_4\right) \cup\left(\alpha_4, \alpha_5\right)$. Then $\sum_{i=1}^5 \alpha_i^2$ is equal to
