Let the function $f(x)=\dfrac{x}{3}+\dfrac{3}{x}+3,\ x\ne0$ be strictly increasing in $(-\infty,\alpha_1)\cup(\alpha_2,\infty)$ and strictly decreasing in $(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)$. Then $\displaystyle \sum_{i=1}^{5}\alpha_i^{2}$ is equal to