5. Given $a, b, c \in \mathbf{R}_{+}$, and $a+3b+c=9$. Then the minimum value of $a+b^{2}+c^{3}$ is $\qquad$

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5. $\frac{243-8 \sqrt{3}}{36}$.

Let $\alpha, \beta>0$. Then $b^{2}+\alpha^{2} \geqslant 2 \alpha b, c^{3}+\beta^{3}+\beta^{3} \geqslant 3 \beta^{2} c$.
Thus $a+b^{2}+c^{3} \geqslant a+2 \alpha b+3 \beta^{2} c-\alpha^{2}-2 \beta^{3}$.
By the given conditions, take $\alpha=\frac{3}{2}, \beta=\frac{\sqrt{3}}{3}$.
Then $a+b^{2}+c^{3} \geqslant 9-\left(\frac{3}{2}\right)^{2}-2\left(\frac{\sqrt{3}}{3}\right)^{3}$
$$
=\frac{243-8 \sqrt{3}}{36} \text {. }
$$

When $b=\frac{3}{2}, c=\frac{\sqrt{3}}{3}, a=\frac{9}{2}-\frac{\sqrt{3}}{3}$, the above expression achieves its minimum value.