5. Solution: $\frac{6\left(x^{3}-8\right) \sqrt{x^{2}+6 x+9}}{\left(x^{2}+2 x+4\right)\left(x^{2}+x-6\right)} \geq x-2 . \Leftrightarrow \frac{6(x-2)\left(x^{2}+2 x+4\right) \sqrt{(x+3)^{2}}}{\left(x^{2}+2 x+4\right)(x-2)(x+3)} \geq x-2 \Leftrightarrow$ $x \neq 2, \quad \frac{6|x+3|}{(x+3)} \geq x-2 \Leftrightarrow\left[\begin{array}{c}\left\{\begin{array}{c}x \in(-3 ; 2) \cup(2 ;+\infty), \\ x \leq 8,\end{array}\right. \\ \left\{\begin{array}{c}x \in(-\infty ;-3), \\ x \leq-4 .\end{array}\right.\end{array} \Leftrightarrow x \in(-\infty ;-4] \cup(-3 ; 2) \cup(2 ; 8]\right.$.
Answer: $x \in (-\infty; -4] \cup (-3; 2) \cup (2; 8]$.