1. The solution set of the inequality
$$
(x+1)^{3}\left(x^{3}+5 x\right)<10(x+1)^{2}+8
$$

is $\qquad$ .
-1. $(-2,1)$.
Let $f(x)=x^{3}+5 x$. It is easy to know that $f(x)$ is an increasing function on $\mathbf{R}$.
When $x>-1$, the original inequality is
$$
\begin{array}{l}
\left(\frac{2}{x+1}\right)^{3}+\frac{10}{x+1}>x^{3}+5 x \\
\Rightarrow f\left(\frac{2}{x+1}\right)>f(x) \Rightarrow \frac{2}{x+1}>x \\
\Rightarrow-1<x<1 ;
\end{array}
$$

When $x=-1$, the original inequality obviously holds;
When $x<-1$, the original inequality is
$$
\begin{array}{l}
\left(\frac{2}{x+1}\right)^{3}+\frac{10}{x+1}<x^{3}+5 x \\
\Rightarrow f\left(\frac{2}{x+1}\right)<f(x) \Rightarrow \frac{2}{x+1}<x \\
\Rightarrow-2<x<-1 .
\end{array}
$$

In summary, the solution set of the original inequality is $(-2,1)$.