8. The function
$$
f(x)=(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right)
$$

has a range of $\qquad$ .
8. $[2+\sqrt{2}, 8]$.

From the given information, the domain of $f(x)$ is $[-1,1]$.
Let $t=\sqrt{1+x}+\sqrt{1-x}$. Then $t^{2}=2+2 \sqrt{1-x^{2}} \in[2,4]$.
Also, $t \geqslant 0$, so $t \in[\sqrt{2}, 2]$, and
$$
f(x)=g(t)=(t+2) \frac{t^{2}}{2}=\frac{1}{2} t^{3}+t^{2} \text {. }
$$

Clearly, $g(t)$ is monotonically increasing on the interval $[\sqrt{2}, 2]$, therefore, the range of $g(t)$, which is also the range of $f(x)$, is $[2+\sqrt{2}, 8]$.