3. Given real numbers $a, b$ satisfy
$$
a^{2}=-1-5 a, 5 b=-1-b^{2} \text {. }
$$

Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is
3. $-5 \pm \sqrt{21}$ or -23.

From the given,
$$
a^{2}+5 a+1=0, b^{2}+5 b+1=0 \text {. }
$$

When $a=b$, $a=\frac{-5 \pm \sqrt{21}}{2}$, then
$$
b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=a+b=2 a=-5 \pm \sqrt{21} \text {. }
$$

When $a \neq b$, $a, b$ are the two real roots of the quadratic equation $x^{2}+5 x+1=0$. At this time, $a+b=-5, a b=1$, which means $a, b$ are both negative. Therefore,
$$
\begin{array}{l}
b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=-\frac{b}{a} \sqrt{a b}-\frac{a}{b} \sqrt{a b} \\
=-\frac{(a+b)^{2}-2 a b}{a b} \sqrt{a b}=-23 .
\end{array}
$$