3. Given that $A D$ is the angle bisector of $\triangle A B C$, $\angle C$ $=90^{\circ}, \frac{A B-A C}{B D-C D}=\sqrt{5}$. Then
$$
\frac{A B+A C}{A D+B C}=
$$
$-$ $\qquad$
3. $25 \sqrt{5}-10 \sqrt{30}$.

As shown in Figure 7, draw $D E \perp A B$ at point $E$. It is easy to see that
$$
\begin{array}{l}
A C=A E, \\
C D=D E .
\end{array}
$$

Let $B E=m, B D=n$, and $D E=k$.
In the right triangle $\triangle B D E$, by the Pythagorean theorem, we have
$$
m^{2}+k^{2}=n^{2} \text {. }
$$

Also, $A B-A C=A B-A E=B E=m$, and by the problem statement,
$$
\frac{m}{n-k}=\sqrt{5} \text {. }
$$

Solving (1) and (2) together, we get $n=1.5 k, m=\frac{\sqrt{5}}{2} k$.
In the right triangle $\triangle A B C$, by the Pythagorean theorem, we have
$$
A C=A E=\sqrt{5} k .
$$

Thus, $A D=\sqrt{A C^{2}+C D^{2}}=\sqrt{6} k$.
$$
\begin{array}{l}
\text { Therefore, } \frac{A B+A C}{A D+B C}=\frac{\left(\sqrt{5} k+\frac{\sqrt{5}}{2} k\right)+\sqrt{5} k}{\sqrt{6} k+2.5 k} \\
=25 \sqrt{5}-10 \sqrt{30} .
\end{array}
$$