In triangle $ABC,$ $\angle A = 90^\circ,$ $AC = 1,$ and $AB = 5.$  Point $D$ lies on ray $\overrightarrow{AC}$ such that $\angle DBC = 2 \angle CBA.$  Compute $AD.$
Let $\theta = \angle CBA.$  Since $\angle DBC = 2 \theta,$ $\angle DBA = 3 \theta.$

[asy]
unitsize(1 cm);

pair A, B, C, D;

A = (0,0);
B = (5,0);
C = (0,1);
D = (0,37/11);

draw(A--B--D---cycle);
draw(B--C);

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, W);
label("$D$", D, NW);
label("$1$", (A + C)/2, W);
label("$5$", (A + B)/2, S);
[/asy]

Note that $\tan \theta = \frac{1}{5}.$  By the triple angle formula,
\[\tan 3 \theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} = \frac{3 (\frac{1}{5}) - (\frac{1}{5})^3}{1 - 3 (\frac{1}{5})^2} = \frac{37}{55}.\]Therefore,
\[AD = AB \tan 3 \theta = 5 \cdot \frac{37}{55} = \boxed{\frac{37}{11}}.\]