In the diagram line segments $AB$ and $CD$ are of length 1 while angles $ABC$ and $CBD$ are $90^\circ$ and $30^\circ$ respectively.  Find $AC$.

[asy]
import geometry;
import graph;

unitsize(1.5 cm);

pair A, B, C, D;

B = (0,0);
D = (3,0);
A = 2*dir(120);
C = extension(B,dir(30),A,D);

draw(A--B--D--cycle);
draw(B--C);
draw(arc(B,0.5,0,30));

label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$30^\circ$", (0.8,0.2));
label("$90^\circ$", (0.1,0.5));

perpendicular(B,NE,C-B);
[/asy]
1. **Define Variables and Angles:**
   Let \( x = AD \), \(\phi = \angle ADB\), and \( a = BC \).

2. **Using Triangle ABD:**
   In \(\triangle ABD\), we have:
   \[
   \sin \phi = \frac{1}{x}
   \]
   Since \(\sin \angle BDC = \sin \phi\), from \(\triangle BDC\), we have:
   \[
   \frac{a}{\sin \phi} = \frac{1}{\sin 30^\circ}
   \]
   Given \(\sin 30^\circ = \frac{1}{2}\), we get:
   \[
   \frac{a}{\sin \phi} = 2 \implies a = 2 \sin \phi = \frac{2}{x}
   \]

3. **Using Cosine Law in Triangle ABC:**
   In \(\triangle ABC\), we apply the cosine law:
   \[
   AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos \angle ABC
   \]
   Given \(AB = 1\), \(BC = \frac{2}{x}\), and \(\angle ABC = 90^\circ\), we have:
   \[
   \cos 90^\circ = 0 \implies AC^2 = 1^2 + \left(\frac{2}{x}\right)^2
   \]
   Simplifying, we get:
   \[
   AC^2 = 1 + \frac{4}{x^2}
   \]

4. **Using Triangle ABD Again:**
   From \(\triangle ABD\), we have:
   \[
   (1 + x)^2 = 1 + \frac{4}{x^2} + \frac{2}{x}
   \]
   Expanding and simplifying:
   \[
   1 + 2x + x^2 = 1 + \frac{4}{x^2} + \frac{2}{x}
   \]
   Multiplying through by \(x^2\) to clear the fraction:
   \[
   x^4 + 2x^3 + x^2 = x^2 + 2x + 4
   \]
   Simplifying further:
   \[
   x^4 + 2x^3 - 2x - 4 = 0
   \]

5. **Factoring the Polynomial:**
   Factoring the polynomial, we get:
   \[
   x(x^3 - 2) + 2(x^3 - 2) = 0 \implies (x^3 - 2)(x + 2) = 0
   \]
   Solving for \(x\):
   \[
   x^3 - 2 = 0 \implies x = \sqrt[3]{2}
   \]
   Since \(x + 2 = 0\) gives \(x = -2\), which is not possible for a length, we discard it.

Thus, \(AD = \sqrt[3]{2}\).

The final answer is \(\boxed{\sqrt[3]{2}}\).