In trapezoid $ ABCD$, $ AB \parallel CD$, $ \angle CAB < 90^\circ$, $ AB \equal{} 5$, $ CD \equal{} 3$, $ AC \equal{} 15$. What are the sum of different integer values of possible $ BD$?

$\textbf{(A)}\ 101  \qquad\textbf{(B)}\ 108  \qquad\textbf{(C)}\ 115  \qquad\textbf{(D)}\ 125  \qquad\textbf{(E)}\ \text{None}$
1. Let \( T \) be the intersection of the diagonals \( AC \) and \( BD \). Since \( AB \parallel CD \), triangles \( \triangle TAB \) and \( \triangle TCD \) are similar by AA similarity criterion (corresponding angles are equal).

2. From the similarity \( \triangle TAB \sim \triangle TCD \), we have the ratio of corresponding sides:
   \[
   \frac{TA}{TC} = \frac{AB}{CD} = \frac{5}{3}
   \]
   Let \( TA = k \) and \( TC = m \). Then:
   \[
   \frac{k}{m} = \frac{5}{3} \implies k = \frac{5}{8} \cdot 15 = \frac{75}{8}
   \]

3. Denote \( x = BD \). Then, by the similarity ratio:
   \[
   TB = \frac{5}{8}x
   \]

4. Using the triangle inequality in \( \triangle TAB \):
   \[
   \frac{75}{8} - 5 < \frac{5}{8}x < 5 + \frac{75}{8}
   \]
   Simplifying the inequalities:
   \[
   \frac{75}{8} - 5 = \frac{75 - 40}{8} = \frac{35}{8} \implies \frac{35}{8} < \frac{5}{8}x \implies 35 < 5x \implies 7 < x
   \]
   \[
   5 + \frac{75}{8} = \frac{40 + 75}{8} = \frac{115}{8} \implies \frac{5}{8}x < \frac{115}{8} \implies 5x < 115 \implies x < 23
   \]
   Therefore:
   \[
   7 < x < 23
   \]

5. Since \( \angle CAB \) must be acute, we use the Pythagorean theorem in \( \triangle TAB \):
   \[
   \left(\frac{5}{8}x\right)^2 < 5^2 + \left(\frac{75}{8}\right)^2
   \]
   Simplifying:
   \[
   \left(\frac{5}{8}x\right)^2 < 25 + \left(\frac{75}{8}\right)^2
   \]
   \[
   \left(\frac{5}{8}x\right)^2 < 25 + \frac{5625}{64}
   \]
   \[
   \left(\frac{5}{8}x\right)^2 < \frac{1600 + 5625}{64} = \frac{7225}{64}
   \]
   \[
   \left(\frac{5}{8}x\right)^2 < \frac{7225}{64} \implies \left(\frac{5x}{8}\right)^2 < \frac{7225}{64} \implies \left(\frac{5x}{8}\right)^2 < \frac{7225}{64}
   \]
   \[
   \left(\frac{5x}{8}\right)^2 < \frac{7225}{64} \implies \left(\frac{5x}{8}\right)^2 < \frac{7225}{64} \implies x^2 < 289 \implies x < 17
   \]
   Therefore:
   \[
   7 < x < 17
   \]

6. The integer values of \( x \) are \( 8, 9, 10, 11, 12, 13, 14, 15, 16 \). The sum of these integers is:
   \[
   8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 108
   \]

The final answer is \(\boxed{108}\)