Rewrite the integrand as $(\tan x{)}^{-11/2}{\sec }^{8}x$. Then ${\sec }^{8}x=(1+{\tan }^{2}x{)}^3{\sec }^{2}x$, so the integral becomes

$\int (\tan x{)}^{-11/2}(1+{\tan }^{2}x{)}^3{\sec }^{2}xdx$.

Substitute $t=\tan x,dt={\sec }^{2}xdx$, expand $(1+t^2{)}^3=1+3t^2+3t^4+t^6$:

$\int (t^{-11/2}+3t^{-7/2}+3t^{-3/2}+t^{1/2})dt=-\frac{2}{9}{t}^{-9/2}-\frac{6}{5}{t}^{-5/2}-6t^{-1/2}+\frac{2}{3}{t}^{3/2}+C$.

Replace $t^{-k/2}=(\cot x{)}^{k/2}$ and $t^{3/2}=(\cot x{)}^{-3/2}$:

$=-\frac{2}{9}(\cot x{)}^{9/2}-\frac{6}{5}(\cot x{)}^{5/2}-\frac{6}{1}(\cot x{)}^{1/2}+\frac{2}{3}(\cot x{)}^{-3/2}+C$.

So $(p_1,q_1)=(2,9),(p_2,q_2)=(6,5),(p_3,q_3)=(6,1),(p_4,q_4)=(2,3)$.

$\frac{15p_1{p}_2{p}_3{p}_4}{q_1{q}_2{q}_3{q}_4}=\frac{15\cdot 2\cdot 6\cdot 6\cdot 2}{9\cdot 5\cdot 1\cdot 3}=\frac{2160}{135}=16$.