This one is not that bad

$\begin{aligned} I & = \int_\frac 1{\sqrt 2}^1 \frac 1{x^5\sqrt{4x^2-1}} \ dx \quad \quad \small \color{#3D99F6}{\text{Let } \sec u = 2x \quad \sec u \tan u \ du = 2 \ dx} \\ & = \int_\frac \pi 4^\frac \pi 3 \frac {16 \sec u \tan u}{\sec^5 u \color{#3D99F6}{\sqrt{\sec^2 u -1}}} \ du \quad \quad \small \color{#3D99F6}{\sqrt{\sec^2 u -1} = \sqrt{\tan^2 u} = \tan u} \\ & = 16 \int_\frac \pi 4^\frac \pi 3 \cos^4 u \ du \\ & = 16 \int_\frac \pi 4^\frac \pi 3 \left(\frac{\cos 2u +1}2 \right)^2 \ du \\ & = 4 \int_\frac \pi 4^\frac \pi 3 \left(\cos^2 2u + 2\cos 2u + 1 \right) \ du \\ & = 4 \int_\frac \pi 4^\frac \pi 3 \left(\frac 12 \left( \cos 4u + 1 \right) +2\cos 2u + 1 \right) \ du \\ & = 2 \int_\frac \pi 4^\frac \pi 3 \left(\cos 4u +4\cos 2u + 3 \right) \ du \\ & = 2 \left(\frac{\sin 4u}4 +2\sin 2u + 3u \right) \bigg|_\frac \pi 4^\frac \pi 3 \\ & = \frac 12 \left(-\frac {\sqrt{3}}2 - 0 \right) + 4 \left(\frac {\sqrt{3}}2 -1 \right) + 6 \left(\frac \pi 3 -\frac \pi 4 \right) \\ & = -4 + \frac {7\sqrt 3}4 + \frac \pi 2 \end{aligned}$

$\implies A+B+C+D = 4+7+3+2 = \boxed{16}$