Decent integral

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {dx}{\left(a^2\sin^2 x + b^2 \cos^2 x \right)^2} & \small \color{#3D99F6} \text{Multiply up and down by }\sec^4 x \\ & = \int_0^\frac \pi 2 \frac {\sec^4 x}{\left(a^2\tan^2 x + b^2 \right)^2}\ dx & \small \color{#3D99F6} \text{Let }t = \tan x \implies dt = \sec^2 x \ dx \\ & = \int_0^\infty \frac {t^2+1}{\left(a^2t^2 + b^2 \right)^2}\ dt \\ & = \frac 1{a^2} \int_0^\infty \frac {a^2t^2+b^2 + a^2 - b^2}{\left(a^2t^2 + b^2 \right)^2}\ dt \\ & = \frac 1{a^2} \int_0^\infty \frac 1{a^2t^2 + b^2} \ dt + \frac {a^2-b^2}{a^2} \color{#3D99F6} \int_0^\infty \frac 1{\left(a^2t^2 + b^2 \right)^2}\ dt & \small \color{#3D99F6} \text{By reduction formula (see note).} \end{aligned}$

$\begin{aligned} \ \ & = \frac 1{a^2} \int_0^\infty \frac 1{a^2t^2 + b^2} \ dt + \frac {a^2-b^2}{a^2} \color{#3D99F6}\left(\frac 1{2b^2} \int_0^\infty \frac 1{a^2t^2 + b^2} \ dt +\cancel{\frac t{2b^2(a^2t^2 + b^2)}\bigg|_0^\infty}^0 \right) \end{aligned}$

$\begin{aligned} \ \ & = \frac {a^2+b^2}{2a^2b^2} \int_0^\infty \frac 1{a^2t^2 + b^2} \ dt \\ & = \frac {a^2+b^2}{2a^2b^2} \int_0^\infty \frac 1{b^2\left({\color{#3D99F6}\frac {a^2}{b^2}t^2} + 1\right)} \ dt & \small \color{#3D99F6} \text{Let }u = \frac ab t \implies du = \frac ab \ dt \\ & = \frac {a^2+b^2}{2a^3b^3} \int_0^\infty \frac 1{\left({\color{#3D99F6} u^2} + 1\right)} \ dt \\ & = \frac {a^2+b^2}{2a^3b^3} \tan^{-1} u \bigg|_0^\infty \\ & = \frac {\pi(a^2+b^2)}{4a^3b^3} \end{aligned}$

$\implies l + n + m = 4+2+3 = \boxed{9}$

---

Reduction formula:

$\small \displaystyle \int \frac 1{(ax^2+b)^n} \ dx = \frac {2n-3}{2b(n-1)} \int \frac 1{(ax^2+b)^{n-1}} \ dx + \frac x{2b(n-1)(ax^2+b)^{n-1}}$

Begin with the standard integral, $\int _{0}^{\pi /2}\frac {dx}{a^2\sin ^2x+b^2\cos ^2x}=\frac {\pi }{2ab}$ .So clearly partial differentiation wrt,a and b yields $\frac {\pi}{2a^2b}=\int _{0}^{\pi /2}\frac {2a\sin ^2x}{(a^2\sin ^2x+b^2\cos ^2x)^2}$ .Similarly for b.Now its easy to divide our eq by $2a$ to get $\int _{0}^{\pi /2}\frac {dx}{(a^2\sin ^2x+b^2\cos ^2x)^2}=\frac {\pi (a^2+b^2)}{4(a^3b^3)}$

We could rather generalise this,let $I _{n}=\int _{0}^{\pi /2}\frac {dx}{(a^2\cos ^2x+b^2\sin ^2x)^n}$ ,using the same principle.