You will find it different

Given that $\begin{aligned} \int \frac {x^2+20}{(x\sin x + 5\cos x)^2} dx & = \frac {-Ax}{\cos x(x\sin x + B\cos x)} + \tan x + C\end{aligned}$

$\begin{aligned} \implies \frac {x^2+20}{(x\sin x + 5\cos x)^2} & = \frac d{dx} \left(\frac {-Ax}{\cos x(x\sin x + B\cos x)} + \tan x + C \right) \\ & = \frac {-A\cos x(x\sin x + B\cos x)+Ax(-\sin x(x\sin x + B\cos x) + \cos x(\sin x + x\cos x - B\sin x))}{\cos^2 x(x\sin x + B\cos x)^2} + \sec^2 x \\ & = \frac {-Ax\sin x\cos x -AB\cos^2 x-Ax^2\sin^2 x - ABx\sin x \cos x + Ax \sin x\cos x + Ax^2\cos^2 x - ABx\sin x\cos x}{\cos^2 x(x\sin x + B\cos x)^2} + \frac 1 {\cos^2 x} \\ & = \frac {Ax^2(\cos^2 x - \sin^2 x) - 2ABx\sin x\cos x -AB\cos^2 x +(x\sin x + B\cos x)^2}{\cos^2 x(x\sin x + B\cos x)^2} \\ & = \frac {Ax^2(2\cos^2 x - 1) - 2ABx\sin x\cos x -AB\cos^2 x +x^2\sin^2 x + 2Bx\sin x \cos x + B^2\cos^2 x}{\cos^2 x(x\sin x + B\cos x)^2} \\ & = \frac {(2A\cos^2 x-A + \sin^2 x)x^2 + 2B(1-A)x\sin x\cos x +B(B-A)\cos^2 x}{\cos^2 x(x\sin x + B\cos x)^2} \\ & = \frac {(2A\cos^2 x-A + 1 - \cos^2 x)x^2 + 2B(1-A)x\sin x\cos x +B(B-A)\cos^2 x}{\cos^2 x(x\sin x + B\cos x)^2} \\ & = \frac {(2A-1)x^2 + (1-A)x^2 \sec^2 x + 2B(1-A)x\tan x +B(B-A)}{(x\sin x + B\cos x)^2} \end{aligned}$

Comparing the coefficients both sides, we have $A=1$ and $B=5$ and $A^B+B^A = 1^5+5^1 = \boxed{6}$

The above integral can be written as: $I= \large \int \frac{x^2+20}{(x \sin x+ 5 \cos x)^2 } dx$

$I= \large \int \dfrac{\color{#3D99F6} (x^8 )\color{#333333} (x^2+20)}{(x^5 \sin x+ 5x^4 \cos x)^{2}} dx$

Note that $\frac{d}{dx} (x^5 \sin x+ 5x^4 \cos x) = (20+x^2)x^{3} \cos x$

Now integration by part can easily make an above type format, (let me know if anyone needs IBP )