Integral problem

\begin{align} & \int e^{x\sin x+\cos x}(\frac{x^4\cos^3x-x\sin^2x+\cos x}{x^2\cos^2x})dx\\ & \hspace{5mm} =\int e^{x\sin x+\cos x}(x^2\cos x-\frac{x\sin^2x-\cos x}{x^2\cos^2x})dx\\ & \hspace{5mm} =\int e^{x\sin x+\cos x}(x^2\cos x-\frac{x\tan^2x-\sec x}{x^2})dx \end{align}

Realize that $\frac{x\tan^2x-\sec x}{x^2}=\frac{d}{dx}\frac{\sec x}{x}$ and that you can make $\frac{\sec x}{x}$ appear elsewhere by factoring $x^2\cos x-1$ into $(x-\frac{\sec x}{x})(x\cos x)$. So the above is equal to:

\begin{align} & \int e^{x\sin x+\cos x} \left((x-\frac{\sec x}{x})(x\cos x)+1-\frac{x\tan^2x-\sec x}{x^2}\right) \, dx\\ &=\int \left[e^{x\sin x+\cos x}\left(x-\frac{\sec x}{x}\right)(x\cos x)+e^{x\sin x+\cos x}\left(1-\frac{x\tan^2x-\sec x}{x^2}\right)\right]dx \end{align}

Now realize that $e^{x\sin x+\cos x}(x\cos x)=\frac{d}{dx}e^{x\sin x+\cos x}$. The above is equal to: $$\int \left[\left(x-\frac{\sec x}{x}\right)\frac{d}{dx}(e^{x\sin x+\cos x})+e^{x\sin x+\cos x}\frac{d}{dx}\left(x-\frac{\sec x}{x}\right)\right]\,dx $$

Now, this looks exactly looks like the product rule with $u=x-\frac{\sec x}{x}$ and $v=e^{x\sin x+\cos x}$. So the integral is equal to $$(x-\frac{\sec x}{x})e^{x\sin x+\cos x}+C$$

(To be honest, I did use WolframAlpha to evaluate the integral and work backward to take the derivative by hand, and then reverse each step, but I don't see any other way of evaluating such a difficult integral by hand...)