How to integrate $ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ $ I want to know how to integrate this function I have tried many things many substitutions but none works.
I even try to expand the numerator by $\sin3x$ and $\cos3x$ properties and tried to convert higher powers to multiple angles

$$ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ =\int \frac{3\sin x -4\sin^3x+4\cos^3x-3\cos x}{ \sin^3 x + \cos^3 x } dx=\int \frac{(\cos x -\sin x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)(1- \sin x\cos x) } dx=\int \frac{(\cos^2 x -\sin^2 x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)^2(1- \sin x\cos x) } dx=\int \frac{\cos 2x (1+ 2\sin 2x)}{ (1+ \sin 2x)(1- 0.5\sin 2x) } dx=0.5\int \frac{1+ 2\sin 2x}{ (1+ \sin 2x)(1- 0.5\sin 2x) } d(\sin 2x)$$ Hope you can finish from here.
