Is it possible to calculate this integral

$$ \mbox{Is it possible to calculate this integral}\quad \int{1 \over \cos^{3}\left(x\right) + \sin^{3}\left(x\right)}\,{\rm d}x\quad {\large ?} $$

I have tried $\dfrac{1}{\cos^3(x)+\sin^3(x)}$=$\dfrac{1}{(\cos(x)+\sin(x))(1-\cos x\sin x)}$ then I made a decomposition. But I'm still stuck. Thank you in advance.

The substitution $u = \tan(\frac{x}{2})$ converts any integrand that is a rational function in the two variables $\cos x$ and $\sin x$ into a rational function in $u,$ which can then be integrated by standard methods. See p. 56 of Hardy's The Integration of Functions of a Single Variable.

Where you have left of $$I=\int\frac1{(\cos x+\sin x)(1-\sin x\cos x)}=\int\frac{\cos x+\sin x}{(1+2\sin x\cos x)(1-\sin x\cos x)}$$

Let $\displaystyle\int(\cos x+\sin x)\ dx=\sin x-\cos x=u\implies u^2=1-2\sin x\cos x$

$$\implies I=\int\frac{2du}{(2-u^2)(1+u^2)}$$

Again, $\displaystyle\frac3{(2-u^2)(1+u^2)}=\frac{(2-u^2)+(1+u^2)}{(1+u^2)(2-u^2)}=\frac1{(1+u^2)}+\frac1{(2-u^2)}$

Finally use this for the second integral and the first one is too simple to be described, right?