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.
 A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&{1 \over \bracks{\cos\pars{x} + \sin\pars{x}}\bracks{1 - \cos\pars{x}\sin\pars{x}}}
={1 \over \bracks{\cos\pars{x} + \tan\pars{\pi/4}\sin\pars{x}}
\bracks{1 - \sin\pars{2x}/2}}
\\[3mm]&={\root{2} \over \cos\pars{x - \pi/4}\bracks{2 - \sin\pars{2x}}}
={\root{2} \over \cos\pars{x - \pi/4}\braces{2 - \sin\pars{2\bracks{x - \pi/4} + \pi/2}}}
\\[3mm]&={\root{2} \over \cos\pars{x - \pi/4}\braces{2 - \cos\pars{2\bracks{x - \pi/4}}}}
\end{align}

With $t \equiv x - \pi/4$:
  \begin{align}
&{1 \over \bracks{\cos\pars{x} + \sin\pars{x}}\bracks{1 - \cos\pars{x}\sin\pars{x}}}
={\root{2} \over \cos\pars{t}\bracks{2 - \cos\pars{2t}}}
={\root{2} \over \cos\pars{t}\braces{2 - \bracks{2\cos^2\pars{t} - 1}}}
\\[3mm]&={\root{2} \over \cos\pars{t}\bracks{3 - 2\cos^2\pars{t}}}
={\root{2} \over 2}\,
{1 \over \cos\pars{t}\bracks{\root{3}/2 - \cos\pars{t}}\bracks{\root{3}/2 + \cos\pars{t}}}
\\[3mm]&={\root{2} \over 2}\bracks{%
{4/3\over \cos\pars{t}} + {3/2 \over \root{3}/2 - \cos\pars{t}} +
 {3/2 \over \root{3}/2 + \cos\pars{t}}}
\\[3mm]&={2\root{2} \over 3}\,{1 \over \cos\pars{t}}
+{3\root{2} \over 4}\bracks{%
{1 \over \root{3}/2 - \cos\pars{t}} + {1 \over \root{3}/2 + \cos\pars{t}}
}
\end{align}

$$
\int{\dd t \over \cos\pars{t}}=\ln\pars{\sec\pars{t} + \tan\pars{t}} +\quad \mbox{a constant}
$$

The remaining integrals can be easily performed with $s \equiv \tan\pars{t/2}$.

A: 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?
