calculus - How to calculate $int frac{sin^{6}(x)}{sin^{6}(x) + cos^{6}(x)} dx?$

How to calculate
$$\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx?$$

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I already know one possible way, that is by :
$$\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx = \int 1 - \frac{\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx$$
$$= x- \int \frac{1}{1+\tan^{6}(x)} dx$$
Then letting $u=\tan(x)$, we must solve
$$\int \frac{1}{(1+u^{6})(1+u^{2})} du$$
We can reduce the denominator and solve it using Partial Fraction technique. This is quite tedious, I wonder if there is a better approach.

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Using same approach, for simpler problem, I get
$$\int \frac{\sin^{3}(x)}{\sin^{3}(x)+\cos^{3}(x)} dx = \frac{x}{2} - \frac{\ln(1+\tan(x))}{6} + \frac{\ln(\tan^{2}(x)- \tan(x)+1)}{3} - \frac{\ln(\sec(x))}{2} + C$$

Answer

Let us take:
$$I=\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx$$
then
$$I=\int \frac{-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
giving $$2I=\int \frac{\sin^{6}(x)-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
This can be written as (using identities like $a^3-b^3$ and $a^3+b^3$)
$$2I= \int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x)(1+\sqrt3\sin(x)\cos(x))}dx+x$$

$$2I=\frac{1}{2}\left(\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))} +\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x))}\right)+x$$
Evaluating the integrals separately by using $u=1+\sqrt3\sin(x)\cos(x)$
for first one gives

$$\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))}=\frac{1}{\sqrt3}\int\frac{(\sin(x)\cos(x)-1)(\sin(x)\cos(x)+1)}{u}du$$
Now use $\sin(x)\cos(x)=\frac{u-1}{\sqrt3}$
which will evaluate the integral as $\frac{u^2}{6\sqrt3}-\frac{2u}{3\sqrt3}-\frac{2\ln(u)}{3\sqrt3}$...Similar approach for other with $v=1-\sqrt3\sin(x)\cos(x)$

The final value is

$$I=\frac{x}{2}-\frac{\sin(x)\cos(x)}{6}+\frac{\ln(1-\sqrt3\sin(x)\cos(x))}{6\sqrt3}-\frac{\ln(1+\sqrt3\sin(x)\cos(x))}{6\sqrt3}+C$$