$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $ $$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $$

is there more efficient and elegant way? Other than the solution below.
a solution:
Let $ I = \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $,
$$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx  = \int \frac{\sin^{8}(x) + \cos^{8}(x) - \cos^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = x - \int \frac{\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$
so we have $2I$ is:
$$ 2I = x + \int \frac{\sin^{8}(x)-\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$
Now,
$$ \sin^{8}(x) - \cos^{8}(x) = \left[ \sin^{4}(x) - \cos^{4}(x) \right] \left[ \sin^{4}(x) + \cos^{4}(x) \right]  $$
$$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right]\left[ \sin^{2}(x) + \cos^{2}(x) \right] \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 \sin^{2}(x)\cos^{2}(x) \right]  $$
$$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] $$
For the denominator:
$$ \sin^{8}(x) + \cos^{8}(x) = \left[ \sin^{4}(x) + \cos^{4}(x) \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$
$$ =  \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$
$$ = \left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4} $$
Now
$$ 2I =x+ \int \frac{ \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] }{\left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4}} dx $$
Let $U = \sin(x) \cos(x)$, then $U'(x) = \cos^{2}(x) - \sin^{2}(x)$, so
$$ 2I = x- \int \frac{\left[1 - 2 U^{2} \right] }{\left[1 - 2 U^{2} \right]^{2} - 2 U^{4}} dU $$
$$ =  x- \int \frac{\left[ 2 U^{2}  - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU $$
And from here I can continue using Partial Fraction:
$$- \int \frac{\left[ 2 U^{2}  - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU = -\frac{1}{2}\int \frac{1}{ (1 - (2 + \sqrt{2})U^{2})} + \frac{1}{(1 - (2 - \sqrt{2})U^{2})} dU $$
and then for each of the 2 terms we use Partial Fraction again. First, let $\alpha = \sqrt{2+\sqrt{2}}U$, $\beta = \sqrt{2-\sqrt{2}}U$, so we have
$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}}\int \frac{1}{ 1 - \alpha^{2}} d \alpha - \frac{1}{2\sqrt{2-\sqrt{2}}}   \int \frac{1}{(1 - \beta^{2})} d \beta $$
then we have
$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}} \left( \int \frac{1/2}{1-\alpha} d \alpha + \int \frac{1/2}{1+\alpha} d\alpha \right) - \frac{1}{2\sqrt{2-\sqrt{2}}} \left(   \int \frac{1/2}{1-\beta } d \beta +  \int \frac{1/2}{1 + \beta } d \beta \right) $$
$$ = -\frac{1}{4\sqrt{2+\sqrt{2}}} \left( -\ln(1-\alpha) + \ln(1+\alpha) \right) - \frac{1}{4\sqrt{2-\sqrt{2}}} \left(   -\ln(1-\beta) + \ln(1+\beta)  \right)  $$

Continue with
$$ 2I = x + \int \frac{\sin^{8}x-\cos^{8}x}{ \sin^{8}x + \cos^{8}x }dx =x+\int \frac{\tan^{8}x-1}{ \tan^{8}x + 1}dx \tag1$$
where, with $t = \tan x$
\begin{align}
&\int \frac{\tan^{8}x-1}{ \tan^{8}x + 1}dx \\
=& \int \frac{(\tan^{4}x+1)(\tan^2x-1)\ \sec^2x}{ (\tan^{4}x -\sqrt2\tan^2x+1) (\tan^{4}x -\sqrt2\tan^2x+1)}dx \\
=&\ \frac12\int \frac{1-t^2}{1-\sqrt2t^2+t^4}
 + \frac{1-t^2}{1+\sqrt2t^2+t^4}\ dt\\
=& -\frac1{2\sqrt{2-\sqrt2}}\coth^{-1}
\frac{t+\frac1t}{\sqrt{2-\sqrt2}}
— \frac1{2\sqrt{2+\sqrt2}}\coth^{-1}
\frac{t+\frac1t}{\sqrt{2+\sqrt2}}
\end{align}
Substitute into (1) to obtain
$$
I = \frac x2 -\frac1{4\sqrt{2-\sqrt2}}\coth^{-1}
\frac{\cot x+{\tan x}}{\sqrt{2-\sqrt2}}-\frac1{4\sqrt{2+\sqrt2}}\coth^{-1}
\frac{\cot x+\tan x}{\sqrt{2+\sqrt2}}\\
$$
