Is there an elementary method to evaluate the indefinite integral $\int \frac{1}{1+\sin ^{6} x} d x?$ Inspired by the post, I want to increase the power of $\sin x$ by $2$ to $6$,
$$
I=\int\frac{1}{1+\sin ^{6} x} d x.
$$
As usual, we multiply both the numerator and denominator by $\sec^6 x$ and get
$$
\begin{aligned}
I &=\int \frac{\sec ^{6} x}{\sec ^{6} x+\tan ^{6} x} d x \\
& \stackrel{t=\tan x}{=} \int\frac{\left(1+t^{2}\right)^{2} d t}{\left(1+t^{2}\right)^{3}+t^{6}}
\end{aligned}
$$
Factorizing the denominator yields $$
I=\int \frac{1+2 t^{2}+t^{4}}{\left(1+t^{2}+t^{4}\right)\left(2 t^{2}+1\right)} d t
$$
Resolving the rational function $$
\frac{1+2 x+x^{2}}{\left(1+x+x^{2}\right)(2 x+1)}=\frac{1}{3(2 x+1)}+\frac{x+2}{3\left(x^{2}+x+1\right)},
$$
into partial fractions yields $$
\int\frac{1+2 t^{2}+t^{4}}{(1 +t^{2}+t^{4})(2t^2+1)} d t=\frac{1}{3} \left(\underbrace{\int\frac{d t}{2 t^{2}+1}}_{K} +\underbrace{\int\frac{t^{2}+2}{t^{4}+t^{2}+1} d t}_{J}\right) 
$$
$$
K=\int \frac{d t}{2 t^{2}+1}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} t) +C_1 =\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+C_1
$$
We now focus on evaluating the last integral.
\begin{aligned}
J &=\int \frac{t^{2}+2}{t^{4}+t^{2}+1} d t \\
&=\int \frac{1+\frac{2}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t \\
&=\int \frac{\frac{3}{2}\left(1+\frac{1}{t^{2}}\right)-\frac{1}{2}\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+1} d t \\
&=\frac{3}{2} \int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+3}-\frac{1}{2} \int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-1} \\
&=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+\frac{1}{4 \sqrt{2}} \ln \left|\frac{t+\frac{1}{t}+1}{t+\frac{1}{t}-1}\right|+C \\
&=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{3} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|+C_2
\end{aligned}
Now we can conclude that
$$
\begin{aligned}
I=& \frac{1}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)\right.\left.-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{2} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|\right]+C
\end{aligned}
$$
My Question
Can we go further with  $n\geq 8$,
$$
I_n= \int\frac{1}{1+\sin ^{n} x} d x?
$$

Answering by mistake the linked post instead of your, my shortest result for $n=4$ is
$$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan
   ^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$
For $n=6$, doing the same
$$\frac {1}{1+\sin^6(x)}=\frac{2}{(a+1) (b+1) (3-\cos (2 x))}-\frac{2}{(a+1) (a-b) (2 a-1+\cos (2 x))}+$$ $$\frac{2}{(b+1) (a-b) (2 b-1+\cos (2 x))}$$ where $(a,b)$ are the complex roots of $x^2-x+1=0$.
Remember that
$$\int \frac {dx}{\cos(2x)+k}=-\frac{1}{\sqrt{1-k^2}}\tanh ^{-1}\left(\frac{(k-1) }{\sqrt{1-k^2}}\tan (x)\right) $$ I suppose that we could continue in this spirit assuming that we know the roots of $x^n+1=0$
