How do we handle the integral $I_{4}=\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{4} d \theta$? After struggling with $I_2,I_3$ in the post, I dare to tackle $I_4$ now.
We first rewrite the integral
$$I_{4}=\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{4} d \theta =\int \frac{\sec ^{2} \theta}{\left(\sec ^{2} \theta+\tan ^{2} \theta\right)^{4}} d(\tan \theta) $$
Letting $x=\tan \theta$ converts $$
\begin{aligned}I_4 &=\int \frac{1+x^{2}}{\left(2 x^{2}+1\right)^{4}} d t \\
&=\frac{1}{2}\left[\int \frac{d t}{\left(2 x^{2}+1\right)^{3}}+\int \frac{d t}{\left(2 x^{2}+1\right)^{4}}\right]
\end{aligned}
$$
By my post, we have an elegant reduction formula:
$$
\begin{aligned}
J_{n} &=\int \frac{d x}{\left(a x^{2}+b\right)^{n}}, \text { where } n \geqslant 2 \\
&=\frac{x}{2 b(n-1)\left(a x^{2}+b\right)^{n-1}}+\frac{2 n-3}{2 b(n-1)} J_{n-1}
\end{aligned}
$$
When $a=2$ and $b=1,$ $$
J_{n}=\frac{x}{2(n-1)\left(2 x^{2}+1\right)^{n-1}}+\frac{2 n-3}{2(n-1)} J_{n-1}
$$
Hence
$$
I_{4}=\frac{1}{2}\left(J_{3}+J_{4}\right)
$$
Let’s start with $$J_1= \int \frac{d x}{2 x^{2}+1}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C_1 $$
$$
J_{2}=\frac{x}{2\left(2 x^{2}+1\right)}+\frac{1}{2} J_{1}=\frac{x}{2\left(2 x^{2}+1\right)}+\frac{1}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C_2
$$
$$
\begin{aligned}
J_{3} &=\frac{x}{4\left(2 x^{2}+1\right)^{2}}+\frac{3}{4} J_{2} \\
&=\frac{x}{4\left(2 x^{2}+1\right)^{2}}+\frac{3 x}{8\left(2 x^{2}+1\right)}+\frac{3}{8 \sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C_3
\end{aligned}
$$
$$
J_{4}=\frac{x}{6\left(2 x^{2}+1\right)^{3}}+\frac{5 x}{24\left(2 x^{2}+1\right)^{2}}+\frac{5 x}{16\left(2 x^{2}+1\right)}+\frac{5}{16 \sqrt{2}} \tan ^{-1}(\sqrt2 x)+C_4
$$
Now we can conclude that $$
\begin{array}{c}
\displaystyle I_{4}=\frac{\tan \theta}{96}\left[\frac{8}{\left(2 \tan ^{2} \theta+1\right)^{3}}+\frac{22}{\left(2 \tan ^{2} \theta+1\right)^{2}}+\frac{33}{2 \tan ^{2} \theta+1}\right] +\frac{11}{32 \sqrt{2}} \tan ^{-1}(\sqrt{2} \tan \theta)+C
\end{array}
$$
Please let me know if there is any mistake.
Can we go further?
Wish you all a happy and healthy New Year 2022!

In general, for any even power, substitute $\tan^2\theta = \frac12\tan^2 x$ to transform the integral to a manageable one
$$I_{2m}= \int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{2m} d \theta
= \frac1{2^{m-\frac12}}\int \cos^{2m}x\>(1+\cos^2x)^{m-1}dx
$$
which, for $m=2$, reduces to
\begin{align}
I_4
= &\>\frac1{2^{3/2}}\int \cos^{4}x\>(1+\cos^2x)\>dx\\
= &\>\frac{1}{2^{5/2}}\left(\frac13\cos^5x +\frac{11}{12}\cos^3x +\frac{11}{8}\cos x \right)\sin x
+ \frac{11}{2^{11/2}}x+C
\end{align}
