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!
