What purely real analytic techniques are there to evaluate $\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\,\mathrm{d}x$? $\newcommand{\d}{\,\mathrm{d}}$Last night, I evaluated the following integral: $$\begin{align}I:&=\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\d x\\&=\int_{-1}^1\frac{1}{(1+x^4)\sqrt{1-x^2}}\d x\\&=\frac{\pi}{2^{3/4}}
(\sin(\pi/8)+\cos(\pi/8))\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Using a "double keyhole" (as I phrase it) contour method involving a management of branch cuts and residues at infinity, here. Although I was happy to have succeeded in this, I wondered afterwards if I would have had any hope of evaluating $I$ with real analytic technique only.
The challenge:

Evaluate $I$ without use of complex analysis or even of complex arithmetic (e.g. for partial fraction decompositions involving $i$)

I posed this to some friends and they came up with the following method which I wanted to share with MSE:

$$\begin{align}I&\overset{x\mapsto\tan x}{=}\int_{-\infty}^\infty\frac{1+x^2}{(1+x^2)^2+x^4}\d x\\&\overset{x\mapsto1/x}{=}2\int_0^\infty\frac{1+x^2}{(1+x^2)^2+1}\d x\\&=2\int_0^\infty\int_0^\infty e^{-t(1+x^2)}\cos(t)\d t\d x\quad\text{Repr. with IBP}\\&=\sqrt{\pi}\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\end{align}$$
$$\begin{align}J:&=\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\\J^2&=\int_0^\infty\int_0^\infty\frac{e^{-(t+x)}\cos(t)\cos(x)}{\sqrt{tx}}\d t\d x\\&\overset{x\mapsto tx}{=}\int_0^\infty\int_0^\infty\frac{e^{-t(1+x)}\cos(t)\cos(tx)}{\sqrt{x}}\d x\d t\\&=\frac{1}{2}\int_0^\infty\frac{1}{\sqrt{x}}\cdot\frac{1+x+x^2}{(1+x)(1+x^2)}\d x\\&\overset{x\mapsto x^2}{=}\frac{1}{2}\int_0^\infty\left(\frac{1+x^2}{1+x^4}+\frac{1}{1+x^2}\right)\d x\\&=\frac{1}{2}\left[\frac{\pi}{4}\csc\left(\frac{\pi}{4}\right)+\frac{\pi}{4}\csc\left(\frac{3\pi}{4}\right)+\frac{\pi}{2}\right]\\&=\frac{\pi}{4}(1+\sqrt{2})\end{align}$$

Referencing this answer by Sangchul.
We conclude:

$$\begin{align}I&=\sqrt{\pi}\cdot\sqrt{J^2}\\&=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{4}(1+\sqrt{2})}\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$

Among those who helped me, who use MSE, I credit @TheSimpliFire and @KStarGamer who are much better at real integration than I am!
My question is less of a question and more of a request for a list - a list of other, purely real, methods to attack this integral. I hope the outcome of this will be an interesting selection of advanced integration techniques that I and others can learn from.
Note 1: I am aware of this posting by Quanto but it uses complex numbers.
Note 2: You must expand the cosine product as a sum of cosines and use the same integral representation (which is classically gotten from complex arithmetic but can be done with integration by parts): $$\int_0^\infty e^{-tx}\cos(t)\d t=\frac{x}{x^2+1},\,x\gt0$$

Working the antiderivative
Using your first step
$$\int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du$$
$$\left(u^2+1\right)^2+u^4=2 \left(u^2-u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)
   \left(u^2+u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)$$ Partial fraction decomposition
$$ \frac{u^2+1}{\left(u^2+1\right)^2+u^4}=$$
$$\frac 1 {2 \sqrt{2 \left(\sqrt{2}-1\right)} }\Bigg[\frac {\left(\sqrt{2}-2\right) u+2
   \sqrt{\sqrt{2}-1} } {2 u^2-2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }+\frac {\left(2-\sqrt{2}\right) u+2
   \sqrt{\sqrt{2}-1}} {2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}  }\Bigg]$$
$$4\int \frac {\left(\sqrt{2}-2\right) u+2
   \sqrt{\sqrt{2}-1} } {2 u^2-2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }\,du=$$ $$\left(\sqrt{2}-2\right) \log \left(2 u^2-2 \sqrt{\sqrt{2}-1}
   u+\sqrt{2}\right)+2 \sqrt{3-2 \sqrt{2}} \left(2+\sqrt{2}\right) \tan
   ^{-1}\left(\frac{2 u-\sqrt{\sqrt{2}-1}}{\sqrt{1+\sqrt{2}}}\right)$$
$$4\int\frac {\left(2-\sqrt{2}\right) u+2
   \sqrt{\sqrt{2}-1}} {2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}  }\,du$$
$$2 \sqrt{3-2 \sqrt{2}} \left(2+\sqrt{2}\right) \tan ^{-1}\left(\frac{2
   u+\sqrt{\sqrt{2}-1}}{\sqrt{1+\sqrt{2}}}\right)-\left(\sqrt{2}-2\right) \log
   \left(2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}\right)$$
$$\color{red}{8 \sqrt{2 \left(\sqrt{2}-1\right)} \int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du=}$$
$$\color{red}{-2 \sqrt{2} \tan ^{-1}\left(\frac{2 \sqrt{\sqrt{2}-1} u}{2 \left(\sqrt{2}-1\right)
   u^2+\sqrt{2}-2}\right)-\left(2-\sqrt{2}\right) \log \left(\frac{2 u
   \left(u-\sqrt{\sqrt{2}-1}\right)+\sqrt{2}}{2 u
   \left(u+\sqrt{\sqrt{2}-1}\right)+\sqrt{2}}\right)}$$ gives the desired result.
