Solve integral $\int\frac{dx}{\sin x+ \cos x+\tan x +\cot x}$ I need to find:

$$\int\frac{1}{\sin x+ \cos x+\tan x +\cot x}\ dx$$

My attempts:
I have tried the conventional substitutions. I have tried the $\tan(x/2)$ substitutions, tried to solve it by quadratic but nothing has worked so far.
 A: Let $t=x-\frac\pi4$ and recognize $2\sin x\cos x= 2\cos^2t-1$
$$I= \int\frac{dx}{\sin x+ \cos x+\tan x +\cot x}
=\frac1{\sqrt2}\int \frac{\cos^2t-\frac12}{\cos^3t-\frac12\cos t+\frac1{\sqrt2}}dt
$$
Factorize the denominator ($y=\cos t$)
$$y^3-\frac12y+\frac1{\sqrt2}
=(y+a)\left(y-\frac {a-ib}2 \right) \left(y-\frac {a+ib}2 \right)
$$
where $a$ given below is the real root and $b =\frac12\sqrt{\frac3{\sqrt2a}-\frac12}$
$$a=  \bigg(\frac1{2\sqrt2}+\frac16\sqrt{\frac{13}3}\bigg)^{1/3}+ \bigg(\frac1{2\sqrt2}-\frac16\sqrt{\frac{13}3}\bigg)^{1/3}
\overset{\cdot}=1.0758
$$
Then, partially-fractionalize the integrand
\begin{align}
I&= \frac1{3+\sqrt2a}\int \frac{\frac1{\sqrt2}}{\cos t+a}
+\Re \frac{a+\sqrt2 -i \frac{a^2}b}{\cos t +\frac{ib-a}2}\ dt\\
\end{align}
and apply the known integral
$\int \frac{1}{\cos t+p}dt=\frac2{\sqrt{p^2-1}}\tan^{-1}\frac{\tan\frac t2}{\sqrt{\frac{p+1}{p-1}}}
$
to obtain
\begin{align}I
&= \frac1{3+\sqrt2a}\bigg(\frac{\sqrt2}{\sqrt{a^2-1}}\tan^{-1}\frac{\tan\frac t2}{\sqrt{\frac{a+1}{a-1}}} 
+4\Re \frac{a+\sqrt2-i\frac{a^2}b}{\sqrt{(ib-a)^2-4}}\tan^{-1}\frac{\tan\frac t2}{\sqrt{\frac{ib-a+2}{ib-a-2}}} \bigg)
\end{align}
A: Partial Solution
$$\begin{aligned}
&\int\frac{\mathrm{d}x}{\sin(x)+\cos(x)+\tan(x)+\cot(x)}
\\&=\int\frac{sc}{s^2c+c^2s+1}\,\mathrm{d}x\quad *
\\&=\int sc\sum_{n\geq0}(-1)^n(s^2c+c^2s)^n\,\mathrm{d}x\quad\text{(Binomial series)}**
\\&=\int sc\sum_{n\geq0}(-1)^n\left(\sum_{0\leq k\leq n}\binom{n}{k}(s^2c)^{k}(c^2s)^{n-k}\right)\mathrm{d}x\quad\text{(Binomial theorem)}
\\&=\sum_{n\geq0}\,\sum_{0\leq k\leq n}(-1)^n\binom{n}{k}\int s^{n+k+1}c^{2n-k+1}\,\mathrm{d}x
\end{aligned}$$
Then the integral in the last expression can be computed iteratively, for $a,b\in\mathbb{N}$, as:
$$\int s^{a}c^{b}\,\mathrm{d}x
=-\frac{s^{a-1}c^{b+1}}{a+b}+\frac{a-1}{a+b}\int s^{a-2}c^{b}\,\mathrm{d}x$$
This solves the integral as a series but doesn't give a closed form and still has the caveat of infinite terms.

* $s=\sin(x),c=\cos(x)$
** The series is the expansion of $\big((s^2c+c^2s)+1\big)^{-1}$, which converges since $|s^2c+c^2s|<1$.
The iterative algorithm for $\int s^ac^b\,\mathrm{d}x$ is here, on Wikipedia.
