Solve the indefinite integral
$$ I=\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}\;dx $$
My Attempt:
$$ \begin{align} I&=\int\frac{1}{\sin x+\cos x+\frac{1}{\sin x \cos x}+\frac{\sin x +\cos x}{\sin x\cos x}}\;dx\\ \\ &=\int\frac{\sin x\cos x}{\left(\sin x+\cos x\right)\left(\sin x\cos x \right)+1+\left(\sin x+\cos x\right)}\;dx \end{align} $$
How can I complete the solution from here?
You want
$\begin{align} \int\frac{dx}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x} &= \int\frac{dx}{\sin x+\cos x+\frac{\sin x}{\cos x} +\frac{\cos x}{\sin x}+\frac1{\sin x}+\frac1{\cos x}}\\ &= \int\frac{\sin x \cos x\ dx}{\sin^2 x \cos x+\cos^2 x \sin x+\sin^2 x +\cos^2 x+\cos x+\sin x}\\ &= \int\frac{\sin x \cos x\ dx}{\sin x \cos x(\sin x+ \cos x)+1+\cos x+\sin x}\\ &= \int\frac{\sin x \cos x\ dx}{(\sin x \cos x+1)(\sin x+ \cos x)+1}\\ \end{align} $.
Applying substitutions $\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2 dt}{1+t^2}$, this becomes
$\begin{align} \int\frac{\sin x \cos x\ dx}{(\sin x \cos x+1)(\sin x+ \cos x)+1} &=\int\frac{\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\frac{2t}{1+t^2}+ \frac{1-t^2}{1+t^2})+1}\\ &=\int\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\ &=\int\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\ &=\int\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\ &=\int\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \quad \text{ (according to Wolfram Alpha)}\\ &=\int\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\ \end{align} $.
According to Wolfram Alpha, that quartic can be factored as the product of two quadratics, but the coefficients look irrational.
I'll leave it at this.
If you can't see anything simpler: when you have a rational function of sin and cos, you can always use the tangent half-angle substitution $t = \tan (x/2)$. Then $$\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2 dt}{1+t^2}.$$ Substituting these converts your integral into the integral of a rational function of $t$, which can then be integrated by partial fractions.
You can use Euler's identities and expand it in terms of complex exponents like (after some simplifications): $$\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}=\frac{e^{\mathrm i x} \left((1+\mathrm i) e^{2 \mathrm i x}-2 \mathrm i e^{\mathrm i x}+(\mathrm i-1)\right)}{e^{4 \mathrm i x}-(1+\mathrm i) e^{3 \mathrm i x}+6 \mathrm i e^{2 \mathrm i x}+(1-\mathrm i) e^{\mathrm i x}-1}$$ So $$ \begin{eqnarray} \int\!\!\frac{1}{\sin x\!+\!\cos x\!+\!\tan x\!+\!\cot x\!+\!\csc x\!+\!\sec x}\!\mathrm dx&=&\frac{1}{\mathrm i}\!\!\int \!\! \frac{(1+\mathrm i) e^{2 \mathrm i x}-2 \mathrm i e^{\mathrm i x}+(\mathrm i-1)}{e^{4 \mathrm i x}-(1+\mathrm i) e^{3 \mathrm i x}+6 \mathrm i e^{2 \mathrm i x}+(1-\mathrm i) e^{\mathrm i x}-1}\!\!\mathrm de^{\mathrm i x}\!\!=\\ &=&\frac{1}{\mathrm i}\!\!\int \!\! \frac{(1+\mathrm i) t^{2}-2 \mathrm i t+(\mathrm i-1)}{t^{4}-(1+\mathrm i) t^{3}+6 \mathrm i t^{2}+(1-\mathrm i) t^{}-1} \!\!\mathrm dt&=&\\ &=&\frac{1}{\mathrm i}\!\!\int \!\! \frac{(t-t_{11})(t-t_{12})}{(t-t_{21})(t-t_{22})(t-t_{32})(t-t_{42})} \!\!\mathrm dt \end{eqnarray} $$ After that you can use partial fraction expansion to simplify the answer (Mathematica gives 4 distinct roots for denominator).
Substitute $x=y+\dfrac\pi4$ :
$$\begin{align*} I &= \int \frac{dx}{\sin x+\cos x+\csc x+\sec x+\tan x+\cot x} \\ &= \int \frac{dy}{\frac{\cos y+\sin y}{\sqrt2} + \frac{\cos y-\sin y}{\sqrt2} + \frac{\sqrt2}{\cos y+\sin y} + \frac{\sqrt2}{\cos y-\sin y} + \frac{\cos y+\sin y}{\cos y-\sin y} + \frac{\cos y-\sin y}{\sin y-\cos y}} \\ &= \int \frac{2\cos^2y-1}{2\sqrt2\,\cos^3y+\sqrt2\,\cos y+2} \, dy \\ \end{align*}$$
By inspection, $\sqrt2 \, c \left(2c^2 + 1\right) + 2 = 0$ at $c=-\dfrac1{\sqrt2}$, and thus we may simplify
$$I = \int \frac{\sqrt2\,\cos y+1}{2\cos^2y - \sqrt2 \, \cos y + 2} \, dy$$
Now substituting $y=2\arctan z \implies \cos y = \dfrac{1-z^2}{1+z^2}$ makes for an easier partial fraction expansion,
$$I = \frac{3\sqrt2-2}7 \int \frac{1-(3+2\sqrt2)z^2}{1+\frac{9+4\sqrt2}7 z^4} \, dz$$
The right side expands to
$$\int \frac{1-az^2}{1+b^2z^4}\,dz = \frac1{2ib} \int \left(\frac{a+ib}{1+ibz^2} - \frac{a-ib}{1-ibz^2}\right) \, dz$$
and this produces a combination of the $\arctan$ and $\operatorname{artanh}$ of a complex argument.
Consider,
$$I=\int\frac{dx}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}$$
Substitute $x-\frac{\pi}{4}=t$
\[ \sin\left(t + \frac{\pi}{4}\right) = \frac{\sin t + \cos t}{\sqrt{2}}, \quad \cos\left(t + \frac{\pi}{4}\right) = \frac{\cos t - \sin t}{\sqrt{2}} \]
\[ \tan\left(t + \frac{\pi}{4}\right) = \frac{\tan t + 1}{1 - \tan t}, \quad \cot\left(t + \frac{\pi}{4}\right) = \frac{\cot t + 1}{1 - \cot t} \] \[ \sec\left(t + \frac{\pi}{4}\right) = \frac{1}{\cos\left(t + \frac{\pi}{4}\right)} = \frac{\sqrt{2}}{\cos t - \sin t} \] \[ \csc\left(t + \frac{\pi}{4}\right) = \frac{1}{\sin\left(t + \frac{\pi}{4}\right)} = \frac{\sqrt{2}}{\sin t + \cos t} \]
After a lot of simplification,
$$=\int \frac{\sqrt2 \cos t -1}{2\cos^2 t-\sqrt2 \cos t+2}\,dt$$
Substitute $tan \frac{t}2=u\implies \cos t = \frac{1-t^2}{1+t^2}, \frac{2}{1+t^2}\,dt=\,du$
$$=-\frac{2+3\sqrt2}{7}\int \frac{u^2-(3-2\sqrt2)}{u^2+\left(\frac{10-4\sqrt2}{7}\right)}\,du$$
For ease of notations, $a=3-2\sqrt2$ and $b=\frac{10-4\sqrt2}{7}$
$$=-\frac{2+3\sqrt2}{7}\int \frac{u^2-a}{u^2+b}\,du$$
$$=-\frac{2+3\sqrt2}{7}\color{red}{\int \frac{u^2}{u^2+b}\,du}+a\cdot\frac{2+3\sqrt2}{7}\color{red}{\int \frac{1}{u^2+b}\,du}$$
These are two standard integrals,
The first integral evaluates to,
$=\frac{1}{4\sqrt{b}\sqrt{2\sqrt{b}}}\ln\left|\frac{\tan^2\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}-\sqrt{2\sqrt{b}}\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)}{\tan^2\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}+\sqrt{2\sqrt{b}}\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)}\right|+\frac{1}{2\sqrt{b}\sqrt{2\sqrt{b}}}\tan^{-1}\left(\frac{\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}-\sqrt{2\sqrt{b}}\cot\left(\frac{x}{2}-\frac{\pi}{8}\right)}{\sqrt{2\sqrt{b}}}\right)$
The second integral evaluates to,
$=-\frac{1}{4\sqrt{b}\sqrt{2\sqrt{b}}}\ln\left|\frac{\tan^2\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}-\sqrt{2\sqrt{b}}\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)}{\tan^2\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}+\sqrt{2\sqrt{b}}\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)}\right|+\frac{1}{2\sqrt{b}\sqrt{2\sqrt{b}}}\tan^{-1}\left(\frac{\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)-\sqrt{b}\cot\left(\frac{x}{2}-\frac{\pi}{8}\right)}{\sqrt{2\sqrt{b}}}\right)$
So combining these results,
$I=-\frac{\sqrt{2}+1}{7\sqrt{b}\sqrt{2\sqrt{b}}}\ln\sqrt{\left|\frac{\tan^2\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}-\sqrt{2\sqrt{b}}\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)}{\tan^2\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}+\sqrt{2\sqrt{b}}\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)}\right|}+\frac{2-\sqrt{2}}{7\sqrt{b}\sqrt{2\sqrt{b}}}\tan^{-1}\left(\frac{\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)+\sqrt{b}-\sqrt{2\sqrt{b}}\cot\left(\frac{x}{2}-\frac{\pi}{8}\right)}{\sqrt{2\sqrt{b}}}\right)$
