I got this problem in my homework exercise:
$$\int \arctan(\sec x + \tan x) dx$$
I simplified it to
$$\int \arctan\left(\dfrac{1+\sin x}{\cos x}\right) dx$$
$$=\int \arctan\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right) dx$$
now, I tried putting $\sqrt{\dfrac{1+\sin x}{1-\sin x}} = \tan t$
Then $x = \arcsin(\sin^2 t-\cos^2 t)$ but it becomes complete mess after that!
$${1+\sin(x)\over \cos(x)}={\sin(x/2)+\cos(x/2)\over \cos(x/2)-\sin(x/2)}={1+\tan(x/2)\over 1-\tan(x/2)}=\tan(\pi/4+x/2)$$
Also keep in mind: $$\tan^{-1}(\tan(z))=\begin{cases}z&-\pi/2\le z\le\pi/2\\ z-\pi&\ \ \ \pi/2\lt z\le \pi\\z+\pi&\ \ -\pi\le z\lt-\pi/2 \end{cases}$$
An direct way to use the integration by parts:
Let $u = \arctan(\sec(x)+\tan(x))$ then $du = \frac{1}{2} dx$
$dv = dx$ then $v = x$.
Then the integral becomes $$ x \arctan(\sec(x)+\tan(x)) - \int \frac{x}{2}dx$$
Using integration by parts, we get:
$$\int \arctan(\sec x + \tan x) dx=x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C$$
By integration by parts we obtain $$\int \arctan(\sec x + \tan x) dx=x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C.$$ Since we have the identity (see Qwerty's answer): $$\sec x + \tan x={1+\sin(x)\over \cos(x)}=\tan(\pi/4+x/2),$$ the primitive can be simplified (!?) to $$x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C=x\arctan(\tan(\pi/4+x/2))-\dfrac{x^2}{4}+C\\=\frac{x^2+\pi x}{4}-\pi x \left\lfloor \frac{x}{2\pi}+\frac{3}{4}\right\rfloor+C.$$
P.S. Note that if we erroneously write that $\arctan(\sec x + \tan x)=\pi/4+x/2$, then we get a different answer $$\int \arctan(\sec x + \tan x) dx=\int (\pi/4+x/2) dx=\frac{x^2+\pi x}{4}+C.$$
sec x + tan x = tan t. then sec x - tan x = cot t. sec x = (cot t + tan t)/2. (sec t)^2 dt = sec x(sec x + tan x)dx = (tan t(tan t + cot t)/2)dx. Substitute dx and simplify.
A bit late to the party but I guess I could provide a general piecewise solution from where Qwerty left off.
From Qwerty you would end up with $\arctan(\tan(x/2+\pi/4))$, but you go further and break down this expression into a piecewise definition for all $x\in\mathbb{R}$.
Recall that $\tan(\theta)$ is invertible on the interval $-\pi/2\lt \theta\lt\pi/2$ so for those values we have $\arctan(\tan(\theta))=\theta$.
Notice that since the interval $(-\pi/2,\pi/2)$ is of length $\pi$, we can reach every other input by shifting the open interval $(-\pi/2,\pi/2)$ by some integer multiple of $\pi$. But given that $\tan(\theta)$ is periodic on $\pi$ (which you can confirm with the formula for $\tan(a+b)$), then $\arctan(\tan(\theta))$ also has a period of $\pi$, therefore the outputs repeat so the graph of $\arctan(\tan(\theta))$ is the line segment of $y=\theta$ that goes from the points $(-\pi/2,-\pi/2)$ to $(\pi/2,\pi/2)$ shifted laterally by $\pi k$ units for $k\in\mathbb{Z}$.
This can be summarized as follows
$$\arctan(\tan(\theta))=\left\{\theta-\pi k: -\frac{\pi}{2}+\pi k\lt \theta\lt\frac{\pi}{2}+\pi k\space, k\in\mathbb{Z}\right\}$$
Replacing $\theta$ with $x/2+\pi/4$ gives the graph $y=x/2+\pi/4-\pi k$ for $k\in\mathbb{Z}$ over the interval $-\pi/2+\pi k\lt x/2+\pi/4\lt\pi/2+\pi k$, which in turn solves to $-3\pi/2+2\pi k\lt x\lt\pi/2+2\pi k$.
So the piecewise definition of $\arctan(\sec(x)+\tan(x))$ becomes
$$\arctan(\sec(x)+\tan(x))=\left\{\frac{x}{2}+\frac{\pi}{4}-\pi k:-\frac{3\pi}{2}+2\pi k\lt x\lt\frac{\pi}{2}+2\pi k\space, k\in\mathbb{Z}\right\}\space (1)$$
so the indefinite integral evaluates to
$$\int\arctan(\sec(x)+\tan(x))dx=\left\{\frac{1}{4}x^2+\frac{1}{4}\pi(1-4k)x+C:-\frac{3\pi}{2}+2\pi k\lt x\lt\frac{\pi}{2}+2\pi k\space, k\in\mathbb{Z}\right\}$$
Or you can use the expression proved in previous answers using IBP and then substitute $(1)$ in place of $\arctan(\sec(x)+\tan(x))$ and you would obtain the exact same answer up to a constant.
