Continuous antiderivative of $\frac{1}{1+\cos^2 x}$ without the floor function.

One important thing is while doing u-substitution, the substitution has to be injective. When we substitute $u=\tan x$ in samjoe’s answer, $\tan x$ is not injective on the whole real line, but is injective in intervals of length $\pi$. That’s why you get the right ‘behavior’ only within intervals but not between them.

I believe that the Geogebra’s answer can be derived by noting that $$\frac1{\pi}(arctan(\cot(\pi x))+\pi x-\pi/2)$$ behaves exactly the same as a floor function.

Use also the summation formula for arctan: $$arctan (u)+arctan (v)=arctan(\frac{u+v}{1-uv})$$

ADDED:

To demonstrate the importance of injectivity of substitutions, consider the integral $$\int^1_{0}xdx$$ which equals $\frac12$.

If we substitute in $u=x^2-x$, we obtain something like $$\int^0_0 \cdots du=0$$

What caused the paradox is $x^2-x$ is not injective in the interval $[0,1]$.

Similarly, there is nothing wrong for $$\int^x_k \frac1{1+\cos ^2x}dx=\int^x_k\frac{\sec^2x}{2+\tan^2x}dx=^{u=\tan x}\int^{arctan(x)}_{arctan(k)}\frac{du}{2+u^2}=\frac1{\sqrt2}arctan(\frac{\tan x}{\sqrt2})+C$$ as long as $\tan x$ is injective in the interval $[k,x]$. If the injectivity is not achieved in the interval, $C$ would change when $x$ goes from an injective interval of $\tan x$ to another.

This agrees with what the OP observed: the floor function thing is a constant in each injective interval of $\tan x$, and changes when going across the intervals. You may consider, the floor function thing is part of $C$.

The choice of $k$ is arbitrary. But when we try to find an antiderivative for all $x$ while $k$ remains fixed, it is impossible to always achieve the injectivity in $[k,x]$. As a trade off, we need to add a floor function to compensate for the silent change of $C$.

ADDED 2:

@samjoe derived the antiderivative $$\frac{\pi}{\sqrt2}\left \lfloor\frac{x+\pi/2}{\pi}\right\rfloor + \frac{1}{\sqrt 2}\arctan\left(\frac{\tan x}{\sqrt2}\right) $$

By noting $$\lfloor x\rfloor=\frac1{\pi}(arctan(\cot(\pi x))+\pi x-\pi/2)$$, the above expression can be rewritten to $$\frac{x}{\sqrt2}+\frac{arctan(-\tan x)}{\sqrt2}+\frac{arctan(\frac{\tan x}{\sqrt2})}{\sqrt2}$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}({arctan(-\tan x)}+arctan(\frac{\tan x}{\sqrt2}))$$ By the summation formula stated above $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{-\tan x+\frac{\tan x}{\sqrt2}}{1+\frac{\tan^2x}{\sqrt2}})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{\tan x(1-\sqrt2)}{\sqrt2+\tan^2x+1-1})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{\tan x(1-\sqrt2)}{\sqrt2+\sec^2x-1})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)\sin x\cos x}{(\sqrt2-1)\cos^2 x+1})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)2\sin x\cos x}{(\sqrt2-1)(2\cos^2 x)+2})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)\sin 2x}{(\sqrt2-1)(\cos 2x +1)+2})$$ $$=\frac{x}{\sqrt2}+\frac1{\sqrt2}arctan(\frac{(1-\sqrt2)\sin 2x}{(\sqrt2-1)\cos 2x+\sqrt2+1})$$ which is exactly what we want.