Verification of the integral: $\int \frac{dx}{a\cos^2x+b\sin(2x)+c\sin^2x}$ I have solved several integrals (of the form below) with given coefficients for $a, b, c$ and then though whether I could generalize that or not. I would like to ask for verification. Problem statement:

Evaluate:
  $$
\int \frac{dx}{a\cos^2x+b\sin(2x)+c\sin^2x}
$$

Below is what I got so far. I've started by factoring out $1\over \cos^2x$ to obtain:
$$
I = \int \frac{1}{\cos^2x}\cdot\frac{dx}{a + 2b\tan x + c\tan^2x}
$$
Now substitute $t = \tan x$:
$$
I = \int \frac{dt}{a + 2bt + ct^2}
$$
Now depending on the coefficients, we might have three cases: either there are 2 distinct real roots, both roots are equal, or there are no roots in $\Bbb R$.
If there are no roots, then we could complete the square and consider a standard integral. Suppose:
$$
ct^2 + 2bt + a = c(t-h)^2 + k \\
ct^2 + 2bt + a = ct^2 + 2ch t+ch^2 + k
$$
After solving a linear system of equations one may obtain:
$$
h = {b\over c}\\
k = {ac + b^2\over c}
$$
Hence the integral becomes:
$$
I = {1\over c}\int \frac{dt}{\left(t - {b\over c}\right)^2 + {ac + b^2 \over c^2}}
$$
Which is a standard integral:
$$
\boxed{I = {1\over \sqrt{ac + b^2}}\arctan\left(\frac{c\tan x + b}{\sqrt{ac-b^2}}\right) + Const}
$$

As for the second case, namely when there are two distinct roots. After solving  a quadratic equation:
$$
t_{1,2} = \frac{-b \pm \sqrt{b^2 - ac}}{c}
$$
Let:
$$
R_1 = \frac{-b + \sqrt{b^2 - ac}}{c}\\
R_2 = \frac{-b - \sqrt{b^2 - ac}}{c}
$$
So the integral becomes:
$$
I = \int \frac{dt}{(t-R_1)(t-R_2)}
$$
Using partial fraction decomposition:
$$
\frac{1}{ct^2 + 2bt + a} = \frac{A}{t - R_1} + \frac{B}{t-R_2} \\
At - AR_2 + Bt - BR_1 = 1\\
\begin{cases}
A + B = 0\\
-AR_2-BR_1 = 1
\end{cases}\\
A = -B\\
B = {1\over (R_2-R_1)}
$$
Therefore:
$$
I = \int \frac{1}{(R_1 - R_2)(t - R_1)}dt - \int \frac{1}{(R_1-R_2)(t-R_2)}dt
$$
Which yields:
$$
\begin{align}
I &= {1\over R_1 - R_2}\ln|t - R_1| - {1\over R_1 - R_2}\ln|t - R_2| \\
&={1\over R_1 - R_2}\ln|\tan x - R_1| - {1\over R_1 - R_2}\ln|\tan x - R_2| \\
&=\boxed{{1\over R_1 - R_2}\ln\left|{\tan x - R_1\over \tan x - R_2}\right| + Const}
\end{align}
$$

When there are two equal roots, then let the root be denoted by $R_1 = R_2 = R$:
$$
I = \int {dt\over (t - R)^2} = -\frac{1}{t - R}
$$
Or simply:
$$
\boxed{I = -\frac{1}{\tan x - R} = \frac{1}{R-\tan x} + Const}
$$

I'm wondering whether I've done this right, could someone please verify the above? Thank you!
 A: Taking into account the small corrections in the comments, this is correct for the cases with real roots. The case with no real roots has a subtle error that often gets glossed over in these sorts of problems.
The issue is that when there are no real roots, $[a \cos(x)^2 + b\sin(2x) + c \sin(x)^2]^{-1}$ is continuous everywhere. This means its antiderivative must be everywhere differentiable, but what you have for its antiderivative is not, since $\tan^{-1}$ jumps down by $\pi$ every time its argument switches from $+\infty$ to $-\infty$ (or up if it's switching the other way because $c < 0$). To restore continuity, we need to add (or subtract, if $c < 0$) $\pi$ to it every time $x$ passes an odd multiple of $\pi/2$. This can be done with the floor function, giving
$$
I = \frac{1}{\sqrt{ac - b^2}}\left(\tan^{-1}\left[\frac{c\tan(x)+b}{\sqrt{ac-b^2}}\right] +\pi\left\lfloor\frac{x}{\pi}+\frac{1}{2}\right\rfloor\operatorname{sgn}[c]\right) + Const,
$$
where $\operatorname{sgn}$ is the sign function. As you can see in this plot for the $a=2,\,b=c=1$ case, the extra term restores continuity of the antiderivative.

