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!

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.

