Could someone suggest how to approach this integral:
$$ \int_0^\Theta {\frac{1}{a+b\cot\left({\pi{x}^c}\right)}} dx $$
where $a$, $b$, $c$ are scaler values, and $0<\Theta<1$. I have tried some ways through substitution but they result in singularities.
For general $c$ it's is probably hopeless to get a closed-form expression. Notice that for $c < 0$ and general the function has singularities accumulating at $0$, so we may as well restrict our attention to $c \geq 0$. We also assume that $b \neq 0$ (otherwise the integrand is constant).
We can find a closed form in the special case $c = 1$: The substitution $x = \frac{1}{\pi} \arctan t$, $dx = \frac{dt}{\pi(1 + t^2)}$ transforms the integral to the rational integral $$\frac{1}{\pi}\int_0^{\tan \pi\Theta} \frac{t\,dt}{(a t + b) (1 + t^2)} ,$$ which can be handled as usual with the method of partial fractions.
In the special case that $c$ is the reciprocal of a positive integer, say, $c = \frac{1}{k}$, the substitution $x = \left(\frac{u}{\pi}\right)^k$ transforms the integral to a constant multiple of $$\int_\ast^\ast \frac{u^{k - 1} \,du}{a + b \cot u} ,$$ which can evaluated in terms of elementary functions and polylogarithms, but even for $k = 2$ an explicit formula will likely be a mess.
For general $c$, since $\Theta$ is small, for some purposes a series approximation might be sufficient: Near $x = 0$ we have (assuming $x \geq 0$ if $c$ is nonintegral) $$\frac{1}{a + b \cot(x^c)} = \frac{\pi}{b} x^c - \frac{\pi^2a}{b^2} x^{2c} + R_1(x),$$ where $R(x) \in O(x^{3 c})$, so for $\Theta$ near $0$ (assuming $\Theta \geq 0$ for nonintegral $c$), $$\int_0^\Theta \frac{dx}{a + b \cot(x^c)} = \frac{\pi}{b (c + 1)}\Theta^{c + 1} - \frac{\pi^2 a}{b^2 (2 c + 1)} \Theta^{2 c + 1} + R_2(\Theta),$$ where $R_2(\Theta) \in O(\Theta^{3 c + 1})$.