Solving used Real Based Methods: $\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt$

In working on integrals for the past couple of months, I've come across different cases of the following integral:

\begin{equation} I\left(x,a,k,n,m\right) = \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt \end{equation}

Where $x,a\in \mathbb{R}^{+}$.

Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.

Interested in special cases too!

The method I took:

First I wanted to bring the 'a' out the front:

\begin{equation} I(x,a,k,n,m) = \int_0^x \frac{t^k}{\left(a\left[\left(a^{-\frac{1}{n}}t\right)^n + 1\right]\right)^m}\:dt = \frac{1}{a^m} \int_0^x \frac{t^k}{\left(\left(a^{-\frac{1}{n}}t\right)^n + 1\right)^m}\:dt \end{equation} Here let $u = a^{-\frac{1}{n}}t$ Thus,

\begin{equation} I(x,a,k,n,m) = \frac{1}{a^m} \int_0^{a^{-\frac{1}{n}}x} \frac{\left(a^{\frac{1}{n}}u\right)^k}{\left(u^n + 1\right)^m}a^{\frac{1}{n}}\:du = a^{\frac{k + 1}{n} - m}\int_0^{a^{-\frac{1}{n}}x} \frac{u^k}{\left(u^n + 1\right)^m}\:du = a^{\frac{k + 1}{n} - m}I(a^{-\frac{1}{n}}x,1,k,n,m) \end{equation}

From here I will use $I$ in place of $I(x,a,k,n,m)$ for ease of typing. The next step is to make the substitution $w = u^n$ to yield:

\begin{equation} I = a^{\frac{k + 1}{n} - m}\int_0^{ax^n} \frac{w^\frac{k}{n}}{\left(w + 1\right)^m}\frac{\:dw}{nw^{\frac{n - 1}{n}}} = \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_0^{ax^n} \frac{w^{\frac{k + 1}{n} - 1}}{\left(w + 1\right)^m}\:dw \end{equation}

Here make the substitution $z = \frac{1}{1 + w}$ to yield:

\begin{align} I &= \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_1^{\frac{1}{1 + ax^n}} z^m \left(\frac{1 - z}{z}\right)^{\frac{k + 1}{n} - 1}\left(-\frac{1}{z^2}\right) \:dz = \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_{\frac{1}{1 + ax^n}}^1 z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz \\ &= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[\int_0^1 z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz - \int_0^{\frac{1}{1 + ax^n}} z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz \ \right] \\ &= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left( \frac{1}{1 + ax^n}; m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right] \end{align}

Where $B(a,b)$ is the Beta Function and $B(x; a,b)$ is the Incomplete Beta Function.

And so, we arrive at:

\begin{equation} \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(\frac{1}{1 + ax^n}; m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right] \end{equation}

Here we observe that for convergence:

\begin{equation} m - \frac{k + 1}{n} \gt 0,\quad \frac{k + 1}{n} \gt 0,\quad n \neq 0 \end{equation}

Note: when $x \rightarrow \infty$ we have:

\begin{equation} \int_0^\infty \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) \end{equation}

Update: Today I realised that we can use this result for another integral:

\begin{equation} \int_0^\infty \frac{\ln(t)}{\left(t^n + 1\right)^m}\:dt \end{equation}

This is achieved through a simple use of Feynman's Trick. Here we consider the case when $x \rightarrow \infty$ and $a = 1$. We see that

\begin{align} \frac{d}{dk}\left[ \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt \right]&= \frac{d}{dk}\left[\frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right] \\ \int_0^\infty \frac{t^k \ln(t)}{\left(t^n + 1\right)^m}\:dt &= \frac{1}{n^2}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\left[\psi^{(0)}\left(\frac{k + 1}{n}\right) - \psi^{(0)}\left(m - \frac{k + 1}{n}\right) \right] \end{align}

Thus, \begin{equation} \lim_{k \rightarrow 0} \int_0^\infty \frac{t^k \ln(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k \rightarrow 0}\frac{1}{n^2}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\left[\psi^{(0)}\left(\frac{k + 1}{n}\right) - \psi^{(0)}\left(m - \frac{k + 1}{n}\right) \right] \end{equation}

And finally:

\begin{equation} \int_0^\infty \frac{ \ln(t)}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n^2}B\left(m - \frac{1}{n}, \frac{1}{n} \right)\left[\psi^{(0)}\left(\frac{1}{n}\right) - \psi^{(0)}\left(m - \frac{1}{n}\right) \right] \end{equation}

Note: In the case where $m = 1$ we arrive:

\begin{align} \int_0^\infty \frac{ \ln(t)}{\left(t^n + 1\right)^1}\:dt &= \frac{1}{n^2}B\left(1 - \frac{1}{n}, \frac{1}{n} \right)\left[\psi^{(0)}\left(\frac{1}{n}\right) - \psi^{(0)}\left(1 - \frac{1}{n}\right) \right] \\ &= \frac{1}{n^2} \Gamma\left(\frac{1}{n} \right)\Gamma\left(1 - \frac{1}{n} \right) \cdot -\pi\cot\left(\frac{\pi}{n}\right) \\ &= \frac{1}{n^2} \frac{\pi}{\sin\left(\frac{\pi}{n}\right)}\cdot -\pi\cot\left(\frac{\pi}{n}\right) \end{align}

Thus:

\begin{equation} \int_0^\infty \frac{ \ln(t)}{t^n + 1}\:dt = -\frac{\pi^2}{n^2} \operatorname{cosec}\left(\frac{\pi}{n} \right)\cot\left(\frac{\pi}{n}\right) \end{equation}

NOT A SOLUTION:

I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:

1. Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
2. Evaluate the integral $ \int _0^{+\infty} \frac{x^m}{(a+bx^n)^p}$