trying to understand the method of evaluating intergral $\int R[x,(\frac{ax+b}{cx+d})^\lambda,(\frac{ax+b}{cx+d})^\mu,...]dx$ I am reading a section from my textbook about the integration of expressions containing radicals, and I can't understand the reason for certain substituion when evaluating the integral

[Here is what the book section says]:
Suppose I am trying to evaluate the integral
$$\int R[x,(\frac{ax+b}{cx+d})^\lambda,(\frac{ax+b}{cx+d})^\mu,...]dx\ \ \ \ \ \ \ (1)$$
where $R$ is a rational function of its arguments, i.e. the quotient of polynomials in these arguments, whilst $\lambda,\mu,...$ are rational numbers.
Now let $m$ be the common denominator of these fractions (this is what the book says, I think "fraction" here means rational number $\lambda,\mu,...$ ???)
Then introduce a new variable $t$, where
$$\frac{ax+b}{cx+d}=t^m\ (????)$$
Evidently, after this, $x,dx/dt$ and expression
$$(\frac{ax+b}{cx+d})^\lambda,(\frac{ax+b}{cx+d})^\mu,...$$
will be rational functions of $t$, thus we reduced (1) to the integral of a rational fraction

**[Question]: I don't understand the intention behind such substituion, why finding the common denominator $m$ and then define $t^m$ ? What is the advantage of such substitution ? Any example ?
 A: I think a solid example would do wonders here.
Let $$R(x, y, z) = \dfrac{ x^2 + xy - z}{x - z}$$ and consider the integral
$$I =\int R\left(\left(\frac{x + 2}{x-1}\right)^{2/3}, \left(\frac{x + 2}{x-1}\right)^{-1/2}, \left(\frac{x + 2}{x-1}\right)^2\right)\,dx\\=\int\dfrac{ \left(\frac{x + 2}{x-1}\right)^{4/3} + \left(\frac{x + 2}{x-1}\right)^{1/6} - \left(\frac{x + 2}{x-1}\right)^2}{\left(\frac{x + 2}{x-1}\right)^{2/3} - \left(\frac{x + 2}{x-1}\right)^2}dx$$
The least common denominator of the exponents is $6$, so we define
$$\frac{x + 2}{x-1} = t^6$$
You can solve this for $x$ (I hadn't planned for this, it just turned out the fraction I chose is its own inverse function):
$$x = \frac{t^6 + 2}{t^6 - 1}$$
so $$dx = \frac{-18t^5}{(t^6-1)^2}dt$$
And we substitute for each of the arguments (and here you can see why we wanted the LCD of the exponents for $m$):
$$\begin{align}\left(\frac{x + 2}{x-1}\right)^{2/3} &= (t^6)^{2/3} = t^4\\\left(\frac{x + 2}{x-1}\right)^{-1/2} &= (t^6)^{-1/2} = t^{-3}\\\left(\frac{x + 2}{x-1}\right)^2&=(t^6)^2 = t^{12}\end{align}$$
Substitute all this back into the original integral, and we get
$$\begin{align}I &= \int R(t^4, t^{-3},t^{12})\frac{-18t^5}{(t^6-1)^2}dt \\&= \int \dfrac{t^8 + t - t^{12}}{t^4 - t^{12}}\frac{-18t^5}{(t^6-1)^2}dt\\&=18\int\dfrac{t^{13}-t^9 - t^2}{(1 - t^8)(t^6-1)^2}dt\end{align}$$
Which, while messy, will fall to the method of partial fractions.
