I have the following definite integral whose integrand includes the multiplication of a binomial and an arbitrary power:
$$\int^x_a {\left(\frac{x-a}{b-a}\right)}^n x^{-3}\operatorname d x, \qquad b>x>a, \qquad n\in \Re, n>0 \tag{1} $$
Now, a solution to a similar integral is provided by Gradshteyn et al.(2015) with some constraints:
$$\int^u_0 {(\frac{x^{\alpha-1}}{(1+cx)^v})} d x \quad \Re(a)>0, \text{ and } |\arg(1+bu)|< \pi \tag 2$$
I can rearrange the terms in the integrand of Equation 1 to make it similar to Equation 2. However, the solution of Gradshteyn et al. (2015) cannot be used on the rearranged equation (Equation 2) as x has still a negative power of 3. On the other hand, if I define a new variable $t$ for Equation 1, $ t={\frac{x-a}{b-a}}, $ the transformed integrand of Equation 1, as given below, can be equivalent to Equation 2 and solved properly:
$$\int^{(x-a)/(b-a)}_0 {\left[\frac{t^{n-1}}{(1+\frac{b-a}{a}t)^{3}}\right]} dt \quad \Re(n)>0 \text{ and } \left|\arg \left(1+\frac{b-a}{a}t\right)\right|=0 < \pi \tag{3}$$
This is interesting because an integral that cannot be solved with a certain method can be solved by the same method if the variable of the integral is transformed. Do you think that such an anomaly is possible or it is an erroneous result because of some calculation mistakes?
I would be glad for your suggestions.
References
I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series, and Products, Elsevier, Amsterdam, 2015.