I am trying to evaluate the following equation: $\int_0^x t^{-1}(t^a+b)^{1/a}dt$
The following is the results from Wolfram Alpha, but I cannot understand how to derive this.
There was a similar question to this problem:
How this integral was evaluated using hypergeometric series?
Following the solution from this post, I have substituted $t=xu^{1/a}$ to get the following result:
$$\int_0^x t^{-1}(t^a+b)^{1/a}dt= b^{1/a}/a\int_{0}^{1}u^{-1}(\left(x^{a}/b\right)u+1)^{1/a}du$$
Setting d = -1/a, e = 0, f = 1, and z = $(-x^a/b)$ the integration part of the RHS of the above equation is equal to the integration part of the RHS of the following expression
$_2 F_1(d,e;f;z)=\frac{1}{B(e,f-e)}\int_{0}^{1}u^{e-1}(1-u)^{f-e-1}(1-zu)^{-d}du$
However, here $B(0,1)=\infty$, so this approach does not work. Do you know why this happens, and is there any other way to explain how the initial integral takes the hypergeometric functional form?
Best Answer
Generally, if Wolfram|Alpha doesn't show the steps it means they aren't interesting.
On the other hand, we can peek inside some other database, for example that of Rubi.
In particular, if we want to calculate:
$$ I_k(x) := \int x^{-1}\left(x^u+v\right)^{1/u}\text{d}x $$
according to Rubi Rule 272, if $\frac{m+1}{n} \in \mathbb{Z}$ then:
$$ \int x^m\left(a+b\,x^n\right)^p\text{d}x = \frac{1}{n}\int y^{\frac{m+1}{n}-1}\left(a+b\,y\right)^p\text{d}y $$
from which:
$$ I_k(y) = \frac{1}{u}\int y^{-1}(y+v)^{1/u}\text{d}y\,. $$
So, according to Rubi Rule 67, if $b\,c\,d \ne 0 \, \land \, n \not\in \mathbb{Z} \, \land \, \left(m \in \mathbb{Z} \, \vee \, -\frac{d}{b\,c}>0\right)$ then:
$$ \int (b\,x)^m(c+d\,x)^n\text{d}x = \frac{(c+d\,x)^{n+1}}{d\,(n+1)\left(-\frac{d}{b\,c}\right)^{m}}\,{}_2F_1\left(-m,\,n+1;\,n+2;\,1+\frac{d\,x}{c}\right) + k $$
we get what we want:
$$ I_k(y) = -\frac{\left(y+v\right)^{1+\frac{1}{u}}}{(1+u)\,v}\,{}_2F_1\left(1,\,1+\frac{1}{u};\,2+\frac{1}{u};\,1+\frac{y}{v}\right) + k $$
i.e.
$$ I_k(x) = -\frac{\left(x^u+v\right)^{1+\frac{1}{u}}}{(1+u)\,v}\,{}_2F_1\left(1,\,1+\frac{1}{u};\,2+\frac{1}{u};\,1+\frac{x^u}{v}\right) + k\,. $$