Complicated integral Please provide any idea how can I solve the following integral? does it have a solution in general? Thanks. 
$$
\int\ (ax+b)^{\alpha}x^{\beta}dx\,.
$$
$$
0<\alpha<1, \beta>1 , 
$$
And beta is a Real value number. 
 A: You may write it in general as a $_2F_1$ hypergeometric series for $r:=\frac ab$ and using :
\begin{align}
\int\ (1+r\,x)^{\alpha}\ x^{\beta-1}dx&=\int x^{\beta-1}\left(1+(-\alpha)\frac{(-rx)}{1!}+(-\alpha)(1-\alpha)\frac{(-rx)^2}{2!}+\cdots\right)dx\\
&=\frac{x^\beta}{\beta}\left(1+\frac{(-\alpha)(\beta)}{(\beta+1)}\frac{(-rx)}{1!}+\frac{(-\alpha)(1-\alpha)(\beta)(\beta+1)}{(\beta+1)(\beta+2)}\frac{(-rx)^2}{2!}+\cdots\right)\\
&=\frac{x^\beta}{\beta}\sum_{j=0}^{\infty}\frac{(-\alpha)_j(\beta)_j}{(\beta+1)_j}\frac{(-rx)^j}{j!}\\
&=\frac{x^\beta}{\beta}\;
_{2}F_{1}\left.\left(\begin{array}{ccc}
-\alpha&&\beta\\
&\beta+1\\
\end{array}\right|\ -r\,x\right)
\end{align}
Not sure that this expression will help much except in specific cases (many formulas are known)...
A: This may be a binomial–Chebyshev integral.
$$ \int x^p \left( c_1 x^q + c_0 \right) ^r dx $$
where p, q, r are rational. If r is a nonnegative integer, just expand it and have fun! Otherwise, set y = xq and then dy = qxq-1dx, leading to
$$ \frac 1 q \int y^ {\frac {p+1} {q} - 1} \left( c_1 y + c_0 \right) ^r dy. $$
As a result, we only have to consider
$$ \int x^{r_1} \left( c_1 x + c_0 \right) ^{r_2} dx $$
where r1 and r2 are rational. Let n1 be the denominator of r1, n2 the one of r2.


*

*When r1 is a positive integer, set z = c1x + c0.

*When r1 is a negative integer, set z = (c1x + c0)1/n2.

*When r2 is a negative integer, set z = x1/n1.

*When r1 + r2 is an integer, set $z = \dfrac {\left( c_1 x + c_0 \right) ^{1/n_1}} {x^{1/n_1}}$.

*Otherwise, the integral is inexpressible in finite terms, according to Liouville's theorem.


For example, consider
$$ \int \sqrt x \left( x+1 \right) ^{5/2} dx. $$
Set $ z = \dfrac {\sqrt{x+1}} {\sqrt x} $, so $ dx = - \dfrac {2z\,dz} {\left( z^2 - 1 \right) ^2} $. This simplifies the integrand into a rational function.
$$ \int \sqrt x \left( x+1 \right) ^{5/2} dx = \int - \frac {2z^6} {\left( z^2 - 1 \right) ^5} dz. $$
