# Integration of $x^p/(c+x^q)$ for $p, q$ rationals

I have the following integral

$$\int_a^b\frac{x^p}{C+x^q}\textrm{dx}$$, where

• C is a non-zero constant,
• $$a>0, b>0$$, and
• $$p,q$$ are rationals.

Is there a general solution for such an integral. I've found a related integral on Wikipedia: https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_(a_+_b_xn)p

But there the exponents are integer and/or fractions. Any reference would be of great help.

Thanks.

For the antiderivative, there is a general solution for any $$(p,q)$$ $$(c\neq 0)$$which write $$\int\frac{x^p}{c+x^q}dx=\frac{x^{p+1} }{c (p+1)} \,_2F_1\left(1,\frac{p+1}{q};\frac{p+1}{q}+1;-\frac{x^q}{c}\right)$$ where appears the gaussian hypergeometric function.

• Thanks for your answer Claude. Can I ask for a reference where I could find these integrals? Sep 24, 2019 at 6:03

Let $$x^q=t$$,Then $$\int \frac{x^p}{c+x^q} dx= \frac{1}{q}\int \frac {t^{p/q+1/q-1}}{c+t} dt= \frac{t^{1+r}}{cq(1+r)} ~2F_1[1,~1+r,~2+r;~\frac{-t}{c}], ~~~r=p/q+1/q-1,$$ $$\Rightarrow I=\frac{x^{1+p}}{c(1+p)} ~_2F_1[1,~1+r,~2+r,-\frac{x^q}{c}]$$ where the hyper geometric funtion $$~_2F_1$$ is expressed as $$~_2F_1[A,B,C;z]=1+\frac{A B}{C} \frac{z}{1!}+\frac{A(A+1)B(B+1)}{C(C+1)}\frac{z^2}{2!}+...$$

Let $$c<0$$ and $$t=|c|x^q$$, or $$x=\sqrt[q]{\dfrac t{|c|}}$$. The integral becomes

$$\int\frac{x^p}{c+x^q}dx=-|c|^{-(p+q+1)/q}\int\frac{t^{p/q}}{1- t}dx,$$

which is of the incomplete Beta type.

When $$c>0$$, a further change of variable $$t:=\dfrac u{1-u}$$ will turn it to the Beta from.