Integrals of the form $\int x^m(a+bx^n)^p dx$

Question

Find $$\int{x^{13/2}(1+x^{5/2})^{1/2}}dx$$

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My attempt:

Usually when I face the form $\int x^m(a+bx^n)^p dx$, I would factor out $x^n$ and proceed.

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Example:

$$I=\int{x^{-11}(1+x^4)^{-1/2}}dx$$

$$I=\int{x^{-13}(1+\frac{1}{x^4})^{-1/2}dx}$$

Let $t=1+\frac{1}{x^4} \implies dt=\frac{-4}{x^5}dx$

$$I=\int{\frac{-1}{4}\frac{(t-1)^2}{\sqrt t}}dt$$

which is now very easy to evaluate.

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Coming back to the original problem,

$$I=\int{x^{13/2}(1+x^{5/2})^{1/2}}dx$$ Proceeding in the usual way,

$$I=\int{x^{31/4}(1+\frac{1}{x^{5/2}})^{1/2}}dx$$

Setting $t=1+\frac{1}{x^{5/2}} \implies dt=\frac{-5}{2x^{7/2}}dx$

$$I=\int{\frac{-2}{5}x^{45/4}\sqrt t}dt$$

$$I=\int{\frac{-2}{5}\frac{\sqrt t}{(t-1)^{9/2}}}dt$$

which is disappointing, since it doesn't yield anything.

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Questions:

1) Why did the usual method not work here? Does it not work for some specific cases? If yes, when?

2) How can the integral be solved?

3) Is there any general approach for solving $\int x^m(a+bx^n)^p dx $ ?

Answer 1

For integrals known as Differential binomial $$\int x^m(a+bx^n)^p dx$$ where $m,n,p$ are rational numbers are following methods:

1. p whole integer. Take $x=z^N$ where $N$ common denominator of $m,n$.
2. $\frac{m+1}{n}$ whole integer. Then we use $a+x^{n}b=z^N$, where $N$ is denominator of $p$.
3. $\frac{m+1}{n}+p$ whole integer. Then we use $ax^{-n}+b=z^N$, where $N$ is denominator of $p$.

In all other cases, the integral of a differential binomial cannot be expressed by elementary functions (P.L. Chebyshev, 1853).

Answer 2

Why not,

$$ x^{ \frac52} = t$$ Then,

$$ x^{ \frac32} dx= \frac{2}{5} dt$$

Our integral was:

$$\int{x^{13/2}(1+x^{5/2})^{1/2}}dx$$

It becomes:

$$ \int t^2 ( 1 + t)^{\frac12} \frac{2dt}{5}$$

Now, integrate by parts:

$$ \frac{2}{3} [ (1+t)^{\frac32} t^2 - \int (2t) (1+t)^{\frac32} dt]$$

And one more integration by part finishes the job.