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

Question

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

My attempt:

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

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.

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.

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 $ ?

See my answer to Integrate $(x^2+1)^\frac{1}{3}$, which cites (with link provided) where in Hardy's book The Integration of Functions of a Single Variable this type of integral is discussed. — Dave L. Renfro, Sep 09 '20 at 18:28
@DaveL.Renfro The reference is really helpful. Thank you so much! — DatBoi, Sep 09 '20 at 18:32

zkutch · Answer 1 · 2020-09-09T18:25:30.387

4

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

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

edited Sep 09 '20 at 18:25

answered Sep 09 '20 at 18:20

zkutch

Thank you for the answer. Can you provide some intuition on why/how this works? I would also like the answer to address the flaw in my method – DatBoi Sep 09 '20 at 18:24
@DatBoi.Your case seems is 2nd. $\frac{\frac{13}{2}+1}{\frac{5}{2}}=3$. No? Also added last line for for completeness of answer. – zkutch Sep 09 '20 at 18:33
Yes, I understood. Thank you! – DatBoi Sep 09 '20 at 18:33
You are welcome. Write, please, if/when some more questions will arise. – zkutch Sep 09 '20 at 18:35
Im not quite sure what you meant by that. – DatBoi Sep 09 '20 at 18:36
ask again when needed - nothing more. – zkutch Sep 09 '20 at 18:37
Sure. Very kind of you :) – DatBoi Sep 09 '20 at 18:39

score 0 · Answer 2 · answered Oct 06 '20 at 13:44

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.