Integration of $\frac{x^{\frac{2}{3}}}{(1+x^{\frac{3}{5}})^2}$

Question

I have to integrate the function $\displaystyle f(x) = \frac{x^{\frac{2}{3}}}{(1+x^{\frac{3}{5}})^2}$ with respect to $x$.

Putting $x^{\frac{3}{5}}=t,$ then $x^{-\frac{2}{3}}dx = \frac{3}{2}dt$.

Am I not able to go further. Could someone help me with this? Thanks

Answer 1

$$x^{2/3} = t ~~~~~~~~~~~ \text{d}t = \frac{2}{3}x^{-1/3}\ \text{d}x = \frac{2}{3} t^{-1/2}\ \text{d}x$$

$$x^{3/5} \to t^{9/10}$$

$$\int\frac{3}{2}\frac{t\sqrt{t}}{(1 + t^{9/10})^2}\ \text{d}t$$

Whose result is

$$\frac{15}{567} \left(\frac{63 t^{7/10}}{t^{9/10}+1}+81 t^{7/10}+56 \log \left(\sqrt[5]{t}-\sqrt[10]{t}+1\right)-112 \log \left(\sqrt[10]{t}+1\right)-112 \sqrt{3} \tan ^{-1}\left(\frac{2 \sqrt[10]{t}-1}{\sqrt{3}}\right)-112 \cos \left(\frac{2 \pi }{9}\right) \log \left(\sqrt[5]{t}-2 \sqrt[10]{t} \cos \left(\frac{\pi }{9}\right)+1\right)+112 \cos \left(\frac{\pi }{9}\right) \log \left(\sqrt[5]{t}+2 \sqrt[10]{t} \sin \left(\frac{\pi }{18}\right)+1\right)-112 \sin \left(\frac{\pi }{18}\right) \log \left(\sqrt[5]{t}+2 \sqrt[10]{t} \cos \left(\frac{2 \pi }{9}\right)+1\right)+224 \cos \left(\frac{\pi }{18}\right) \tan ^{-1}\left(\csc \left(\frac{2 \pi }{9}\right) \left(\sqrt[10]{t}+\cos \left(\frac{2 \pi }{9}\right)\right)\right)+224 \sin \left(\frac{\pi }{9}\right) \tan ^{-1}\left(\sqrt[10]{t} \sec \left(\frac{\pi }{18}\right)+\tan \left(\frac{\pi }{18}\right)\right)-224 \sin \left(\frac{2 \pi }{9}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{9}\right) \left(\sqrt[10]{t}-\cos \left(\frac{\pi }{9}\right)\right)\right)\right)$$

And going back to $x$:

$$\frac{5}{{189 \left(x^{3/5}+1\right)}}\left(144 x^{7/15}+81 x^{16/15}-112 x^{3/5} \log \left(\sqrt[15]{x}+1\right)+56 x^{3/5} \log \left(x^{2/15}-\sqrt[15]{x}+1\right)+56 \log \left(x^{2/15}-\sqrt[15]{x}+1\right)-112 \sqrt{3} \left(x^{3/5}+1\right) \tan ^{-1}\left(\frac{2 \sqrt[15]{x}-1}{\sqrt{3}}\right)-112 x^{3/5} \cos \left(\frac{2 \pi }{9}\right) \log \left(x^{2/15}-2 \sqrt[15]{x} \cos \left(\frac{\pi }{9}\right)+1\right)-112 \cos \left(\frac{2 \pi }{9}\right) \log \left(x^{2/15}-2 \sqrt[15]{x} \cos \left(\frac{\pi }{9}\right)+1\right)+112 x^{3/5} \cos \left(\frac{\pi }{9}\right) \log \left(x^{2/15}+2 \sqrt[15]{x} \sin \left(\frac{\pi }{18}\right)+1\right)-112 x^{3/5} \sin \left(\frac{\pi }{18}\right) \log \left(x^{2/15}+2 \sqrt[15]{x} \cos \left(\frac{2 \pi }{9}\right)+1\right)+112 \cos \left(\frac{\pi }{9}\right) \log \left(x^{2/15}+2 \sqrt[15]{x} \sin \left(\frac{\pi }{18}\right)+1\right)-112 \sin \left(\frac{\pi }{18}\right) \log \left(x^{2/15}+2 \sqrt[15]{x} \cos \left(\frac{2 \pi }{9}\right)+1\right)+224 \left(x^{3/5}+1\right) \cos \left(\frac{\pi }{18}\right) \tan ^{-1}\left(\csc \left(\frac{2 \pi }{9}\right) \left(\sqrt[15]{x}+\cos \left(\frac{2 \pi }{9}\right)\right)\right)+224 x^{3/5} \sin \left(\frac{\pi }{9}\right) \tan ^{-1}\left(\sqrt[15]{x} \sec \left(\frac{\pi }{18}\right)+\tan \left(\frac{\pi }{18}\right)\right)-224 x^{3/5} \sin \left(\frac{2 \pi }{9}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{9}\right) \left(\sqrt[15]{x}-\cos \left(\frac{\pi }{9}\right)\right)\right)-112 \log \left(\sqrt[15]{x}+1\right)+224 \sin \left(\frac{\pi }{9}\right) \tan ^{-1}\left(\sqrt[15]{x} \sec \left(\frac{\pi }{18}\right)+\tan \left(\frac{\pi }{18}\right)\right)-224 \sin \left(\frac{2 \pi }{9}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{9}\right) \left(\sqrt[15]{x}-\cos \left(\frac{\pi }{9}\right)\right)\right)\right)$$

Answer 2

Hint. Let $t:=x^{1/15}$ then you obtain the integral of a rational function $$\int\frac{x^{\frac{2}{3}}}{(1+x^{\frac{3}{5}})^2}dx= 15\int\frac{t^{24}}{(1+t^{9})^2}dt.$$ It is quite bad. Is this the complete and correct statement of your exercise?

Answer 3

As you see from answers, you can get nice monsters.

Making the problem more general, considering $$I=\int \frac{x^a}{\left(1+x^b\right)^2}\,dx$$ the general result write $$b\,I=x^{a+1} \left(\frac{b-a-1 }{a+1}\, _2F_1\left(1,\frac{a+1}{b};\frac{a+b+1}{b};-x^b\right)+\frac{1}{1+x^b}\right )$$ where appears the gaussian or ordinary hypergeometric function.

Edit

I suppose that the problem is to integrate $$\frac{x^{2/3}}{\left(1+x^\color{red}{{5/3}}\right)^2}$$ which is simple.

Probably, another typo in a textbook.

Update

If you start from Robert Z's answer, the problem is not too bad since $$\frac{1}{\left(1+t^9\right)^2}=\sum_{n=0}^\infty (-1)^{n-1}n \,t^{9(n-1)}$$ which makes

$$\int\frac{t^{24}}{(1+t^{9})^2}\,dt=\sum_{n=0}^\infty\frac{(-1)^{n-1} n }{9 n+16}t^{9 n+16}$$ Go back to $x$ and continue enjoying.