How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$?

Question

I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$

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This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ we will use the substitution: $$ a+bx^k = u^{\frac{1}{n}}$$

Therefore,

$$ u^3 = 1 + x^2 \iff x = \sqrt{u^3 +1} \text{ (Mistake Here. Check the comments) } $$ $$\iff dx = \frac{3u^2}{2\sqrt{u^3}+1}$$

So the new integral is,

$$ \int x^5 (1+x^2)^{\frac{2}{3}} dx = \frac{3}{2} \int u^4 (u^3+1)^{\frac{7}{6}} du$$

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Instead of simplifying the integral, the substitution did nothing by keeping it at the same form, with different values on the variables $m,k,n$.

I tried to substitute once again and it doesn't seem to lead in any known paths, anytime soon.

Any ideas on how this could be evaluated?

Answer 1

HINT: Let $1+x^2=t^3\implies 2xdx=3t^2dt$ or $xdx=\frac{3}{2}t^2dt$ $$\int x^5(1+x^2)^{2/3}dx=\int (x^2)^2(1+x^2)^{2/3}xdx$$

$$=\int (t^3-1)^2(t^3)^{2/3}\ \frac{3t^2}{2}dt$$ $$=\frac32\int(t^3-1)^2t^4 dt $$ $$=\frac{3}{2}\int (t^{10}-2t^7+t^4)dt$$

Answer 2

Do the substitution $u^{3/5} = 1+ x^2$. Then $x = \sqrt{u^{3/5}-1}$, so $\mathrm{d}x = \frac{3\,u^{-2/5}}{5\,\sqrt{u^{3/5}-1}}\,\mathrm d u$ and $$ ∫ x^5 \,(1+x^2)^{2/3}\,\mathrm d x = \frac{3}{5}∫ (u^{3/5}-1)^2 \,\mathrm d x. $$

Answer 3

Take $x^5 \times (1 + x^2)^{\frac{2}{3}} = (x^{15} \times (1+x^2)^2)^{\frac{1}{3}}$ or some other form of that sort and integrate.

Thus, you'll be integrating $(x^{19} + 2x^{17} + x^{15})^{\frac{1}{3}}$

Applying substitution method here may work.

Answer 4

Alternatively, denote $x^2=t\Rightarrow 2xdx=dt$: $$\int x^5 (1+x^2)^{\frac{2}{3}}dx=\frac12\int t^2 (1+t)^{\frac{2}{3}}dt=\\ \frac12\int (t^2-1+1)(1+t)^{\frac23}dt=\int(t-1)(1+t)^\frac53dt+\int(1+t)^\frac23dt=\\ \frac12\int(t+1-2)(1+t)^{\frac53}dt+\frac12\int(1+t)^{\frac23}dt=\\ \frac12\int(1+t)^{\frac83}dt-\int (1+t)^{\frac53}dt+\frac12\int(1+t)^{\frac23}dt=\\ \frac3{22}(1+t)^{\frac{11}3}-\frac3{8}(1+t)^{\frac{8}3}+\frac3{10}(1+t)^{\frac{5}3}+C$$ Now, plug $t=x^2$ and you are done. Wolfram alpha answer and comparison.

Answer 5

Generalization

For

$$I=\int x^{m-1+rm}(x^m+c)^p\ dx$$

Set $x^m+c=y,dy=?$

$$mI=\int(y-c)^ry^p\ dy$$

Now use Binomial expansion