TITLE: Two tough functions to integrate: $f(x)={\left({\frac{A}{x^\alpha}+\sqrt{B+\frac{C}{x^{2\alpha}}}}\right)}^{\frac{1}{3}}$ and $g(x)=\frac{1}{f(x)}$
QUESTION [8 upvotes]: I’m trying to solve two (in my opinion, tough) integrals which appear in part of my problem. I tried different ways but in the end I failed. See them below, please.
$${\rm{integral}}\,1 = \int {{{\left( {\frac{A}{{{x^\alpha }}}\, + \sqrt {B + \frac{C}{{{x^{2\alpha }}}}\,} } \right)}^{\frac{1}{3}}}} dx ,$$
and
$${\rm{integral}}\,2 = \int {\frac{1}{{{{\left( {\frac{A}{{{x^\alpha }}}\, + \sqrt {B + \frac{{C\,}}{{{x^{2\alpha }}}}} } \right)}^{\frac{1}{3}}}}}} dx,$$
where $\alpha$ is a positive integer ($\alpha \ge 2$). How can I solve them? I was wondering if someone could help me integrate these functions. Any help is appreciated. Much thanks.
Edit: Don't you think that if we set $\alpha =2 $, the integral might be easier to solve? Having this, I think I can solve the general case with $\alpha \ge 2$.

REPLY [2 votes]: $\color{brown}{\textbf{Simple cases.}}$
If $\;\underline{B=0},\;$ then the integration looks trivial.
If $\;\underline{C=0},\;$ then, by Wolfram Alpha,


$\color{brown}{\textbf{The first integral.}}$
$\color{green}{\mathbf{Case\;\alpha=3.}}$
Since
$$f_3(x) = \dfrac1x\sqrt[\large 3]{A+\sqrt{Bx^6+C}}
= \dfrac16\dfrac{6Bx^5}{Bx^6+C-C}\sqrt[\large3]{A+\sqrt{Bx^6+C}},\tag{1.1}$$
then the substitution
$$t^2=Bx^6+C,\quad 6Bx^5\,\text dx=2t\,\text dt,\quad C=p^2,\tag{1.2}$$
presents the integral in the form of
$$F_3(t) = \dfrac{F_{30}(p,t)+F_{30}(-p,t)}6,\tag{1.3}$$
where the integral
$$F_{30}(p,t)=\int \dfrac{\sqrt[\large 3]{A+t}}{t-p}\,\text dt\tag{1.4}$$
has the closed form of

$\color{green}{\mathbf{Case\;\alpha\not=3.}}$
Since
$$f(x) = x^{\large-^\alpha/_3} \sqrt[\large 3]{A+\sqrt{Bx^{2\alpha}+C}}, \tag{2.1}$$
then the substitution
$$x=y^{\large\frac3{3-\alpha}},\quad \text dx=y^{\large\frac{\alpha}{3-\alpha}}\text dy,\quad \beta =\dfrac{6\alpha}{3-\alpha}\tag{2.2}$$
presents the first integral in the form of
$$F_\beta(y) = \int\sqrt[\large 3]{A+\sqrt{C+By^{\beta}}\,}\,\text dy.\tag{2.3}$$
If $\;\underline{A^2>|C+By^\beta|},\;$ then
$$F_\beta(y) = \sqrt[\large 3]A \sum\limits_{k=0}^{\infty}\dbinom{^1/_3}k\, \dfrac{C^{\large^k/_2}}{A^k} \int \left(1+\dfrac BCy^{\beta}\right)^{\large^k/_2}\,\text dy$$
$$ = \sqrt[\large 3]A \sum\limits_{k=0}^{\infty}\dbinom{^1/_3}k\, \dfrac{C^{\large^k/_2}}{A^k}\, y\, \operatorname{_2F_1}\left(\dfrac 1\beta, -\dfrac k2, \dfrac{1+\beta}\beta, -\dfrac BCy^\beta\right),$$
$$F_\alpha(x) = \sqrt[\large 3]{Ax^{3-\alpha}} \sum\limits_{k=0}^{\infty}\dbinom{^1/_3}k\, \dfrac{C^{\large^k/_2}}{A^k}\,  \operatorname{_2F_1}\left(\dfrac{3-\alpha}{6\alpha}, -\dfrac k2, \dfrac{3+5\alpha}{6\alpha}, -\dfrac BC x^{2\alpha}\right).\tag{2.4}$$

If $\;\underline{A^2<|C+By^\beta|},\;$ then
$$F_\beta(y) = \sum\limits_{k=0}^{\infty}\dbinom{^1/_3}k\,\dfrac{A^k}{C^{\large^k/_2}} \int \left(1+\dfrac BCy^{\beta}\right)^{\large-^k/_2}\sqrt[\large 6]{C+By^\beta} \,\text dy$$
$$ = \sqrt[\large 6]C \sum\limits_{k=0}^{\infty}\dbinom{^1/_3}k\, \dfrac{A^k}{C^{\large^k/_2}}\, y\, \operatorname{_2F_1}\left(\dfrac 1\beta, \dfrac k2-\dfrac16, \dfrac{1+\beta}\beta, -\dfrac BCy^\beta\right),$$
$$F_\alpha(x) = \sqrt[\large 6]{Cx^{6-2\alpha}} \sum\limits_{k=0}^{\infty} \dbinom{^1/_3}k\, \dfrac{A^k}{C^{\large^k/_2}}\, \operatorname{_2F_1}\left(\dfrac{3-\alpha}{6\alpha}, \dfrac k2-\dfrac16, \dfrac{3+5\alpha}{6\alpha}, -\dfrac BC x^{2\alpha}\right).\tag{2.5}$$
$\color{brown}{\textbf{The second integral.}}$
$\color{green}{\mathbf{Case\;\alpha=-3.}}$
Since
$$g_{-3}(x) = \dfrac1{x\sqrt[\large 3]{A+\sqrt{Bx^{-6}+C}}}
= -\dfrac16\dfrac{-6Bx^{-7}}{Bx^{-6}+C-C}
\dfrac1{\sqrt[\large 3]{A+\sqrt{Bx^6+C}}},\tag{3.1}$$
then the substitution
$$s^2=Bx^{-6}+C,\quad -6Bx^{-7}\,\text dx=2s\,\text ds,\quad C=p^2,\tag{3.2}$$
presents the integral in the form of
$$G_{-3}(s) = -\dfrac{F_{30}(p,s)+F_{30}(-p,s)}6,\tag{3.3}$$
where the integral $\;F_{30}(p,t)\;$ is defined by $(1.4).$
$\color{green}{\mathbf{Case\;\alpha\not=-3.}}$
Since
$$g(x) = \dfrac{x^{\large^\alpha/_3}}
{\sqrt[\large3]{A+\sqrt{Bx^{2\alpha}+C}}},\tag{4.1}$$
then the substitution
$$x=z^{\large\frac3{3+\alpha}},\quad \text dx=^{-\large\frac{\alpha}{3+\alpha}}\text dz,\quad \gamma =\dfrac{6\alpha}{3+\alpha}\tag{4.2}$$
presents the second integral in the form of
$$G_\gamma(z) = \int \dfrac {\text dz}{\sqrt[\large 3]{A+\sqrt{C+Bz^{\gamma}}}}.\tag{4.3}$$
If $\;\underline{A^2>|C+Bz^\gamma|},\;$ then
$$G_\gamma(z) = \dfrac1{\sqrt[\large 3]A} \sum\limits_{k=0}^{\infty}\dbinom{-^1/_3}k\, \dfrac{C^{\large^k/_2}}{A^k} \int \left(1+\dfrac BCz^{\gamma}\right)^{\large^k/_2}\,\text dz$$
$$ = \dfrac1{\sqrt[\large 3]A} \sum\limits_{k=0}^{\infty}\dbinom{-^1/_3}k\, \dfrac{C^{\large^k/_2}}{A^k}\, z\, \operatorname{_2F_1}\left(\dfrac 1\gamma, -\dfrac k2, \dfrac{1+\gamma}\gamma, -\dfrac BCz^\gamma\right),$$
$$G_\alpha(x) = \sqrt[\large 3]{\dfrac{x^{3+\alpha}}A} \sum\limits_{k=0}^{\infty}\dbinom{-^1/_3}k\, \dfrac{C^{\large^k/_2}}{A^k}\,  \operatorname{_2F_1}\left(\dfrac{3+\alpha}{6\alpha}, -\dfrac k2, \dfrac{3+7\alpha}{6\alpha}, -\dfrac BC x^{2\alpha}\right).\tag{4.4}$$
If $\;\underline{A^2<|C+Bz^\gamma|},\;$ then
$$G_\gamma(z) = \sum\limits_{k=0}^{\infty}\dbinom{-^1/_3}k\,\dfrac{A^k}{C^{\large^k/_2}} \int \left(1+\dfrac BCz^{\gamma}\right)^{\large-^k/_2}\dfrac1{\sqrt[\large 6]{C+Bz^\gamma}} \,\text dz$$
$$ = \dfrac1{\sqrt[\large 6]C} \sum\limits_{k=0}^{\infty}\dbinom{-^1/_3}k\, \dfrac{A^k}{C^{\large^k/_2}}\, z\, \operatorname{_2F_1}\left(\dfrac 1\gamma, \dfrac k2+\dfrac16, \dfrac{1+\gamma}\gamma, -\dfrac BCz^\gamma\right),$$
$$G_\alpha(x) = \sqrt[\large 6]{\dfrac C{z^{6+2\alpha}}} \sum\limits_{k=0}^{\infty} \dbinom{-^1/_3}k\, \dfrac{A^k}{C^{\large^k/_2}}\, \operatorname{_2F_1}\left(\dfrac{3+\alpha}{6\alpha}, \dfrac k2+\dfrac16, \dfrac{3+7\alpha}{6\alpha}, -\dfrac BC x^{2\alpha}\right).\tag{4.5}$$
$\color{brown}{\textbf{Summary.}}$
Therefore, both the given integrals can be presented in the closed form or (previously) as 1D series of the hypergeometric functions.