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

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 $\alpha \ge 2$.

Answer 1

$\color{brown}{\textbf{Simple cases.}}$

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.

Answer 2

Here will be an evaluation of:

$${ \rm{integral}}\,1,2 = \int {{{\left( {\frac{A}{{{x^\alpha }}}\, + \sqrt {B + \frac{C}{{{x^{2\alpha }}}}\,} } \right)}^{\pm\frac{1}{3}}}} dx ,$$

Note that if $$(a+b)^v=a^v\left(1+\frac ba\right)^v$$ then the radius of convergence for a Binomial Series would be $\left|\frac ba\right|<1$

There is also a slightly generalized version for $(a+b)^v$ instead of $(1+b)^v$. Here it is applied to your problem and simplified into an Incomplete Beta function. I will use $\alpha=a$ for easier typing:

$$\int {{{\left( {\frac{A}{{{x^a}}}\, + \sqrt {B + \frac{C}{{{x^{2a}}}}\,} } \right)}^{\pm\frac{1}{3}}}} dx =\int\sum_{n=0}^\infty \binom{\pm\frac13}n A^n x^{-2an} \sqrt{B+Cx^{-2a}}^{\pm\frac13-n}dx= \sum_{n=0}^\infty \binom{\frac13}n A^n B^{\pm\frac16} \int x^{-2an}\sqrt B^{-n}\left(1+\frac CB x^{-2a}\right)^{\pm\frac16-\frac n2}dx $$

so $B\ne0$ which almost looks like an Incomplete Beta function:

$$\text B_z(a,b)=\int_0^z t^{a-1} (1-t)^{b-1}dt$$

It can be shown that you can use that you can use the Gauss Hypergeometric function to integrate:

$$\int\sum_{n=0}^\infty \binom{\pm\frac13}n A^n x^{-2an} \sqrt{B+Cx^{-2a}}^{\pm\frac13-n}dx =c+\sum_{n=0}^\infty \binom{\pm\frac13} n\frac{A^n}{(1-2an)x^{2an-1}} \big(B+C x^{-2a}\big)^{\pm\frac16-\frac n2}\left(\frac{C x^{-2a}}B+1\right)^{\frac n2\mp\frac16}\,_2\text F_1\left(n-\frac 1{2a},\frac n2\mp\frac16,n-\frac1{2a}+1,-\frac{C x^{-2a}}B\right)$$

The Gauss Hypergeometric function also can be written as an Incomplete Beta function assuming an analytic continuation of the Incomplete Beta function to make the domain larger.

$$\frac k{z^k}\text B_z(k,b)=\,_2\text F_1(k,1-b,k+1,z), k= n-\frac 1{2a},1-b= \frac n2\mp\frac16 ,z= -\frac{C x^{-2a}}B\ne0$$

Therefore:

$$c+\sum_{n=0}^\infty \binom{\pm\frac13} n\frac{A^n}{(1-2an)x^{2an-1}} \big(B+C x^{-2a}\big)^{\pm\frac16-\frac n2}\left(\frac{C x^{-2a}}B+1\right)^{\frac n2\mp\frac16}\,_2\text F_1\left(n-\frac 1{2a},\frac n2\mp\frac16,n-\frac1{2a}+1,-\frac{C x^{-2a}}B\right)=c+\sum_{n=0}^\infty \binom{\pm\frac13} n\frac{A^n}{(1-2an)x^{2an-1}} \big(B+C x^{-2a}\big)^{\pm\frac16-\frac n2}\left(\frac{C x^{-2a}}B+1\right)^{\frac n2\mp\frac16}\frac{n-\frac 1{2a}}{\left(-\frac{C x^{-2a}}B\right)^{n-\frac 1{2a}}}\text B_{-\frac{C x^{-2a}}B }\left(n-\frac 1{2a},1-\left(\frac n2\mp\frac16\right)\right)= \boxed{c+\sum_{n=0}^\infty \binom{\pm\frac13} n\frac{A^n}{(1-2an)x^{2an-1}} \big(B+C x^{-2a}\big)^{\pm\frac16-\frac n2}\left(\frac{C x^{-2a}}B+1\right)^{\frac n2\mp\frac16}\left(n-\frac 1{2a}\right){\left(-\frac{C x^{-2a}}B\right)^{\frac 1{2a}-n}}\text B_{-\frac{C x^{-2a}}B }\left(n-\frac 1{2a},\left\{\frac76\text{ or }\frac56\right\} -\frac n2\right) }$$

The $\{\text{or}\}$ is due to the $\pm$ in the argument. The above result gives the largest domain whereas the following simplifications restrict the domain from canceling and more:

$$c+\sum_{n=0}^\infty \binom{\pm\frac13} n\frac{A^n}{(1-2an)x^{2an-1}} \big(B+C x^{-2a}\big)^{\pm\frac16-\frac n2}\left(\frac{C x^{-2a}}B+1\right)^{\frac n2\mp\frac16}\left(n-\frac 1{2a}\right){\left(-\frac{C x^{-2a}}B\right)^{\frac 1{2a}-n}}\text B_{-\frac{C x^{-2a}}B }\left(n-\frac 1{2a},\left\{\frac76\text{ or }\frac56\right\} -\frac n2\right) = \boxed{c+\sqrt[a]iB^{\pm\frac16-\frac1{2a}}C^\frac1{2a}\sum_{n=0}^\infty \binom{\pm\frac13} n\frac{(-A)^n B^{\frac n2}}{C^n(1-2an)}\text B_{-{C x^{-2a}}}\left(n-\frac 1{2a},\left\{\frac76\text{ or }\frac56\right\} -\frac n2\right)} $$

If I think of some other important simplifications, then I will add them in. If you have any, then please tell me.

Note that various research papers have defined Incomplete Beta function series like:

An Extension of the Mittag-Leffler Function and Its Associated Properties

and

A New Extension of Mittag-Leffler function

but these only are for an Incomplete Beta function with the sum index in one argument and not both like in this answer. Please correct me and give me feedback!