\(\int \frac {x (A+B x^3)}{\sqrt [3]{a+b x^3}} \, dx\) [339]

Optimal result

Integrand size = 20, antiderivative size = 86 \[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {B x^2 \left (a+b x^3\right )^{2/3}}{4 b}+\frac {(2 A b-a B) x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{4 b \sqrt [3]{a+b x^3}} \] Output:

1/4*B*x^2*(b*x^3+a)^(2/3)/b+1/4*(2*A*b-B*a)*x^2*(1+b*x^3/a)^(1/3)*hypergeo 
m([1/3, 2/3],[5/3],-b*x^3/a)/b/(b*x^3+a)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \left (5 A x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+2 B x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )\right )}{10 \sqrt [3]{a+b x^3}} \] Input:

Integrate[(x*(A + B*x^3))/(a + b*x^3)^(1/3),x]
 

Output:

((1 + (b*x^3)/a)^(1/3)*(5*A*x^2*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3) 
/a)] + 2*B*x^5*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3)/a)]))/(10*(a + b 
*x^3)^(1/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {959, 889, 888}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(2 A b-a B) \int \frac {x}{\sqrt [3]{b x^3+a}}dx}{2 b}+\frac {B x^2 \left (a+b x^3\right )^{2/3}}{4 b}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {\sqrt [3]{\frac {b x^3}{a}+1} (2 A b-a B) \int \frac {x}{\sqrt [3]{\frac {b x^3}{a}+1}}dx}{2 b \sqrt [3]{a+b x^3}}+\frac {B x^2 \left (a+b x^3\right )^{2/3}}{4 b}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {x^2 \sqrt [3]{\frac {b x^3}{a}+1} (2 A b-a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{4 b \sqrt [3]{a+b x^3}}+\frac {B x^2 \left (a+b x^3\right )^{2/3}}{4 b}\)

Input:

Int[(x*(A + B*x^3))/(a + b*x^3)^(1/3),x]
 

Output:

(B*x^2*(a + b*x^3)^(2/3))/(4*b) + ((2*A*b - a*B)*x^2*(1 + (b*x^3)/a)^(1/3) 
*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(4*b*(a + b*x^3)^(1/3))
 

Defintions of rubi rules used

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 889

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I 
ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(c*x) 
^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0 
] &&  !(ILtQ[p, 0] || GtQ[a, 0])
 

rule 959

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p 
+ 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p 
 + 1) + 1))   Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

Maple [F]

\[\int \frac {x \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x\]

Input:

int(x*(B*x^3+A)/(b*x^3+a)^(1/3),x)
 

Output:

int(x*(B*x^3+A)/(b*x^3+a)^(1/3),x)
 

Fricas [F]

\[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="fricas")
 

Output:

integral((B*x^4 + A*x)/(b*x^3 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.55 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {A x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} + \frac {B x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {8}{3}\right )} \] Input:

integrate(x*(B*x**3+A)/(b*x**3+a)**(1/3),x)
 

Output:

A*x**2*gamma(2/3)*hyper((1/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a 
**(1/3)*gamma(5/3)) + B*x**5*gamma(5/3)*hyper((1/3, 5/3), (8/3,), b*x**3*e 
xp_polar(I*pi)/a)/(3*a**(1/3)*gamma(8/3))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*x/(b*x^3 + a)^(1/3), x)
 

Giac [F]

\[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*x/(b*x^3 + a)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {x\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \] Input:

int((x*(A + B*x^3))/(a + b*x^3)^(1/3),x)
 

Output:

int((x*(A + B*x^3))/(a + b*x^3)^(1/3), x)
 

Reduce [F]

\[ \int \frac {x \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a \] Input:

int(x*(B*x^3+A)/(b*x^3+a)^(1/3),x)
 

Output:

int(x**4/(a + b*x**3)**(1/3),x)*b + int(x/(a + b*x**3)**(1/3),x)*a