\(\int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx\) [400]

Optimal result

Integrand size = 15, antiderivative size = 120 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=-\frac {2 a x \left (a+b x^3\right )^{2/3}}{9 b^2}+\frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}+\frac {2 a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{7/3}}-\frac {a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{9 b^{7/3}} \] Output:

-2/9*a*x*(b*x^3+a)^(2/3)/b^2+1/6*x^4*(b*x^3+a)^(2/3)/b+2/27*a^2*arctan(1/3 
*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(7/3)-1/9*a^2*ln(-b^(1 
/3)*x+(b*x^3+a)^(1/3))/b^(7/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.40 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-4 a x+3 b x^4\right )}{18 b^2}+\frac {2 a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{9 \sqrt {3} b^{7/3}}-\frac {2 a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{27 b^{7/3}}+\frac {a^2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{27 b^{7/3}} \] Input:

Integrate[x^6/(a + b*x^3)^(1/3),x]
 

Output:

((a + b*x^3)^(2/3)*(-4*a*x + 3*b*x^4))/(18*b^2) + (2*a^2*ArcTan[(Sqrt[3]*b 
^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/(9*Sqrt[3]*b^(7/3)) - (2*a^2 
*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(27*b^(7/3)) + (a^2*Log[b^(2/3)*x^ 
2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(27*b^(7/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {843, 843, 769}

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.

\(\displaystyle \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {2 a \int \frac {x^3}{\sqrt [3]{b x^3+a}}dx}{3 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {2 a \left (\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{3 b}\right )}{3 b}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {2 a \left (\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{3 b}\right )}{3 b}\)

Input:

Int[x^6/(a + b*x^3)^(1/3),x]
 

Output:

(x^4*(a + b*x^3)^(2/3))/(6*b) - (2*a*((x*(a + b*x^3)^(2/3))/(3*b) - (a*(Ar 
cTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) - Lo 
g[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))))/(3*b)))/(3*b)
 

Defintions of rubi rules used

rule 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
 

rule 843

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

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.20

method	result	size
pseudoelliptic	\(\frac {9 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {4}{3}} x^{4}-12 a x \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {1}{3}}-4 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a^{2}-4 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2}+2 \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2}}{54 b^{\frac {7}{3}}}\)	\(144\)

Input:

int(x^6/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
 

Output:

1/54*(9*(b*x^3+a)^(2/3)*b^(4/3)*x^4-12*a*x*(b*x^3+a)^(2/3)*b^(1/3)-4*3^(1/ 
2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)*a^2-4*ln((- 
b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a^2+2*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3 
)*x+(b*x^3+a)^(2/3))/x^2)*a^2)/b^(7/3)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.32 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=\left [\frac {6 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} b x^{3} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} x\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} + 2 \, a\right ) - 4 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 2 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (3 \, b^{2} x^{4} - 4 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{3}}, -\frac {12 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} x - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{x}\right ) + 4 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 2 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, b^{2} x^{4} - 4 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{3}}\right ] \] Input:

integrate(x^6/(b*x^3+a)^(1/3),x, algorithm="fricas")
 

Output:

[1/54*(6*sqrt(1/3)*a^2*b*sqrt((-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a)^(1 
/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x 
^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2/3)*x)*sqrt((-b)^(1/3)/b) + 2*a) - 4*a^2*( 
-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + 2*a^2*(-b)^(2/3)*log 
(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2 
) + 3*(3*b^2*x^4 - 4*a*b*x)*(b*x^3 + a)^(2/3))/b^3, -1/54*(12*sqrt(1/3)*a^ 
2*b*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^(1 
/3))*sqrt(-(-b)^(1/3)/b)/x) + 4*a^2*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 
+ a)^(1/3))/x) - 2*a^2*(-b)^(2/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)* 
(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 3*(3*b^2*x^4 - 4*a*b*x)*(b*x^3 + 
a)^(2/3))/b^3]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.31 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate(x**6/(b*x**3+a)**(1/3),x)
                                                                                    
                                                                                    
 

Output:

x**7*gamma(7/3)*hyper((1/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a* 
*(1/3)*gamma(10/3))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.53 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=-\frac {2 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {7}{3}}} + \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{27 \, b^{\frac {7}{3}}} - \frac {2 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{27 \, b^{\frac {7}{3}}} + \frac {\frac {7 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}}{18 \, {\left (b^{4} - \frac {2 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}}\right )}} \] Input:

integrate(x^6/(b*x^3+a)^(1/3),x, algorithm="maxima")
 

Output:

-2/27*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^( 
1/3))/b^(7/3) + 1/27*a^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^ 
3 + a)^(2/3)/x^2)/b^(7/3) - 2/27*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b 
^(7/3) + 1/18*(7*(b*x^3 + a)^(2/3)*a^2*b/x^2 - 4*(b*x^3 + a)^(5/3)*a^2/x^5 
)/(b^4 - 2*(b*x^3 + a)*b^3/x^3 + (b*x^3 + a)^2*b^2/x^6)
 

Giac [F]

\[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {x^{6}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^6/(b*x^3+a)^(1/3),x, algorithm="giac")
 

Output:

integrate(x^6/(b*x^3 + a)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {x^6}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \] Input:

int(x^6/(a + b*x^3)^(1/3),x)
 

Output:

int(x^6/(a + b*x^3)^(1/3), x)
 

Reduce [F]

\[ \int \frac {x^6}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \] Input:

int(x^6/(b*x^3+a)^(1/3),x)
 

Output:

int(x**6/(a + b*x**3)**(1/3),x)