\(\int \frac {x^3 (A+B x^3)}{\sqrt [3]{a+b x^3}} \, dx\) [332]

Optimal result

Integrand size = 22, antiderivative size = 143 \[ \int \frac {x^3 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {(3 A b-2 a B) x \left (a+b x^3\right )^{2/3}}{9 b^2}+\frac {B x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {a (3 A b-2 a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{7/3}}+\frac {a (3 A b-2 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{7/3}} \] Output:

1/9*(3*A*b-2*B*a)*x*(b*x^3+a)^(2/3)/b^2+1/6*B*x^4*(b*x^3+a)^(2/3)/b-1/27*a 
*(3*A*b-2*B*a)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2) 
/b^(7/3)+1/18*a*(3*A*b-2*B*a)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(7/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {3 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3} \left (6 A b-4 a B+3 b B x^3\right )+2 \sqrt {3} a (-3 A b+2 a B) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 a (-3 A b+2 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+a (-3 A b+2 a B) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{7/3}} \] Input:

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

Output:

(3*b^(1/3)*x*(a + b*x^3)^(2/3)*(6*A*b - 4*a*B + 3*b*B*x^3) + 2*Sqrt[3]*a*( 
-3*A*b + 2*a*B)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3 
))] - 2*a*(-3*A*b + 2*a*B)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + a*(-3*A 
*b + 2*a*B)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2 
/3)])/(54*b^(7/3))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {959, 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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 769

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

Input:

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

Output:

(B*x^4*(a + b*x^3)^(2/3))/(6*b) + ((3*A*b - 2*a*B)*((x*(a + b*x^3)^(2/3))/ 
(3*b) - (a*(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3] 
*b^(1/3)) - Log[-(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]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(268\) vs. \(2(116)=232\).

Time = 1.27 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.88

method	result	size
pseudoelliptic	\(\frac {9 B \,b^{\frac {4}{3}} x^{4} \left (b \,x^{3}+a \right )^{\frac {2}{3}}+18 A \,b^{\frac {4}{3}} x \left (b \,x^{3}+a \right )^{\frac {2}{3}}-12 B a x \,b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {2}{3}}+6 A \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}\, a b -4 B \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}\, a^{2}+6 A \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a b -3 A \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a b -4 B \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2}+2 B \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2}}{54 b^{\frac {7}{3}}}\)	\(269\)

Input:

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

Output:

1/54*(9*B*b^(4/3)*x^4*(b*x^3+a)^(2/3)+18*A*b^(4/3)*x*(b*x^3+a)^(2/3)-12*B* 
a*x*b^(1/3)*(b*x^3+a)^(2/3)+6*A*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^ 
(1/3))/b^(1/3)/x)*3^(1/2)*a*b-4*B*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a 
)^(1/3))/b^(1/3)/x)*3^(1/2)*a^2+6*A*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a*b 
-3*A*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*b-4 
*B*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a^2+2*B*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 = 0.13 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.98 \[ \int \frac {x^3 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\left [-\frac {3 \, \sqrt {\frac {1}{3}} {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 2 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (2 \, B a^{2} - 3 \, A a b\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, B b^{2} x^{4} - 2 \, {\left (2 \, B a b - 3 \, A b^{2}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{3}}, -\frac {2 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (2 \, B a^{2} - 3 \, A a b\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {6 \, \sqrt {\frac {1}{3}} {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}} - 3 \, {\left (3 \, B b^{2} x^{4} - 2 \, {\left (2 \, B a b - 3 \, A b^{2}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{3}}\right ] \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.65 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.56 \[ \int \frac {x^3 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {A x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} + \frac {B x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

A*x**4*gamma(4/3)*hyper((1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a 
**(1/3)*gamma(7/3)) + B*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (116) = 232\).

Time = 0.14 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.29 \[ \int \frac {x^3 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {1}{18} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} - \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{2}}\right )} A - \frac {1}{54} \, {\left (\frac {4 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} - \frac {2 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {7}{3}}} + \frac {4 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}} - \frac {3 \, {\left (\frac {7 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{4} - \frac {2 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}}}\right )} B \] Input:

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

Output:

1/18*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^( 
1/3))/b^(4/3) - a*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^ 
(2/3)/x^2)/b^(4/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) - 6*( 
b*x^3 + a)^(2/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^2))*A - 1/54*(4*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) - 
2*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) + 4*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7/3) - 3*(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))*B
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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