\(\int x^3 (a+b x^3)^{2/3} (A+B x^3) \, dx\) [313]

Optimal result

Integrand size = 22, antiderivative size = 178 \[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {a (9 A b-4 a B) x \left (a+b x^3\right )^{2/3}}{81 b^2}+\frac {(9 A b-4 a B) x^4 \left (a+b x^3\right )^{2/3}}{54 b}+\frac {B x^4 \left (a+b x^3\right )^{5/3}}{9 b}-\frac {a^2 (9 A b-4 a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{81 \sqrt {3} b^{7/3}}+\frac {a^2 (9 A b-4 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{162 b^{7/3}} \] Output:

1/81*a*(9*A*b-4*B*a)*x*(b*x^3+a)^(2/3)/b^2+1/54*(9*A*b-4*B*a)*x^4*(b*x^3+a 
)^(2/3)/b+1/9*B*x^4*(b*x^3+a)^(5/3)/b-1/243*a^2*(9*A*b-4*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)+1/162*a^2*(9*A*b-4 
*B*a)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(7/3)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.19 \[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {3 \sqrt [3]{b} \left (a+b x^3\right )^{2/3} \left (-8 a^2 B x+6 a b x \left (3 A+B x^3\right )+9 b^2 x^4 \left (3 A+2 B x^3\right )\right )+2 \sqrt {3} a^2 (-9 A b+4 a B) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 a^2 (-9 A b+4 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+a^2 (-9 A b+4 a B) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{486 b^{7/3}} \] Input:

Integrate[x^3*(a + b*x^3)^(2/3)*(A + B*x^3),x]
 

Output:

(3*b^(1/3)*(a + b*x^3)^(2/3)*(-8*a^2*B*x + 6*a*b*x*(3*A + B*x^3) + 9*b^2*x 
^4*(3*A + 2*B*x^3)) + 2*Sqrt[3]*a^2*(-9*A*b + 4*a*B)*ArcTan[(Sqrt[3]*b^(1/ 
3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] - 2*a^2*(-9*A*b + 4*a*B)*Log[-(b^ 
(1/3)*x) + (a + b*x^3)^(1/3)] + a^2*(-9*A*b + 4*a*B)*Log[b^(2/3)*x^2 + b^( 
1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(486*b^(7/3))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {959, 811, 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.

\(\displaystyle \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(9 A b-4 a B) \int x^3 \left (b x^3+a\right )^{2/3}dx}{9 b}+\frac {B x^4 \left (a+b x^3\right )^{5/3}}{9 b}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(9 A b-4 a B) \left (\frac {1}{3} a \int \frac {x^3}{\sqrt [3]{b x^3+a}}dx+\frac {1}{6} x^4 \left (a+b x^3\right )^{2/3}\right )}{9 b}+\frac {B x^4 \left (a+b x^3\right )^{5/3}}{9 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(9 A b-4 a B) \left (\frac {1}{3} a \left (\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{3 b}\right )+\frac {1}{6} x^4 \left (a+b x^3\right )^{2/3}\right )}{9 b}+\frac {B x^4 \left (a+b x^3\right )^{5/3}}{9 b}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {(9 A b-4 a B) \left (\frac {1}{3} a \left (\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{3 b}\right )+\frac {1}{6} x^4 \left (a+b x^3\right )^{2/3}\right )}{9 b}+\frac {B x^4 \left (a+b x^3\right )^{5/3}}{9 b}\)

Input:

Int[x^3*(a + b*x^3)^(2/3)*(A + B*x^3),x]
 

Output:

(B*x^4*(a + b*x^3)^(5/3))/(9*b) + ((9*A*b - 4*a*B)*((x^4*(a + b*x^3)^(2/3) 
)/6 + (a*((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))/(9*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 811

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

Time = 1.41 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(-\frac {-6 a \left (\frac {B \,x^{3}}{3}+A \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} x \,b^{\frac {4}{3}}-9 \left (\frac {2 B \,x^{3}}{3}+A \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{4} b^{\frac {7}{3}}+a^{2} \left (\frac {8 \left (b \,x^{3}+a \right )^{\frac {2}{3}} x B \,b^{\frac {1}{3}}}{3}+\left (A b -\frac {4 B a}{9}\right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )\right )}{54 b^{\frac {7}{3}}}\)	\(180\)

Input:

int(x^3*(b*x^3+a)^(2/3)*(B*x^3+A),x,method=_RETURNVERBOSE)
 

Output:

-1/54*(-6*a*(1/3*B*x^3+A)*(b*x^3+a)^(2/3)*x*b^(4/3)-9*(2/3*B*x^3+A)*(b*x^3 
+a)^(2/3)*x^4*b^(7/3)+a^2*(8/3*(b*x^3+a)^(2/3)*x*B*b^(1/3)+(A*b-4/9*B*a)*( 
-2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)+ln( 
(b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*ln((-b^(1/3 
)*x+(b*x^3+a)^(1/3))/x))))/b^(7/3)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.71 \[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\left [-\frac {3 \, \sqrt {\frac {1}{3}} {\left (4 \, B a^{3} b - 9 \, A a^{2} 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 (4 \, B a^{3} - 9 \, A a^{2} 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 (4 \, B a^{3} - 9 \, A a^{2} 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 (18 \, B b^{3} x^{7} + 3 \, {\left (2 \, B a b^{2} + 9 \, A b^{3}\right )} x^{4} - 2 \, {\left (4 \, B a^{2} b - 9 \, A a b^{2}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{486 \, b^{3}}, -\frac {2 \, {\left (4 \, B a^{3} - 9 \, A a^{2} 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 (4 \, B a^{3} - 9 \, A a^{2} 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 (4 \, B a^{3} b - 9 \, A a^{2} 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 (18 \, B b^{3} x^{7} + 3 \, {\left (2 \, B a b^{2} + 9 \, A b^{3}\right )} x^{4} - 2 \, {\left (4 \, B a^{2} b - 9 \, A a b^{2}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{486 \, b^{3}}\right ] \] Input:

integrate(x^3*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="fricas")
 

Output:

[-1/486*(3*sqrt(1/3)*(4*B*a^3*b - 9*A*a^2*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*(4*B*a^3 - 9*A*a^2*b)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - 
(4*B*a^3 - 9*A*a^2*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*(18*B*b^3*x^7 + 3*(2*B*a*b^2 + 9*A*b^3)*x 
^4 - 2*(4*B*a^2*b - 9*A*a*b^2)*x)*(b*x^3 + a)^(2/3))/b^3, -1/486*(2*(4*B*a 
^3 - 9*A*a^2*b)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (4*B*a^3 
 - 9*A*a^2*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)*(4*B*a^3*b - 9*A*a^2*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*(18*B*b^3*x 
^7 + 3*(2*B*a*b^2 + 9*A*b^3)*x^4 - 2*(4*B*a^2*b - 9*A*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 = 5.62 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.47 \[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {A a^{\frac {2}{3}} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {B a^{\frac {2}{3}} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate(x**3*(b*x**3+a)**(2/3)*(B*x**3+A),x)
 

Output:

A*a**(2/3)*x**4*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), b*x**3*exp_polar(I*p 
i)/a)/(3*gamma(7/3)) + B*a**(2/3)*x**7*gamma(7/3)*hyper((-2/3, 7/3), (10/3 
,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (147) = 294\).

Time = 0.14 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.30 \[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {1}{54} \, {\left (\frac {2 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} + \frac {3 \, {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}}\right )} A - \frac {1}{243} \, {\left (\frac {4 \, \sqrt {3} a^{3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} - \frac {2 \, a^{3} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {7}{3}}} + \frac {4 \, a^{3} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}} + \frac {3 \, {\left (\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} b^{2}}{x^{2}} + \frac {11 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{3} b}{x^{5}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a^{3}}{x^{8}}\right )}}{b^{5} - \frac {3 \, {\left (b x^{3} + a\right )} b^{4}}{x^{3}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b^{3}}{x^{6}} - \frac {{\left (b x^{3} + a\right )}^{3} b^{2}}{x^{9}}}\right )} B \] Input:

integrate(x^3*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="maxima")
 

Output:

1/54*(2*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b 
^(1/3))/b^(4/3) - 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^(4/3) + 2*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) 
 + 3*((b*x^3 + a)^(2/3)*a^2*b/x^2 + 2*(b*x^3 + a)^(5/3)*a^2/x^5)/(b^3 - 2* 
(b*x^3 + a)*b^2/x^3 + (b*x^3 + a)^2*b/x^6))*A - 1/243*(4*sqrt(3)*a^3*arcta 
n(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 2*a^3*l 
og(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^3*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7/3) + 3*(2*(b*x^3 + a)^(2/ 
3)*a^3*b^2/x^2 + 11*(b*x^3 + a)^(5/3)*a^3*b/x^5 - 4*(b*x^3 + a)^(8/3)*a^3/ 
x^8)/(b^5 - 3*(b*x^3 + a)*b^4/x^3 + 3*(b*x^3 + a)^2*b^3/x^6 - (b*x^3 + a)^ 
3*b^2/x^9))*B
 

Giac [F]

\[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{3} \,d x } \] Input:

integrate(x^3*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*(b*x^3 + a)^(2/3)*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int x^3\,\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{2/3} \,d x \] Input:

int(x^3*(A + B*x^3)*(a + b*x^3)^(2/3),x)
 

Output:

int(x^3*(A + B*x^3)*(a + b*x^3)^(2/3), x)
 

Reduce [F]

\[ \int x^3 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {10 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} x +33 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{4}+18 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{7}-10 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{3}}{162 b} \] Input:

int(x^3*(b*x^3+a)^(2/3)*(B*x^3+A),x)
 

Output:

(10*(a + b*x**3)**(2/3)*a**2*x + 33*(a + b*x**3)**(2/3)*a*b*x**4 + 18*(a + 
 b*x**3)**(2/3)*b**2*x**7 - 10*int((a + b*x**3)**(2/3)/(a + b*x**3),x)*a** 
3)/(162*b)