\(\int \frac {x^4 (A+B x^3)}{(a+b x^3)^{2/3}} \, dx\) [349]

Optimal result

Integrand size = 22, antiderivative size = 146 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {(6 A b-5 a B) x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {B x^5 \sqrt [3]{a+b x^3}}{6 b}+\frac {a (6 A b-5 a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}+\frac {a (6 A b-5 a B) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{8/3}} \] Output:

1/18*(6*A*b-5*B*a)*x^2*(b*x^3+a)^(1/3)/b^2+1/6*B*x^5*(b*x^3+a)^(1/3)/b+1/2 
7*a*(6*A*b-5*B*a)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1 
/2)/b^(8/3)+1/18*a*(6*A*b-5*B*a)*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(8/3)
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.26 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {3 b^{2/3} x^2 \sqrt [3]{a+b x^3} \left (6 A b-5 a B+3 b B x^3\right )-2 \sqrt {3} a (-6 A b+5 a B) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 a (-6 A b+5 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+a (-6 A b+5 a B) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{8/3}} \] Input:

Integrate[(x^4*(A + B*x^3))/(a + b*x^3)^(2/3),x]
 

Output:

(3*b^(2/3)*x^2*(a + b*x^3)^(1/3)*(6*A*b - 5*a*B + 3*b*B*x^3) - 2*Sqrt[3]*a 
*(-6*A*b + 5*a*B)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1 
/3))] - 2*a*(-6*A*b + 5*a*B)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + a*(-6 
*A*b + 5*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^(8/3))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {959, 843, 853}

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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 853

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

Input:

Int[(x^4*(A + B*x^3))/(a + b*x^3)^(2/3),x]
 

Output:

(B*x^5*(a + b*x^3)^(1/3))/(6*b) + ((6*A*b - 5*a*B)*((x^2*(a + b*x^3)^(1/3) 
)/(3*b) - (2*a*(-(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(S 
qrt[3]*b^(2/3))) - Log[b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*b^(2/3))))/(3*b)) 
)/(6*b)
 

Defintions of rubi rules used

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 853

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp 
[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, 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. \(272\) vs. \(2(119)=238\).

Time = 1.26 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.87

method	result	size
pseudoelliptic	\(\frac {9 B \,b^{\frac {5}{3}} x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+18 A \,b^{\frac {5}{3}} x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}}-15 B a \,x^{2} b^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}-12 A \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}\, a b +10 B \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}\, a^{2}+12 A \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a b -6 A \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a b -10 B \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2}+5 B \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2}}{54 b^{\frac {8}{3}}}\)	\(273\)

Input:

int(x^4*(B*x^3+A)/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)
 

Output:

1/54*(9*B*b^(5/3)*x^5*(b*x^3+a)^(1/3)+18*A*b^(5/3)*x^2*(b*x^3+a)^(1/3)-15* 
B*a*x^2*b^(2/3)*(b*x^3+a)^(1/3)-12*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+10*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+12*A*ln((-b^(1/3)*x+(b*x^3+a)^(1/3)) 
/x)*a*b-6*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-10*B*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a^2+5*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^(8/3)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.49 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {6 \, \sqrt {\frac {1}{3}} {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} {\left (b^{2}\right )}^{\frac {1}{6}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x}\right ) - 2 \, {\left (5 \, B a^{2} - 6 \, A a b\right )} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + {\left (5 \, B a^{2} - 6 \, A a b\right )} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (3 \, B b^{3} x^{5} - {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{4}} \] Input:

integrate(x^4*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="fricas")
 

Output:

1/54*(6*sqrt(1/3)*(5*B*a^2*b - 6*A*a*b^2)*(b^2)^(1/6)*arctan(sqrt(1/3)*((b 
^2)^(1/3)*b*x + 2*(b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x)) - 2* 
(5*B*a^2 - 6*A*a*b)*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b) 
/x) + (5*B*a^2 - 6*A*a*b)*(b^2)^(2/3)*log(((b^2)^(1/3)*b*x^2 + (b*x^3 + a) 
^(1/3)*(b^2)^(2/3)*x + (b*x^3 + a)^(2/3)*b)/x^2) + 3*(3*B*b^3*x^5 - (5*B*a 
*b^2 - 6*A*b^3)*x^2)*(b*x^3 + a)^(1/3))/b^4
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.55 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {A x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {8}{3}\right )} + \frac {B x^{8} \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {11}{3}\right )} \] Input:

integrate(x**4*(B*x**3+A)/(b*x**3+a)**(2/3),x)
 

Output:

A*x**5*gamma(5/3)*hyper((2/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*a 
**(2/3)*gamma(8/3)) + B*x**8*gamma(8/3)*hyper((2/3, 8/3), (11/3,), b*x**3* 
exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(11/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (119) = 238\).

Time = 0.12 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.24 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {1}{54} \, B {\left (\frac {10 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {8}{3}}} + \frac {5 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {8}{3}}} - \frac {10 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {8}{3}}} + \frac {3 \, {\left (\frac {8 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b}{x} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}}{x^{4}}\right )}}{b^{4} - \frac {2 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}}}\right )} - \frac {1}{9} \, A {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {5}{3}}} + \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {5}{3}}} - \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {5}{3}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{{\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x}\right )} \] Input:

integrate(x^4*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="maxima")
 

Output:

1/54*B*(10*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x 
)/b^(1/3))/b^(8/3) + 5*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^(8/3) - 10*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b 
^(8/3) + 3*(8*(b*x^3 + a)^(1/3)*a^2*b/x - 5*(b*x^3 + a)^(4/3)*a^2/x^4)/(b^ 
4 - 2*(b*x^3 + a)*b^3/x^3 + (b*x^3 + a)^2*b^2/x^6)) - 1/9*A*(2*sqrt(3)*a*a 
rctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(5/3) + a*l 
og(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(5/3) 
- 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(5/3) + 3*(b*x^3 + a)^(1/3)*a/ 
((b^2 - (b*x^3 + a)*b/x^3)*x))
 

Giac [F]

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

integrate(x^4*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*x^4/(b*x^3 + a)^(2/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {x^4\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \] Input:

int((x^4*(A + B*x^3))/(a + b*x^3)^(2/3),x)
 

Output:

int((x^4*(A + B*x^3))/(a + b*x^3)^(2/3), x)
 

Reduce [F]

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

int(x^4*(B*x^3+A)/(b*x^3+a)^(2/3),x)
 

Output:

int(x**7/(a + b*x**3)**(2/3),x)*b + int(x**4/(a + b*x**3)**(2/3),x)*a