\(\int (a+b x^3)^{2/3} (c+d x^3) \, dx\) [109]

Optimal result

Integrand size = 19, antiderivative size = 141 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx=\frac {(6 b c-a d) x \left (a+b x^3\right )^{2/3}}{18 b}+\frac {d x \left (a+b x^3\right )^{5/3}}{6 b}+\frac {a (6 b c-a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{4/3}}-\frac {a (6 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{4/3}} \] Output:

1/18*(-a*d+6*b*c)*x*(b*x^3+a)^(2/3)/b+1/6*d*x*(b*x^3+a)^(5/3)/b+1/27*a*(-a 
*d+6*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(4 
/3)-1/18*a*(-a*d+6*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.28 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx=\frac {3 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3} \left (6 b c+2 a d+3 b d x^3\right )-2 \sqrt {3} a (-6 b c+a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+2 a (-6 b c+a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )-a (-6 b c+a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{4/3}} \] Input:

Integrate[(a + b*x^3)^(2/3)*(c + d*x^3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {913, 748, 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.

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

\(\Big \downarrow \) 913

\(\displaystyle \frac {(6 b c-a d) \int \left (b x^3+a\right )^{2/3}dx}{6 b}+\frac {d x \left (a+b x^3\right )^{5/3}}{6 b}\)

\(\Big \downarrow \) 748

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

\(\Big \downarrow \) 769

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

Input:

Int[(a + b*x^3)^(2/3)*(c + d*x^3),x]
 

Output:

(d*x*(a + b*x^3)^(5/3))/(6*b) + ((6*b*c - a*d)*((x*(a + b*x^3)^(2/3))/3 + 
(2*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))/(6*b)
 

Defintions of rubi rules used

rule 748

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p 
+ 1)), x] + Simp[a*n*(p/(n*p + 1))   Int[(a + b*x^n)^(p - 1), x], x] /; Fre 
eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat 
or[p + 1/n], Denominator[p]])
 

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 913

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

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {3 \left (\frac {d \,x^{3}}{2}+c \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} x \,b^{\frac {4}{3}}+a \left (d x \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {1}{3}}-\frac {\left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) \left (a d -6 b c \right )}{6}\right )}{9 b^{\frac {4}{3}}}\)	\(152\)

Input:

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

Output:

1/9/b^(4/3)*(3*(1/2*d*x^3+c)*(b*x^3+a)^(2/3)*x*b^(4/3)+a*(d*x*(b*x^3+a)^(2 
/3)*b^(1/3)-1/6*(-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))*(a*d-6*b*c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 424, normalized size of antiderivative = 3.01 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx=\left [-\frac {3 \, \sqrt {\frac {1}{3}} {\left (6 \, a b^{2} c - a^{2} b d\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) + 2 \, {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, b^{2} d x^{4} + 2 \, {\left (3 \, b^{2} c + a b d\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{2}}, -\frac {2 \, {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - {\left (6 \, a b c - a^{2} d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {6 \, \sqrt {\frac {1}{3}} {\left (6 \, a b^{2} c - a^{2} b d\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}} - 3 \, {\left (3 \, b^{2} d x^{4} + 2 \, {\left (3 \, b^{2} c + a b d\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b^{2}}\right ] \] Input:

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

Output:

[-1/54*(3*sqrt(1/3)*(6*a*b^2*c - a^2*b*d)*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*(6* 
a*b*c - a^2*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (6*a*b*c 
- a^2*d)*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*d*x^4 + 2*(3*b^2*c + a*b*d)*x)*(b*x^3 + a)^(2/3 
))/b^2, -1/54*(2*(6*a*b*c - a^2*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^( 
1/3))/x) - (6*a*b*c - a^2*d)*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)*(6*a*b^2*c - a^2*b*d)*ar 
ctan(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3) - 3* 
(3*b^2*d*x^4 + 2*(3*b^2*c + a*b*d)*x)*(b*x^3 + a)^(2/3))/b^2]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.95 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.58 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx=\frac {a^{\frac {2}{3}} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{\frac {2}{3}} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:

integrate((b*x**3+a)**(2/3)*(d*x**3+c),x)
 

Output:

a**(2/3)*c*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/ 
a)/(3*gamma(4/3)) + a**(2/3)*d*x**4*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), 
b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (114) = 228\).

Time = 0.11 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.28 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx=-\frac {1}{9} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x^{2}}\right )} c + \frac {1}{54} \, {\left (\frac {2 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} + \frac {3 \, {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}\right )}}{b^{3} - \frac {2 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b}{x^{6}}}\right )} d \] Input:

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

Output:

-1/9*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^( 
1/3))/b^(1/3) - a*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^ 
(2/3)/x^2)/b^(1/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3) + 3*( 
b*x^3 + a)^(2/3)*a/((b - (b*x^3 + a)/x^3)*x^2))*c + 1/54*(2*sqrt(3)*a^2*ar 
ctan(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))*d
 

Giac [F]

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

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

Output:

integrate((b*x^3 + a)^(2/3)*(d*x^3 + c), x)
 

Mupad [F(-1)]

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

int((a + b*x^3)^(2/3)*(c + d*x^3),x)
 

Output:

int((a + b*x^3)^(2/3)*(c + d*x^3), x)
 

Reduce [F]

\[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a d x +6 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b c x +3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b d \,x^{4}-2 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{2} d +12 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a b c}{18 b} \] Input:

int((b*x^3+a)^(2/3)*(d*x^3+c),x)
 

Output:

(2*(a + b*x**3)**(2/3)*a*d*x + 6*(a + b*x**3)**(2/3)*b*c*x + 3*(a + b*x**3 
)**(2/3)*b*d*x**4 - 2*int((a + b*x**3)**(2/3)/(a + b*x**3),x)*a**2*d + 12* 
int((a + b*x**3)**(2/3)/(a + b*x**3),x)*a*b*c)/(18*b)