\(\int \frac {(a+b x^3)^{2/3} (A+B x^3)}{x^3} \, dx\) [315]

Optimal result

Integrand size = 22, antiderivative size = 141 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^3} \, dx=\frac {(3 A b+2 a B) x \left (a+b x^3\right )^{2/3}}{6 a}-\frac {A \left (a+b x^3\right )^{5/3}}{2 a x^2}+\frac {(3 A b+2 a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} \sqrt [3]{b}}-\frac {(3 A b+2 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{b}} \] Output:

1/6*(3*A*b+2*B*a)*x*(b*x^3+a)^(2/3)/a-1/2*A*(b*x^3+a)^(5/3)/a/x^2+1/9*(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^(1 
/3)-1/6*(3*A*b+2*B*a)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(1/3)
 

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^3} \, dx=\frac {1}{18} \left (\frac {3 \left (a+b x^3\right )^{2/3} \left (-3 A+2 B x^3\right )}{x^2}+\frac {2 \sqrt {3} (3 A b+2 a B) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b}}-\frac {2 (3 A b+2 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b}}+\frac {(3 A b+2 a B) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b}}\right ) \] Input:

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

Output:

((3*(a + b*x^3)^(2/3)*(-3*A + 2*B*x^3))/x^2 + (2*Sqrt[3]*(3*A*b + 2*a*B)*A 
rcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/b^(1/3) - (2 
*(3*A*b + 2*a*B)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/b^(1/3) + ((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) 
])/b^(1/3))/18
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 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.

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 748

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

\(\Big \downarrow \) 769

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

Input:

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

Output:

-1/2*(A*(a + b*x^3)^(5/3))/(a*x^2) + ((3*A*b + 2*a*B)*((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))/(2*a)
 

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 955

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(-\frac {\frac {3 b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-\frac {2 B \,x^{3}}{3}+A \right )}{2}+\left (\arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) \left (A b +\frac {2 B a}{3}\right ) x^{2}}{3 b^{\frac {1}{3}} x^{2}}\)	\(133\)

Input:

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

Output:

-1/3*(3/2*b^(1/3)*(b*x^3+a)^(2/3)*(-2/3*B*x^3+A)+(arctan(1/3*3^(1/2)*(2*(b 
*x^3+a)^(1/3)/b^(1/3)+x)/x)*3^(1/2)+ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-1/2 
*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*b+2/3 
*B*a)*x^2)/b^(1/3)/x^2
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.77 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.60 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^3} \, dx=\frac {A a^{\frac {2}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} + \frac {B a^{\frac {2}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \] Input:

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

Output:

A*a**(2/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), b*x**3*exp_polar(I*pi)/ 
a)/(3*x**2*gamma(1/3)) + B*a**(2/3)*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,) 
, b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (114) = 228\).

Time = 0.12 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.81 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^3} \, dx=-\frac {1}{6} \, {\left (2 \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right ) - b^{\frac {2}{3}} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 2 \, b^{\frac {2}{3}} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )} A - \frac {1}{9} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x^{2}}\right )} B \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^3} \, dx=\frac {-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a +2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}+10 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a b \,x^{2}}{6 x^{2}} \] Input:

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

Output:

( - 3*(a + b*x**3)**(2/3)*a + 2*(a + b*x**3)**(2/3)*b*x**3 + 10*int((a + b 
*x**3)**(2/3)/(a + b*x**3),x)*a*b*x**2)/(6*x**2)