\(\int \frac {\sqrt [3]{a+b x^3} (A+B x^3)}{x^2} \, dx\) [297]

Optimal result

Integrand size = 22, antiderivative size = 139 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^2} \, dx=\frac {(3 A b+a B) x^2 \sqrt [3]{a+b x^3}}{3 a}-\frac {A \left (a+b x^3\right )^{4/3}}{a x}-\frac {(3 A b+a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}-\frac {(3 A b+a B) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^2} \, dx=\frac {6 b^{2/3} \sqrt [3]{a+b x^3} \left (-3 A+B x^3\right )-2 \sqrt {3} (3 A b+a B) x \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 (3 A b+a B) x \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+(3 A b+a B) x \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 b^{2/3} x} \] Input:

Integrate[((a + b*x^3)^(1/3)*(A + B*x^3))/x^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 129, 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, 811, 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.

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 853

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

Input:

Int[((a + b*x^3)^(1/3)*(A + B*x^3))/x^2,x]
 

Output:

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

Defintions of rubi rules used

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 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 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 = 1.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {-6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {2}{3}} \left (-\frac {B \,x^{3}}{3}+A \right )+\left (A b +\frac {B a}{3}\right ) \left (2 \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) x}{6 b^{\frac {2}{3}} x}\)	\(132\)

Input:

int((b*x^3+a)^(1/3)*(B*x^3+A)/x^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

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

integrate((b*x^3+a)^(1/3)*(B*x^3+A)/x^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.54 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^2} \, dx=\frac {A \sqrt [3]{a} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} + \frac {B \sqrt [3]{a} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \] Input:

integrate((b*x**3+a)**(1/3)*(B*x**3+A)/x**2,x)
 

Output:

A*a**(1/3)*gamma(-1/3)*hyper((-1/3, -1/3), (2/3,), b*x**3*exp_polar(I*pi)/ 
a)/(3*x*gamma(2/3)) + B*a**(1/3)*x**2*gamma(2/3)*hyper((-1/3, 2/3), (5/3,) 
, b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (114) = 228\).

Time = 0.13 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.82 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^2} \, dx=\frac {1}{6} \, {\left (2 \, \sqrt {3} b^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right ) + b^{\frac {1}{3}} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 2 \, b^{\frac {1}{3}} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )} A + \frac {1}{18} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}} + \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {2}{3}}} - \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {2}{3}}} - \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{{\left (b - \frac {b x^{3} + a}{x^{3}}\right )} x}\right )} B \] Input:

integrate((b*x^3+a)^(1/3)*(B*x^3+A)/x^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x^3+a)^(1/3)*(B*x^3+A)/x^2,x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*(b*x^3 + a)^(1/3)/x^2, x)
 

Mupad [F(-1)]

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

int(((A + B*x^3)*(a + b*x^3)^(1/3))/x^2,x)
 

Output:

int(((A + B*x^3)*(a + b*x^3)^(1/3))/x^2, x)
 

Reduce [F]

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

int((b*x^3+a)^(1/3)*(B*x^3+A)/x^2,x)
 

Output:

( - 3*(a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3 + 4*int(((a + b*x 
**3)**(1/3)*x)/(a + b*x**3),x)*a*b*x)/(3*x)