\(\int \frac {x}{(a+\frac {b}{x^3})^2} \, dx\) [447]

Optimal result

Integrand size = 11, antiderivative size = 147 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {x^2}{2 a^2}+\frac {b x^2}{3 a^2 \left (b+a x^3\right )}+\frac {5 b^{2/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{8/3}}+\frac {5 b^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{8/3}}-\frac {5 b^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{8/3}} \] Output:

1/2*x^2/a^2+1/3*b*x^2/a^2/(a*x^3+b)+5/9*b^(2/3)*arctan(1/3*(b^(1/3)-2*a^(1 
/3)*x)*3^(1/2)/b^(1/3))*3^(1/2)/a^(8/3)+5/9*b^(2/3)*ln(b^(1/3)+a^(1/3)*x)/ 
a^(8/3)-5/18*b^(2/3)*ln(b^(2/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/a^(8/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {9 a^{2/3} x^2+\frac {6 a^{2/3} b x^2}{b+a x^3}+10 \sqrt {3} b^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )+10 b^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )-5 b^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{8/3}} \] Input:

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

Output:

(9*a^(2/3)*x^2 + (6*a^(2/3)*b*x^2)/(b + a*x^3) + 10*Sqrt[3]*b^(2/3)*ArcTan 
[(1 - (2*a^(1/3)*x)/b^(1/3))/Sqrt[3]] + 10*b^(2/3)*Log[b^(1/3) + a^(1/3)*x 
] - 5*b^(2/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2])/(18*a^(8/3))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {795, 817, 843, 821, 16, 1142, 25, 27, 1082, 217, 1103}

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.

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

\(\Big \downarrow \) 795

\(\displaystyle \int \frac {x^7}{\left (a x^3+b\right )^2}dx\)

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 795

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* 
(b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
 

rule 817

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*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, 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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.37

method	result	size
risch	\(\frac {x^{2}}{2 a^{2}}+\frac {b \,x^{2}}{3 a^{2} \left (a \,x^{3}+b \right )}-\frac {5 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{3}+b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{9 a^{3}}\)	\(55\)
default	\(\frac {x^{2}}{2 a^{2}}-\frac {b \left (-\frac {x^{2}}{3 \left (a \,x^{3}+b \right )}-\frac {5 \ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {5 \ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{a^{2}}\)	\(120\)

Input:

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

Output:

1/2/a^2*x^2+1/3*b*x^2/a^2/(a*x^3+b)-5/9/a^3*b*sum(1/_R*ln(x-_R),_R=RootOf( 
_Z^3*a+b))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.11 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {9 \, a x^{5} + 15 \, b x^{2} - 10 \, \sqrt {3} {\left (a x^{3} + b\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} - \sqrt {3} b}{3 \, b}\right ) - 5 \, {\left (a x^{3} + b\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} + b \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 10 \, {\left (a x^{3} + b\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{3} x^{3} + a^{2} b\right )}} \] Input:

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

Output:

1/18*(9*a*x^5 + 15*b*x^2 - 10*sqrt(3)*(a*x^3 + b)*(b^2/a^2)^(1/3)*arctan(1 
/3*(2*sqrt(3)*a*x*(b^2/a^2)^(1/3) - sqrt(3)*b)/b) - 5*(a*x^3 + b)*(b^2/a^2 
)^(1/3)*log(b*x^2 - a*x*(b^2/a^2)^(2/3) + b*(b^2/a^2)^(1/3)) + 10*(a*x^3 + 
 b)*(b^2/a^2)^(1/3)*log(b*x + a*(b^2/a^2)^(2/3)))/(a^3*x^3 + a^2*b)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.39 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {b x^{2}}{3 a^{3} x^{3} + 3 a^{2} b} + \operatorname {RootSum} {\left (729 t^{3} a^{8} - 125 b^{2}, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a^{5}}{25 b} + x \right )} \right )\right )} + \frac {x^{2}}{2 a^{2}} \] Input:

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

Output:

b*x**2/(3*a**3*x**3 + 3*a**2*b) + RootSum(729*_t**3*a**8 - 125*b**2, Lambd 
a(_t, _t*log(81*_t**2*a**5/(25*b) + x))) + x**2/(2*a**2)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {b x^{2}}{3 \, {\left (a^{3} x^{3} + a^{2} b\right )}} + \frac {x^{2}}{2 \, a^{2}} - \frac {5 \, \sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} \left (\frac {b}{a}\right )^{\frac {1}{3}}} - \frac {5 \, b \log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{3} \left (\frac {b}{a}\right )^{\frac {1}{3}}} + \frac {5 \, b \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a^{3} \left (\frac {b}{a}\right )^{\frac {1}{3}}} \] Input:

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

Output:

1/3*b*x^2/(a^3*x^3 + a^2*b) + 1/2*x^2/a^2 - 5/9*sqrt(3)*b*arctan(1/3*sqrt( 
3)*(2*x - (b/a)^(1/3))/(b/a)^(1/3))/(a^3*(b/a)^(1/3)) - 5/18*b*log(x^2 - x 
*(b/a)^(1/3) + (b/a)^(2/3))/(a^3*(b/a)^(1/3)) + 5/9*b*log(x + (b/a)^(1/3)) 
/(a^3*(b/a)^(1/3))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {x^{2}}{2 \, a^{2}} + \frac {b x^{2}}{3 \, {\left (a x^{3} + b\right )} a^{2}} + \frac {5 \, \left (-\frac {b}{a}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2}} + \frac {5 \, \sqrt {3} \left (-a^{2} b\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4}} - \frac {5 \, \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{4}} \] Input:

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

Output:

1/2*x^2/a^2 + 1/3*b*x^2/((a*x^3 + b)*a^2) + 5/9*(-b/a)^(2/3)*log(abs(x - ( 
-b/a)^(1/3)))/a^2 + 5/9*sqrt(3)*(-a^2*b)^(2/3)*arctan(1/3*sqrt(3)*(2*x + ( 
-b/a)^(1/3))/(-b/a)^(1/3))/a^4 - 5/18*(-a^2*b)^(2/3)*log(x^2 + x*(-b/a)^(1 
/3) + (-b/a)^(2/3))/a^4
 

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {x^2}{2\,a^2}+\frac {5\,b^{2/3}\,\ln \left (a^{1/3}\,x+b^{1/3}\right )}{9\,a^{8/3}}+\frac {b\,x^2}{3\,\left (a^3\,x^3+b\,a^2\right )}+\frac {5\,b^{2/3}\,\ln \left (\frac {25\,b^2\,x}{9\,a^3}+\frac {25\,b^{7/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{9\,a^{10/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}}-\frac {5\,b^{2/3}\,\ln \left (\frac {25\,b^2\,x}{9\,a^3}+\frac {25\,b^{7/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{9\,a^{10/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}} \] Input:

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

Output:

x^2/(2*a^2) + (5*b^(2/3)*log(a^(1/3)*x + b^(1/3)))/(9*a^(8/3)) + (b*x^2)/( 
3*(a^2*b + a^3*x^3)) + (5*b^(2/3)*log((25*b^2*x)/(9*a^3) + (25*b^(7/3)*((3 
^(1/2)*1i)/2 - 1/2)^2)/(9*a^(10/3)))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(8/3)) - 
 (5*b^(2/3)*log((25*b^2*x)/(9*a^3) + (25*b^(7/3)*((3^(1/2)*1i)/2 + 1/2)^2) 
/(9*a^(10/3)))*((3^(1/2)*1i)/2 + 1/2))/(9*a^(8/3))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.25 \[ \int \frac {x}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {-10 \sqrt {3}\, \mathit {atan} \left (\frac {2 a^{\frac {1}{3}} x -b^{\frac {1}{3}}}{b^{\frac {1}{3}} \sqrt {3}}\right ) a b \,x^{3}-10 \sqrt {3}\, \mathit {atan} \left (\frac {2 a^{\frac {1}{3}} x -b^{\frac {1}{3}}}{b^{\frac {1}{3}} \sqrt {3}}\right ) b^{2}+9 b^{\frac {1}{3}} a^{\frac {5}{3}} x^{5}+15 b^{\frac {4}{3}} a^{\frac {2}{3}} x^{2}-5 \,\mathrm {log}\left (a^{\frac {2}{3}} x^{2}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}}\right ) a b \,x^{3}-5 \,\mathrm {log}\left (a^{\frac {2}{3}} x^{2}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}}\right ) b^{2}+10 \,\mathrm {log}\left (a^{\frac {1}{3}} x +b^{\frac {1}{3}}\right ) a b \,x^{3}+10 \,\mathrm {log}\left (a^{\frac {1}{3}} x +b^{\frac {1}{3}}\right ) b^{2}}{18 b^{\frac {1}{3}} a^{\frac {8}{3}} \left (a \,x^{3}+b \right )} \] Input:

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

Output:

( - 10*sqrt(3)*atan((2*a**(1/3)*x - b**(1/3))/(b**(1/3)*sqrt(3)))*a*b*x**3 
 - 10*sqrt(3)*atan((2*a**(1/3)*x - b**(1/3))/(b**(1/3)*sqrt(3)))*b**2 + 9* 
b**(1/3)*a**(2/3)*a*x**5 + 15*b**(1/3)*a**(2/3)*b*x**2 - 5*log(a**(2/3)*x* 
*2 - b**(1/3)*a**(1/3)*x + b**(2/3))*a*b*x**3 - 5*log(a**(2/3)*x**2 - b**( 
1/3)*a**(1/3)*x + b**(2/3))*b**2 + 10*log(a**(1/3)*x + b**(1/3))*a*b*x**3 
+ 10*log(a**(1/3)*x + b**(1/3))*b**2)/(18*b**(1/3)*a**(2/3)*a**2*(a*x**3 + 
 b))