3.4.35 \(\int \frac {x^3}{(a+b x^3)^2} \, dx\) [335]

3.4.35.1 Optimal result
3.4.35.2 Mathematica [A] (verified)
3.4.35.3 Rubi [A] (verified)
3.4.35.4 Maple [C] (verified)
3.4.35.5 Fricas [A] (verification not implemented)
3.4.35.6 Sympy [A] (verification not implemented)
3.4.35.7 Maxima [A] (verification not implemented)
3.4.35.8 Giac [A] (verification not implemented)
3.4.35.9 Mupad [B] (verification not implemented)

3.4.35.1 Optimal result

Integrand size = 13, antiderivative size = 134 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=-\frac {x}{3 b \left (a+b x^3\right )}-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{4/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{2/3} b^{4/3}} \]

output
-1/3*x/b/(b*x^3+a)+1/9*ln(a^(1/3)+b^(1/3)*x)/a^(2/3)/b^(4/3)-1/18*ln(a^(2/ 
3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(2/3)/b^(4/3)-1/9*arctan(1/3*(a^(1/3)- 
2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(2/3)/b^(4/3)*3^(1/2)
 
3.4.35.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=\frac {-\frac {6 \sqrt [3]{b} x}{a+b x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{2/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{2/3}}}{18 b^{4/3}} \]

input
Integrate[x^3/(a + b*x^3)^2,x]
 
output
((-6*b^(1/3)*x)/(a + b*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3) 
)/Sqrt[3]])/a^(2/3) + (2*Log[a^(1/3) + b^(1/3)*x])/a^(2/3) - Log[a^(2/3) - 
 a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/a^(2/3))/(18*b^(4/3))
 
3.4.35.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {817, 750, 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.

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

\(\Big \downarrow \) 817

\(\displaystyle \frac {\int \frac {1}{b x^3+a}dx}{3 b}-\frac {x}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 750

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

input
Int[x^3/(a + b*x^3)^2,x]
 
output
-1/3*x/(b*(a + b*x^3)) + (Log[a^(1/3) + b^(1/3)*x]/(3*a^(2/3)*b^(1/3)) + ( 
-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) - Log[a^( 
2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(2/3)))/(3*b)
 

3.4.35.3.1 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 750
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*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 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 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]
 
3.4.35.4 Maple [C] (verified)

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

Time = 3.65 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.32

method result size
risch \(-\frac {x}{3 b \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{2}}\) \(43\)
default \(-\frac {x}{3 b \left (b \,x^{3}+a \right )}+\frac {\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{3 b}\) \(112\)

input
int(x^3/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
 
output
-1/3*x/b/(b*x^3+a)+1/9/b^2*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
 
3.4.35.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.92 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=\left [-\frac {6 \, a^{2} b x - 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) + {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}, -\frac {6 \, a^{2} b x - 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) + {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}\right ] \]

input
integrate(x^3/(b*x^3+a)^2,x, algorithm="fricas")
 
output
[-1/18*(6*a^2*b*x - 3*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt(-(a^2*b)^(1/3)/b) 
*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x - a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^ 
2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) + (b* 
x^3 + a)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) - 
2*(b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)))/(a^2*b^3*x^3 + a^3 
*b^2), -1/18*(6*a^2*b*x - 6*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt((a^2*b)^(1/ 
3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^ 
(1/3)/b)/a^2) + (b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + 
(a^2*b)^(1/3)*a) - 2*(b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3))) 
/(a^2*b^3*x^3 + a^3*b^2)]
 
3.4.35.6 Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.29 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=- \frac {x}{3 a b + 3 b^{2} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{2} b^{4} - 1, \left ( t \mapsto t \log {\left (9 t a b + x \right )} \right )\right )} \]

input
integrate(x**3/(b*x**3+a)**2,x)
 
output
-x/(3*a*b + 3*b**2*x**3) + RootSum(729*_t**3*a**2*b**4 - 1, Lambda(_t, _t* 
log(9*_t*a*b + x)))
 
3.4.35.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=-\frac {x}{3 \, {\left (b^{2} x^{3} + a b\right )}} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {\log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

input
integrate(x^3/(b*x^3+a)^2,x, algorithm="maxima")
 
output
-1/3*x/(b^2*x^3 + a*b) + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3) 
)/(a/b)^(1/3))/(b^2*(a/b)^(2/3)) - 1/18*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2 
/3))/(b^2*(a/b)^(2/3)) + 1/9*log(x + (a/b)^(1/3))/(b^2*(a/b)^(2/3))
 
3.4.35.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.97 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=-\frac {\left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b} - \frac {x}{3 \, {\left (b x^{3} + a\right )} b} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{2}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b^{2}} \]

input
integrate(x^3/(b*x^3+a)^2,x, algorithm="giac")
 
output
-1/9*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b) - 1/3*x/((b*x^3 + a)*b) 
 + 1/9*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/ 
b)^(1/3))/(a*b^2) + 1/18*(-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^ 
(2/3))/(a*b^2)
 
3.4.35.9 Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {x^3}{\left (a+b x^3\right )^2} \, dx=\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )}{9\,a^{2/3}\,b^{4/3}}-\frac {x}{3\,b\,\left (b\,x^3+a\right )}+\frac {\ln \left (b\,x+\frac {a^{1/3}\,b^{2/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{2/3}\,b^{4/3}}-\frac {\ln \left (b\,x-\frac {a^{1/3}\,b^{2/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{2/3}\,b^{4/3}} \]

input
int(x^3/(a + b*x^3)^2,x)
 
output
log(b^(1/3)*x + a^(1/3))/(9*a^(2/3)*b^(4/3)) - x/(3*b*(a + b*x^3)) + (log( 
b*x + (a^(1/3)*b^(2/3)*(3^(1/2)*1i - 1))/2)*(3^(1/2)*1i - 1))/(18*a^(2/3)* 
b^(4/3)) - (log(b*x - (a^(1/3)*b^(2/3)*(3^(1/2)*1i + 1))/2)*(3^(1/2)*1i + 
1))/(18*a^(2/3)*b^(4/3))