\(\int \frac {1}{x \sqrt [3]{1-x^3} (1+x^3)} \, dx\) [824]

Optimal result

Integrand size = 22, antiderivative size = 137 \[ \int \frac {1}{x \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {\arctan \left (\frac {1+2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log (x)}{2}+\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}} \] Output:

1/3*arctan(1/3*(1+2*(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)-1/6*arctan(1/3*(1+2^( 
2/3)*(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)-1/2*ln(x)+1/12*ln(x^3+1)*2^( 
2/3)+1/2*ln(1-(-x^3+1)^(1/3))-1/4*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{12} \left (4 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )-2\ 2^{2/3} \sqrt {3} \arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )+4 \log \left (-1+\sqrt [3]{1-x^3}\right )-2\ 2^{2/3} \log \left (-2+2^{2/3} \sqrt [3]{1-x^3}\right )-2 \log \left (1+\sqrt [3]{1-x^3}+\left (1-x^3\right )^{2/3}\right )+2^{2/3} \log \left (2+2^{2/3} \sqrt [3]{1-x^3}+\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )\right ) \] Input:

Integrate[1/(x*(1 - x^3)^(1/3)*(1 + x^3)),x]
 

Output:

(4*Sqrt[3]*ArcTan[(1 + 2*(1 - x^3)^(1/3))/Sqrt[3]] - 2*2^(2/3)*Sqrt[3]*Arc 
Tan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]] + 4*Log[-1 + (1 - x^3)^(1/3)] - 
 2*2^(2/3)*Log[-2 + 2^(2/3)*(1 - x^3)^(1/3)] - 2*Log[1 + (1 - x^3)^(1/3) + 
 (1 - x^3)^(2/3)] + 2^(2/3)*Log[2 + 2^(2/3)*(1 - x^3)^(1/3) + 2^(1/3)*(1 - 
 x^3)^(2/3)])/12
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {948, 97, 67, 16, 1082, 217, 1083, 217}

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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 97

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

\(\Big \downarrow \) 67

\(\displaystyle \frac {1}{3} \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{1-x^3}}d\sqrt [3]{1-x^3}+\frac {3 \int \frac {1}{\sqrt [3]{2}-\sqrt [3]{1-x^3}}d\sqrt [3]{1-x^3}}{2 \sqrt [3]{2}}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{1-x^3}+1}d\sqrt [3]{1-x^3}-\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{2} \sqrt [3]{1-x^3}+2^{2/3}}d\sqrt [3]{1-x^3}-\frac {1}{2} \log \left (x^3\right )+\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{1-x^3}+1}d\sqrt [3]{1-x^3}-\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{2} \sqrt [3]{1-x^3}+2^{2/3}}d\sqrt [3]{1-x^3}-\frac {1}{2} \log \left (x^3\right )+\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{3} \left (\frac {3 \int \frac {1}{-x^6-3}d\left (2^{2/3} \sqrt [3]{1-x^3}+1\right )}{\sqrt [3]{2}}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{1-x^3}+1}d\sqrt [3]{1-x^3}-\frac {1}{2} \log \left (x^3\right )+\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\right )\)

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[1/(x*(1 - x^3)^(1/3)*(1 + x^3)),x]
 

Output:

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

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 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[p]
 

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 948

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

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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

Maple [A] (verified)

Time = 3.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {2}{3}} \ln \left (\left (-x^{3}+1\right )^{\frac {1}{3}}-2^{\frac {1}{3}}\right )}{6}+\frac {2^{\frac {2}{3}} \ln \left (\left (-x^{3}+1\right )^{\frac {2}{3}}+2^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )}{12}-\frac {\arctan \left (\frac {\left (1+2^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) 2^{\frac {2}{3}} \sqrt {3}}{6}+\frac {\ln \left (-1+\left (-x^{3}+1\right )^{\frac {1}{3}}\right )}{3}-\frac {\ln \left (\left (-x^{3}+1\right )^{\frac {2}{3}}+\left (-x^{3}+1\right )^{\frac {1}{3}}+1\right )}{6}+\frac {\arctan \left (\frac {\left (1+2 \left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(145\)

Input:

int(1/x/(-x^3+1)^(1/3)/(x^3+1),x,method=_RETURNVERBOSE)
 

Output:

-1/6*2^(2/3)*ln((-x^3+1)^(1/3)-2^(1/3))+1/12*2^(2/3)*ln((-x^3+1)^(2/3)+2^( 
1/3)*(-x^3+1)^(1/3)+2^(2/3))-1/6*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^( 
1/2))*2^(2/3)*3^(1/2)+1/3*ln(-1+(-x^3+1)^(1/3))-1/6*ln((-x^3+1)^(2/3)+(-x^ 
3+1)^(1/3)+1)+1/3*arctan(1/3*(1+2*(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.83 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.99 \[ \int \frac {1}{x \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx =\text {Too large to display} \] Input:

integrate(1/x/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="fricas")
 

Output:

1/12*2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))*log(1/8*(I*sqrt(3)*(-1)^( 
1/3) - (-1)^(1/3))^3 - 3/4*2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2 + 
 3*(-x^3 + 1)^(1/3) + 1) - 1/24*(2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3 
)) - 2*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2))*log 
(3/8*2^(2/3)*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2 
)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) + 3/8*2^(1/3)*(I*sqrt(3)*(-1)^(1/3) 
- (-1)^(1/3))^2 + 3*(-x^3 + 1)^(1/3)) - 1/24*(2^(2/3)*(I*sqrt(3)*(-1)^(1/3 
) - (-1)^(1/3)) + 2*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^( 
1/3))^2))*log(-3/8*2^(2/3)*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - 
 (-1)^(1/3))^2)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) + 3/8*2^(1/3)*(I*sqrt( 
3)*(-1)^(1/3) - (-1)^(1/3))^2 + 3*(-x^3 + 1)^(1/3)) + 1/3*sqrt(3)*arctan(2 
/3*sqrt(3)*(-x^3 + 1)^(1/3) + 1/3*sqrt(3)) + 1/3*log(-1/24*(I*sqrt(3)*(-1) 
^(1/3) - (-1)^(1/3))^3 + (-x^3 + 1)^(1/3) - 4/3) - 1/6*log((-x^3 + 1)^(2/3 
) + (-x^3 + 1)^(1/3) + 1)
 

Sympy [F]

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

integrate(1/x/(-x**3+1)**(1/3)/(x**3+1),x)
 

Output:

Integral(1/(x*(-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)*(x**2 - x + 1)), x)
 

Maxima [F]

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

integrate(1/x/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="maxima")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) - \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:

integrate(1/x/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="giac")
 

Output:

-1/6*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1 
/3))) + 1/12*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^( 
2/3)) - 1/6*2^(2/3)*log(abs(-2^(1/3) + (-x^3 + 1)^(1/3))) + 1/3*sqrt(3)*ar 
ctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3) + 1)) - 1/6*log((-x^3 + 1)^(2/3) + (- 
x^3 + 1)^(1/3) + 1) + 1/3*log(abs((-x^3 + 1)^(1/3) - 1))
 

Mupad [B] (verification not implemented)

Time = 3.55 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.87 \[ \int \frac {1}{x \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {\ln \left (6-6\,{\left (1-x^3\right )}^{1/3}\right )}{3}+\ln \left ({\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^3\,\left (1458\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2-135\,{\left (1-x^3\right )}^{1/3}\right )-{\left (1-x^3\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (-{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^3\,\left (1458\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2-135\,{\left (1-x^3\right )}^{1/3}\right )-{\left (1-x^3\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {2^{2/3}\,\ln \left (\frac {3\,{\left (1-x^3\right )}^{1/3}}{2}-\frac {3\,2^{1/3}}{2}\right )}{6}+\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\ln \left (\frac {3\,{\left (1-x^3\right )}^{1/3}}{2}-\frac {3\,{\left (-1\right )}^{2/3}\,2^{1/3}}{2}\right )}{6}-\frac {{\left (-1\right )}^{1/3}\,2^{2/3}\,\ln \left (-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^3\,\left (135\,{\left (1-x^3\right )}^{1/3}-\frac {81\,{\left (-1\right )}^{2/3}\,2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )}{432}-{\left (1-x^3\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \] Input:

int(1/(x*(1 - x^3)^(1/3)*(x^3 + 1)),x)
                                                                                    
                                                                                    
 

Output:

log(6 - 6*(1 - x^3)^(1/3))/3 + log(((3^(1/2)*1i)/6 - 1/6)^3*(1458*((3^(1/2 
)*1i)/6 - 1/6)^2 - 135*(1 - x^3)^(1/3)) - (1 - x^3)^(1/3))*((3^(1/2)*1i)/6 
 - 1/6) - log(- ((3^(1/2)*1i)/6 + 1/6)^3*(1458*((3^(1/2)*1i)/6 + 1/6)^2 - 
135*(1 - x^3)^(1/3)) - (1 - x^3)^(1/3))*((3^(1/2)*1i)/6 + 1/6) - (2^(2/3)* 
log((3*(1 - x^3)^(1/3))/2 - (3*2^(1/3))/2))/6 + ((-1)^(1/3)*2^(2/3)*log((3 
*(1 - x^3)^(1/3))/2 - (3*(-1)^(2/3)*2^(1/3))/2))/6 - ((-1)^(1/3)*2^(2/3)*l 
og(- ((3^(1/2)*1i + 1)^3*(135*(1 - x^3)^(1/3) - (81*(-1)^(2/3)*2^(1/3)*(3^ 
(1/2)*1i + 1)^2)/4))/432 - (1 - x^3)^(1/3))*(3^(1/2)*1i + 1))/12
 

Reduce [F]

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

int(1/x/(-x^3+1)^(1/3)/(x^3+1),x)
 

Output:

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