\(\int \frac {1}{x^2 (1-x^3+x^6)} \, dx\) [178]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 416 \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=-\frac {1}{x}+\frac {\left (i-\sqrt {3}\right ) \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}-\frac {\left (3-i \sqrt {3}\right ) \log \left (\sqrt [3]{1-i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (3+i \sqrt {3}\right ) \log \left (\sqrt [3]{1+i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \log \left (\left (1-i \sqrt {3}\right )^{2/3}+\sqrt [3]{2 \left (1-i \sqrt {3}\right )} x+2^{2/3} x^2\right )}{18\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3+i \sqrt {3}\right ) \log \left (\left (1+i \sqrt {3}\right )^{2/3}+\sqrt [3]{2 \left (1+i \sqrt {3}\right )} x+2^{2/3} x^2\right )}{18\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}} \]

[Out]

-1/x+1/6*arctan(1/3*(1+2*2^(1/3)*x/(1-I*3^(1/2))^(1/3))*3^(1/2))*(I-3^(1/2))*2^(1/3)/(1-I*3^(1/2))^(1/3)-1/18*
ln(-2^(1/3)*x+(1-I*3^(1/2))^(1/3))*(3-I*3^(1/2))*2^(1/3)/(1-I*3^(1/2))^(1/3)+1/36*ln(2^(2/3)*x^2+2^(1/3)*x*(1-
I*3^(1/2))^(1/3)+(1-I*3^(1/2))^(2/3))*(3-I*3^(1/2))*2^(1/3)/(1-I*3^(1/2))^(1/3)-1/18*ln(-2^(1/3)*x+(1+I*3^(1/2
))^(1/3))*(3+I*3^(1/2))*2^(1/3)/(1+I*3^(1/2))^(1/3)+1/36*ln(2^(2/3)*x^2+2^(1/3)*x*(1+I*3^(1/2))^(1/3)+(1+I*3^(
1/2))^(2/3))*(3+I*3^(1/2))*2^(1/3)/(1+I*3^(1/2))^(1/3)-1/6*arctan(1/3*(1+2*2^(1/3)*x/(1+I*3^(1/2))^(1/3))*3^(1
/2))*(3^(1/2)+I)*2^(1/3)/(1+I*3^(1/2))^(1/3)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1382, 1524, 298, 31, 648, 631, 210, 642} \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=\frac {\left (-\sqrt {3}+i\right ) \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (\sqrt {3}+i\right ) \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \log \left (2^{2/3} x^2+\sqrt [3]{2 \left (1-i \sqrt {3}\right )} x+\left (1-i \sqrt {3}\right )^{2/3}\right )}{18\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3+i \sqrt {3}\right ) \log \left (2^{2/3} x^2+\sqrt [3]{2 \left (1+i \sqrt {3}\right )} x+\left (1+i \sqrt {3}\right )^{2/3}\right )}{18\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}-\frac {1}{x}-\frac {\left (3-i \sqrt {3}\right ) \log \left (-\sqrt [3]{2} x+\sqrt [3]{1-i \sqrt {3}}\right )}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (3+i \sqrt {3}\right ) \log \left (-\sqrt [3]{2} x+\sqrt [3]{1+i \sqrt {3}}\right )}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}} \]

[In]

Int[1/(x^2*(1 - x^3 + x^6)),x]

[Out]

-x^(-1) + ((I - Sqrt[3])*ArcTan[(1 + (2*x)/((1 - I*Sqrt[3])/2)^(1/3))/Sqrt[3]])/(3*2^(2/3)*(1 - I*Sqrt[3])^(1/
3)) - ((I + Sqrt[3])*ArcTan[(1 + (2*x)/((1 + I*Sqrt[3])/2)^(1/3))/Sqrt[3]])/(3*2^(2/3)*(1 + I*Sqrt[3])^(1/3))
- ((3 - I*Sqrt[3])*Log[(1 - I*Sqrt[3])^(1/3) - 2^(1/3)*x])/(9*2^(2/3)*(1 - I*Sqrt[3])^(1/3)) - ((3 + I*Sqrt[3]
)*Log[(1 + I*Sqrt[3])^(1/3) - 2^(1/3)*x])/(9*2^(2/3)*(1 + I*Sqrt[3])^(1/3)) + ((3 - I*Sqrt[3])*Log[(1 - I*Sqrt
[3])^(2/3) + (2*(1 - I*Sqrt[3]))^(1/3)*x + 2^(2/3)*x^2])/(18*2^(2/3)*(1 - I*Sqrt[3])^(1/3)) + ((3 + I*Sqrt[3])
*Log[(1 + I*Sqrt[3])^(2/3) + (2*(1 + I*Sqrt[3]))^(1/3)*x + 2^(2/3)*x^2])/(18*2^(2/3)*(1 + I*Sqrt[3])^(1/3))

Rule 31

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

Rule 210

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 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1382

Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a +
 b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1))), x] - Dist[1/(a*d^n*(m + 1)), Int[(d*x)^(m + n)*(b*(m + n*(p + 1) +
 1) + c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[n2, 2
*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntegerQ[p]

Rule 1524

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Dist[e/
2 - (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[n2
, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{x}+\int \frac {x \left (1-x^3\right )}{1-x^3+x^6} \, dx \\ & = -\frac {1}{x}+\frac {1}{6} \left (-3+i \sqrt {3}\right ) \int \frac {x}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}+x^3} \, dx-\frac {1}{6} \left (3+i \sqrt {3}\right ) \int \frac {x}{-\frac {1}{2}-\frac {i \sqrt {3}}{2}+x^3} \, dx \\ & = -\frac {1}{x}-\frac {\left (3-i \sqrt {3}\right ) \int \frac {1}{-\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}+x} \, dx}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \int \frac {-\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}+x}{\left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{2/3}+\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )} x+x^2} \, dx}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (3+i \sqrt {3}\right ) \int \frac {1}{-\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}+x} \, dx}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3+i \sqrt {3}\right ) \int \frac {-\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}+x}{\left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{2/3}+\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )} x+x^2} \, dx}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}} \\ & = -\frac {1}{x}-\frac {\left (3-i \sqrt {3}\right ) \log \left (\sqrt [3]{1-i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (3+i \sqrt {3}\right ) \log \left (\sqrt [3]{1+i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \int \frac {\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}+2 x}{\left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{2/3}+\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )} x+x^2} \, dx}{18\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {1}{12} \left (-3+i \sqrt {3}\right ) \int \frac {1}{\left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{2/3}+\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )} x+x^2} \, dx-\frac {1}{12} \left (3+i \sqrt {3}\right ) \int \frac {1}{\left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{2/3}+\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )} x+x^2} \, dx+\frac {\left (3+i \sqrt {3}\right ) \int \frac {\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}+2 x}{\left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{2/3}+\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )} x+x^2} \, dx}{18\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}} \\ & = -\frac {1}{x}-\frac {\left (3-i \sqrt {3}\right ) \log \left (\sqrt [3]{1-i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (3+i \sqrt {3}\right ) \log \left (\sqrt [3]{1+i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \log \left (\left (1-i \sqrt {3}\right )^{2/3}+\sqrt [3]{2 \left (1-i \sqrt {3}\right )} x+2^{2/3} x^2\right )}{18\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3+i \sqrt {3}\right ) \log \left (\left (1+i \sqrt {3}\right )^{2/3}+\sqrt [3]{2 \left (1+i \sqrt {3}\right )} x+2^{2/3} x^2\right )}{18\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right )}{3\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3+i \sqrt {3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right )}{3\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}} \\ & = -\frac {1}{x}+\frac {\left (i-\sqrt {3}\right ) \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1-i \sqrt {3}\right )}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (i+\sqrt {3}\right ) \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {1}{2} \left (1+i \sqrt {3}\right )}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}-\frac {\left (3-i \sqrt {3}\right ) \log \left (\sqrt [3]{1-i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}-\frac {\left (3+i \sqrt {3}\right ) \log \left (\sqrt [3]{1+i \sqrt {3}}-\sqrt [3]{2} x\right )}{9\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \log \left (\left (1-i \sqrt {3}\right )^{2/3}+\sqrt [3]{2 \left (1-i \sqrt {3}\right )} x+2^{2/3} x^2\right )}{18\ 2^{2/3} \sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3+i \sqrt {3}\right ) \log \left (\left (1+i \sqrt {3}\right )^{2/3}+\sqrt [3]{2 \left (1+i \sqrt {3}\right )} x+2^{2/3} x^2\right )}{18\ 2^{2/3} \sqrt [3]{1+i \sqrt {3}}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.15 \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=-\frac {1}{x}-\frac {1}{3} \text {RootSum}\left [1-\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^3}{-\text {$\#$1}+2 \text {$\#$1}^4}\&\right ] \]

[In]

Integrate[1/(x^2*(1 - x^3 + x^6)),x]

[Out]

-x^(-1) - RootSum[1 - #1^3 + #1^6 & , (-Log[x - #1] + Log[x - #1]*#1^3)/(-#1 + 2*#1^4) & ]/3

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.08

method result size
risch \(-\frac {1}{x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (27 \textit {\_Z}^{6}+9 \textit {\_Z}^{3}+1\right )}{\sum }\textit {\_R} \ln \left (-3 \textit {\_R}^{2}+x \right )\right )}{3}\) \(35\)
default \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-\textit {\_Z}^{3}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{5}-\textit {\_R}^{2}}\right )}{3}-\frac {1}{x}\) \(50\)

[In]

int(1/x^2/(x^6-x^3+1),x,method=_RETURNVERBOSE)

[Out]

-1/x+1/3*sum(_R*ln(-3*_R^2+x),_R=RootOf(27*_Z^6+9*_Z^3+1))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=\frac {18^{\frac {2}{3}} {\left (\sqrt {-3} x - x\right )} {\left (i \, \sqrt {3} - 3\right )}^{\frac {1}{3}} \log \left (18^{\frac {1}{3}} {\left (i \, \sqrt {3} - 3\right )}^{\frac {2}{3}} {\left (\sqrt {-3} + 1\right )} + 12 \, x\right ) + 18^{\frac {2}{3}} {\left (\sqrt {-3} x - x\right )} {\left (-i \, \sqrt {3} - 3\right )}^{\frac {1}{3}} \log \left (18^{\frac {1}{3}} {\left (-i \, \sqrt {3} - 3\right )}^{\frac {2}{3}} {\left (\sqrt {-3} + 1\right )} + 12 \, x\right ) - 18^{\frac {2}{3}} {\left (\sqrt {-3} x + x\right )} {\left (i \, \sqrt {3} - 3\right )}^{\frac {1}{3}} \log \left (-18^{\frac {1}{3}} {\left (i \, \sqrt {3} - 3\right )}^{\frac {2}{3}} {\left (\sqrt {-3} - 1\right )} + 12 \, x\right ) - 18^{\frac {2}{3}} {\left (\sqrt {-3} x + x\right )} {\left (-i \, \sqrt {3} - 3\right )}^{\frac {1}{3}} \log \left (-18^{\frac {1}{3}} {\left (-i \, \sqrt {3} - 3\right )}^{\frac {2}{3}} {\left (\sqrt {-3} - 1\right )} + 12 \, x\right ) + 2 \cdot 18^{\frac {2}{3}} x {\left (i \, \sqrt {3} - 3\right )}^{\frac {1}{3}} \log \left (6 \, x - 18^{\frac {1}{3}} {\left (i \, \sqrt {3} - 3\right )}^{\frac {2}{3}}\right ) + 2 \cdot 18^{\frac {2}{3}} x {\left (-i \, \sqrt {3} - 3\right )}^{\frac {1}{3}} \log \left (6 \, x - 18^{\frac {1}{3}} {\left (-i \, \sqrt {3} - 3\right )}^{\frac {2}{3}}\right ) - 108}{108 \, x} \]

[In]

integrate(1/x^2/(x^6-x^3+1),x, algorithm="fricas")

[Out]

1/108*(18^(2/3)*(sqrt(-3)*x - x)*(I*sqrt(3) - 3)^(1/3)*log(18^(1/3)*(I*sqrt(3) - 3)^(2/3)*(sqrt(-3) + 1) + 12*
x) + 18^(2/3)*(sqrt(-3)*x - x)*(-I*sqrt(3) - 3)^(1/3)*log(18^(1/3)*(-I*sqrt(3) - 3)^(2/3)*(sqrt(-3) + 1) + 12*
x) - 18^(2/3)*(sqrt(-3)*x + x)*(I*sqrt(3) - 3)^(1/3)*log(-18^(1/3)*(I*sqrt(3) - 3)^(2/3)*(sqrt(-3) - 1) + 12*x
) - 18^(2/3)*(sqrt(-3)*x + x)*(-I*sqrt(3) - 3)^(1/3)*log(-18^(1/3)*(-I*sqrt(3) - 3)^(2/3)*(sqrt(-3) - 1) + 12*
x) + 2*18^(2/3)*x*(I*sqrt(3) - 3)^(1/3)*log(6*x - 18^(1/3)*(I*sqrt(3) - 3)^(2/3)) + 2*18^(2/3)*x*(-I*sqrt(3) -
 3)^(1/3)*log(6*x - 18^(1/3)*(-I*sqrt(3) - 3)^(2/3)) - 108)/x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.06 \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=\operatorname {RootSum} {\left (19683 t^{6} + 243 t^{3} + 1, \left ( t \mapsto t \log {\left (- 27 t^{2} + x \right )} \right )\right )} - \frac {1}{x} \]

[In]

integrate(1/x**2/(x**6-x**3+1),x)

[Out]

RootSum(19683*_t**6 + 243*_t**3 + 1, Lambda(_t, _t*log(-27*_t**2 + x))) - 1/x

Maxima [F]

\[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=\int { \frac {1}{{\left (x^{6} - x^{3} + 1\right )} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(x^6-x^3+1),x, algorithm="maxima")

[Out]

-1/x - integrate((x^4 - x)/(x^6 - x^3 + 1), x)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 829 vs. \(2 (272) = 544\).

Time = 0.31 (sec) , antiderivative size = 829, normalized size of antiderivative = 1.99 \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=\text {Too large to display} \]

[In]

integrate(1/x^2/(x^6-x^3+1),x, algorithm="giac")

[Out]

1/9*(sqrt(3)*cos(4/9*pi)^5 - 10*sqrt(3)*cos(4/9*pi)^3*sin(4/9*pi)^2 + 5*sqrt(3)*cos(4/9*pi)*sin(4/9*pi)^4 - 5*
cos(4/9*pi)^4*sin(4/9*pi) + 10*cos(4/9*pi)^2*sin(4/9*pi)^3 - sin(4/9*pi)^5 + 2*sqrt(3)*cos(4/9*pi)^2 - 2*sqrt(
3)*sin(4/9*pi)^2 - 4*cos(4/9*pi)*sin(4/9*pi))*arctan(1/2*((-I*sqrt(3) - 1)*cos(4/9*pi) + 2*x)/((1/2*I*sqrt(3)
+ 1/2)*sin(4/9*pi))) + 1/9*(sqrt(3)*cos(2/9*pi)^5 - 10*sqrt(3)*cos(2/9*pi)^3*sin(2/9*pi)^2 + 5*sqrt(3)*cos(2/9
*pi)*sin(2/9*pi)^4 - 5*cos(2/9*pi)^4*sin(2/9*pi) + 10*cos(2/9*pi)^2*sin(2/9*pi)^3 - sin(2/9*pi)^5 + 2*sqrt(3)*
cos(2/9*pi)^2 - 2*sqrt(3)*sin(2/9*pi)^2 - 4*cos(2/9*pi)*sin(2/9*pi))*arctan(1/2*((-I*sqrt(3) - 1)*cos(2/9*pi)
+ 2*x)/((1/2*I*sqrt(3) + 1/2)*sin(2/9*pi))) - 1/9*(sqrt(3)*cos(1/9*pi)^5 - 10*sqrt(3)*cos(1/9*pi)^3*sin(1/9*pi
)^2 + 5*sqrt(3)*cos(1/9*pi)*sin(1/9*pi)^4 + 5*cos(1/9*pi)^4*sin(1/9*pi) - 10*cos(1/9*pi)^2*sin(1/9*pi)^3 + sin
(1/9*pi)^5 - 2*sqrt(3)*cos(1/9*pi)^2 + 2*sqrt(3)*sin(1/9*pi)^2 - 4*cos(1/9*pi)*sin(1/9*pi))*arctan(-1/2*((-I*s
qrt(3) - 1)*cos(1/9*pi) - 2*x)/((1/2*I*sqrt(3) + 1/2)*sin(1/9*pi))) + 1/18*(5*sqrt(3)*cos(4/9*pi)^4*sin(4/9*pi
) - 10*sqrt(3)*cos(4/9*pi)^2*sin(4/9*pi)^3 + sqrt(3)*sin(4/9*pi)^5 + cos(4/9*pi)^5 - 10*cos(4/9*pi)^3*sin(4/9*
pi)^2 + 5*cos(4/9*pi)*sin(4/9*pi)^4 + 4*sqrt(3)*cos(4/9*pi)*sin(4/9*pi) + 2*cos(4/9*pi)^2 - 2*sin(4/9*pi)^2)*l
og((-I*sqrt(3)*cos(4/9*pi) - cos(4/9*pi))*x + x^2 + 1) + 1/18*(5*sqrt(3)*cos(2/9*pi)^4*sin(2/9*pi) - 10*sqrt(3
)*cos(2/9*pi)^2*sin(2/9*pi)^3 + sqrt(3)*sin(2/9*pi)^5 + cos(2/9*pi)^5 - 10*cos(2/9*pi)^3*sin(2/9*pi)^2 + 5*cos
(2/9*pi)*sin(2/9*pi)^4 + 4*sqrt(3)*cos(2/9*pi)*sin(2/9*pi) + 2*cos(2/9*pi)^2 - 2*sin(2/9*pi)^2)*log((-I*sqrt(3
)*cos(2/9*pi) - cos(2/9*pi))*x + x^2 + 1) + 1/18*(5*sqrt(3)*cos(1/9*pi)^4*sin(1/9*pi) - 10*sqrt(3)*cos(1/9*pi)
^2*sin(1/9*pi)^3 + sqrt(3)*sin(1/9*pi)^5 - cos(1/9*pi)^5 + 10*cos(1/9*pi)^3*sin(1/9*pi)^2 - 5*cos(1/9*pi)*sin(
1/9*pi)^4 - 4*sqrt(3)*cos(1/9*pi)*sin(1/9*pi) + 2*cos(1/9*pi)^2 - 2*sin(1/9*pi)^2)*log((I*sqrt(3)*cos(1/9*pi)
+ cos(1/9*pi))*x + x^2 + 1) - 1/x

Mupad [B] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \left (1-x^3+x^6\right )} \, dx=\frac {\ln \left (x-\frac {2^{1/3}\,3^{2/3}\,{\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )}^{2/3}}{6}\right )\,{\left (-36+\sqrt {3}\,12{}\mathrm {i}\right )}^{1/3}}{18}-\frac {1}{x}+\frac {\ln \left (x-\frac {{\left (-36-\sqrt {3}\,12{}\mathrm {i}\right )}^{2/3}}{12}\right )\,{\left (-36-\sqrt {3}\,12{}\mathrm {i}\right )}^{1/3}}{18}-\frac {2^{2/3}\,\ln \left (x-\frac {2^{1/3}\,{\left (-3-\sqrt {3}\,1{}\mathrm {i}\right )}^{2/3}\,{\left (3^{1/3}-3^{5/6}\,1{}\mathrm {i}\right )}^2}{24}\right )\,{\left (-3-\sqrt {3}\,1{}\mathrm {i}\right )}^{1/3}\,\left (3^{1/3}-3^{5/6}\,1{}\mathrm {i}\right )}{36}-\frac {2^{2/3}\,\ln \left (x-\frac {2^{1/3}\,{\left (-3-\sqrt {3}\,1{}\mathrm {i}\right )}^{2/3}\,{\left (3^{1/3}+3^{5/6}\,1{}\mathrm {i}\right )}^2}{24}\right )\,{\left (-3-\sqrt {3}\,1{}\mathrm {i}\right )}^{1/3}\,\left (3^{1/3}+3^{5/6}\,1{}\mathrm {i}\right )}{36}-\frac {2^{2/3}\,\ln \left (x-\frac {2^{1/3}\,{\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )}^{2/3}\,{\left (3^{1/3}-3^{5/6}\,1{}\mathrm {i}\right )}^2}{24}\right )\,{\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )}^{1/3}\,\left (3^{1/3}-3^{5/6}\,1{}\mathrm {i}\right )}{36}-\frac {2^{2/3}\,\ln \left (x-\frac {2^{1/3}\,{\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )}^{2/3}\,{\left (3^{1/3}+3^{5/6}\,1{}\mathrm {i}\right )}^2}{24}\right )\,{\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )}^{1/3}\,\left (3^{1/3}+3^{5/6}\,1{}\mathrm {i}\right )}{36} \]

[In]

int(1/(x^2*(x^6 - x^3 + 1)),x)

[Out]

(log(x - (2^(1/3)*3^(2/3)*(3^(1/2)*1i - 3)^(2/3))/6)*(3^(1/2)*12i - 36)^(1/3))/18 - 1/x + (log(x - (- 3^(1/2)*
12i - 36)^(2/3)/12)*(- 3^(1/2)*12i - 36)^(1/3))/18 - (2^(2/3)*log(x - (2^(1/3)*(- 3^(1/2)*1i - 3)^(2/3)*(3^(1/
3) - 3^(5/6)*1i)^2)/24)*(- 3^(1/2)*1i - 3)^(1/3)*(3^(1/3) - 3^(5/6)*1i))/36 - (2^(2/3)*log(x - (2^(1/3)*(- 3^(
1/2)*1i - 3)^(2/3)*(3^(1/3) + 3^(5/6)*1i)^2)/24)*(- 3^(1/2)*1i - 3)^(1/3)*(3^(1/3) + 3^(5/6)*1i))/36 - (2^(2/3
)*log(x - (2^(1/3)*(3^(1/2)*1i - 3)^(2/3)*(3^(1/3) - 3^(5/6)*1i)^2)/24)*(3^(1/2)*1i - 3)^(1/3)*(3^(1/3) - 3^(5
/6)*1i))/36 - (2^(2/3)*log(x - (2^(1/3)*(3^(1/2)*1i - 3)^(2/3)*(3^(1/3) + 3^(5/6)*1i)^2)/24)*(3^(1/2)*1i - 3)^
(1/3)*(3^(1/3) + 3^(5/6)*1i))/36