\(\int \frac {1}{x (1-x^3+x^6)} \, dx\) [177]

Optimal result

Integrand size = 16, antiderivative size = 41 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=-\frac {\arctan \left (\frac {1-2 x^3}{\sqrt {3}}\right )}{3 \sqrt {3}}+\log (x)-\frac {1}{6} \log \left (1-x^3+x^6\right ) \]

[Out]

ln(x)-1/6*ln(x^6-x^3+1)-1/9*arctan(1/3*(-2*x^3+1)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1371, 719, 29, 648, 632, 210, 642} \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=-\frac {\arctan \left (\frac {1-2 x^3}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{6} \log \left (x^6-x^3+1\right )+\log (x) \]

[In]

Int[1/(x*(1 - x^3 + x^6)),x]

[Out]

-1/3*ArcTan[(1 - 2*x^3)/Sqrt[3]]/Sqrt[3] + Log[x] - Log[1 - x^3 + x^6]/6

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 719

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

Rule 1371

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x \left (1-x+x^2\right )} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1-x}{1-x+x^2} \, dx,x,x^3\right ) \\ & = \log (x)+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^3\right )-\frac {1}{6} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^3\right ) \\ & = \log (x)-\frac {1}{6} \log \left (1-x^3+x^6\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^3\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {1-2 x^3}{\sqrt {3}}\right )}{3 \sqrt {3}}+\log (x)-\frac {1}{6} \log \left (1-x^3+x^6\right ) \\ \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 = 55, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=\log (x)-\frac {1}{3} \text {RootSum}\left [1-\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^3}{-1+2 \text {$\#$1}^3}\&\right ] \]

[In]

Integrate[1/(x*(1 - x^3 + x^6)),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80

method	result	size
risch	\(\ln \left (x \right )-\frac {\ln \left (x^{6}-x^{3}+1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{3}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{9}\)	\(33\)
default	\(-\frac {\ln \left (x^{6}-x^{3}+1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{3}-1\right ) \sqrt {3}}{3}\right )}{9}+\ln \left (x \right )\)	\(35\)

[In]

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

[Out]

ln(x)-1/6*ln(x^6-x^3+1)+1/9*3^(1/2)*arctan(2/3*(x^3-1/2)*3^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) + \log \left (x\right ) \]

[In]

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

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) - 1/6*log(x^6 - x^3 + 1) + log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{6} - x^{3} + 1 \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{3}}{3} - \frac {\sqrt {3}}{3} \right )}}{9} \]

[In]

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

[Out]

log(x) - log(x**6 - x**3 + 1)/6 + sqrt(3)*atan(2*sqrt(3)*x**3/3 - sqrt(3)/3)/9

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) + \frac {1}{3} \, \log \left (x^{3}\right ) \]

[In]

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

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) - 1/6*log(x^6 - x^3 + 1) + 1/3*log(x^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) - 1/6*log(x^6 - x^3 + 1) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (1-x^3+x^6\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^6-x^3+1\right )}{6}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}-\frac {2\,\sqrt {3}\,x^3}{3}\right )}{9} \]

[In]

int(1/(x*(x^6 - x^3 + 1)),x)

[Out]

log(x) - log(x^6 - x^3 + 1)/6 - (3^(1/2)*atan(3^(1/2)/3 - (2*3^(1/2)*x^3)/3))/9