3.47 \(\int \frac {1}{-2+x^6} \, dx\)

Optimal. Leaf size=138 \[ \frac {\log \left (x^2-\sqrt [6]{2} x+\sqrt [3]{2}\right )}{12\ 2^{5/6}}-\frac {\log \left (x^2+\sqrt [6]{2} x+\sqrt [3]{2}\right )}{12\ 2^{5/6}}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{5/6} x}{\sqrt {3}}\right )}{2\ 2^{5/6} \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{5/6} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2\ 2^{5/6} \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}} \]

[Out]

-1/6*arctanh(1/2*x*2^(5/6))*2^(1/6)+1/24*ln(2^(1/3)-2^(1/6)*x+x^2)*2^(1/6)-1/24*ln(2^(1/3)+2^(1/6)*x+x^2)*2^(1
/6)-1/12*arctan(-1/3*3^(1/2)+1/3*2^(5/6)*x*3^(1/2))*2^(1/6)*3^(1/2)-1/12*arctan(1/3*3^(1/2)+1/3*2^(5/6)*x*3^(1
/2))*2^(1/6)*3^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {210, 634, 618, 204, 628, 206} \[ \frac {\log \left (x^2-\sqrt [6]{2} x+\sqrt [3]{2}\right )}{12\ 2^{5/6}}-\frac {\log \left (x^2+\sqrt [6]{2} x+\sqrt [3]{2}\right )}{12\ 2^{5/6}}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{5/6} x}{\sqrt {3}}\right )}{2\ 2^{5/6} \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2^{5/6} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2\ 2^{5/6} \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}} \]

Antiderivative was successfully verified.

[In]

Int[(-2 + x^6)^(-1),x]

[Out]

ArcTan[1/Sqrt[3] - (2^(5/6)*x)/Sqrt[3]]/(2*2^(5/6)*Sqrt[3]) - ArcTan[1/Sqrt[3] + (2^(5/6)*x)/Sqrt[3]]/(2*2^(5/
6)*Sqrt[3]) - ArcTanh[x/2^(1/6)]/(3*2^(5/6)) + Log[2^(1/3) - 2^(1/6)*x + x^2]/(12*2^(5/6)) - Log[2^(1/3) + 2^(
1/6)*x + x^2]/(12*2^(5/6))

Rule 204

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

Rule 206

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

Rule 210

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt[-(a/b
), n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r +
s*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 618

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 628

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 634

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]

Rubi steps

\begin {align*} \int \frac {1}{-2+x^6} \, dx &=-\frac {\int \frac {\sqrt [6]{2}-\frac {x}{2}}{\sqrt [3]{2}-\sqrt [6]{2} x+x^2} \, dx}{3\ 2^{5/6}}-\frac {\int \frac {\sqrt [6]{2}+\frac {x}{2}}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx}{3\ 2^{5/6}}-\frac {\int \frac {1}{\sqrt [3]{2}-x^2} \, dx}{3\ 2^{2/3}}\\ &=-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}+\frac {\int \frac {-\sqrt [6]{2}+2 x}{\sqrt [3]{2}-\sqrt [6]{2} x+x^2} \, dx}{12\ 2^{5/6}}-\frac {\int \frac {\sqrt [6]{2}+2 x}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx}{12\ 2^{5/6}}-\frac {\int \frac {1}{\sqrt [3]{2}-\sqrt [6]{2} x+x^2} \, dx}{4\ 2^{2/3}}-\frac {\int \frac {1}{\sqrt [3]{2}+\sqrt [6]{2} x+x^2} \, dx}{4\ 2^{2/3}}\\ &=-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} x+x^2\right )}{12\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} x+x^2\right )}{12\ 2^{5/6}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2^{5/6} x\right )}{2\ 2^{5/6}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{5/6} x\right )}{2\ 2^{5/6}}\\ &=\frac {\tan ^{-1}\left (\frac {1-2^{5/6} x}{\sqrt {3}}\right )}{2\ 2^{5/6} \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1+2^{5/6} x}{\sqrt {3}}\right )}{2\ 2^{5/6} \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} x+x^2\right )}{12\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} x+x^2\right )}{12\ 2^{5/6}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 122, normalized size = 0.88 \[ -\frac {-\log \left (2^{2/3} x^2-2^{5/6} x+2\right )+\log \left (2^{2/3} x^2+2^{5/6} x+2\right )-2 \log \left (2-2^{5/6} x\right )+2 \log \left (2^{5/6} x+2\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2^{5/6} x-1}{\sqrt {3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2^{5/6} x+1}{\sqrt {3}}\right )}{12\ 2^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-2 + x^6)^(-1),x]

[Out]

-1/12*(2*Sqrt[3]*ArcTan[(-1 + 2^(5/6)*x)/Sqrt[3]] + 2*Sqrt[3]*ArcTan[(1 + 2^(5/6)*x)/Sqrt[3]] - 2*Log[2 - 2^(5
/6)*x] + 2*Log[2 + 2^(5/6)*x] - Log[2 - 2^(5/6)*x + 2^(2/3)*x^2] + Log[2 + 2^(5/6)*x + 2^(2/3)*x^2])/2^(5/6)

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fricas [A]  time = 0.44, size = 175, normalized size = 1.27 \[ \frac {1}{96} \cdot 32^{\frac {5}{6}} \sqrt {3} \arctan \left (-\frac {1}{3} \cdot 32^{\frac {1}{6}} \sqrt {3} x + \frac {1}{12} \cdot 32^{\frac {1}{6}} \sqrt {3} \sqrt {16 \, x^{2} + 32^{\frac {5}{6}} x + 8 \cdot 4^{\frac {2}{3}}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{96} \cdot 32^{\frac {5}{6}} \sqrt {3} \arctan \left (-\frac {1}{3} \cdot 32^{\frac {1}{6}} \sqrt {3} x + \frac {1}{12} \cdot 32^{\frac {1}{6}} \sqrt {3} \sqrt {16 \, x^{2} - 32^{\frac {5}{6}} x + 8 \cdot 4^{\frac {2}{3}}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{384} \cdot 32^{\frac {5}{6}} \log \left (16 \, x^{2} + 32^{\frac {5}{6}} x + 8 \cdot 4^{\frac {2}{3}}\right ) + \frac {1}{384} \cdot 32^{\frac {5}{6}} \log \left (16 \, x^{2} - 32^{\frac {5}{6}} x + 8 \cdot 4^{\frac {2}{3}}\right ) - \frac {1}{192} \cdot 32^{\frac {5}{6}} \log \left (16 \, x + 32^{\frac {5}{6}}\right ) + \frac {1}{192} \cdot 32^{\frac {5}{6}} \log \left (16 \, x - 32^{\frac {5}{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*32^(5/6)*sqrt(3)*arctan(-1/3*32^(1/6)*sqrt(3)*x + 1/12*32^(1/6)*sqrt(3)*sqrt(16*x^2 + 32^(5/6)*x + 8*4^(2
/3)) - 1/3*sqrt(3)) + 1/96*32^(5/6)*sqrt(3)*arctan(-1/3*32^(1/6)*sqrt(3)*x + 1/12*32^(1/6)*sqrt(3)*sqrt(16*x^2
- 32^(5/6)*x + 8*4^(2/3)) + 1/3*sqrt(3)) - 1/384*32^(5/6)*log(16*x^2 + 32^(5/6)*x + 8*4^(2/3)) + 1/384*32^(5/
6)*log(16*x^2 - 32^(5/6)*x + 8*4^(2/3)) - 1/192*32^(5/6)*log(16*x + 32^(5/6)) + 1/192*32^(5/6)*log(16*x - 32^(
5/6))

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giac [A]  time = 1.31, size = 114, normalized size = 0.83 \[ -\frac {1}{12} \, \sqrt {3} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {5}{6}} {\left (2 \, x + 2^{\frac {1}{6}}\right )}\right ) - \frac {1}{12} \, \sqrt {3} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {5}{6}} {\left (2 \, x - 2^{\frac {1}{6}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {1}{6}} \log \left (x^{2} + 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{24} \cdot 2^{\frac {1}{6}} \log \left (x^{2} - 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) - \frac {1}{12} \cdot 2^{\frac {1}{6}} \log \left ({\left | x + 2^{\frac {1}{6}} \right |}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \log \left ({\left | x - 2^{\frac {1}{6}} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*2^(1/6)*arctan(1/6*sqrt(3)*2^(5/6)*(2*x + 2^(1/6))) - 1/12*sqrt(3)*2^(1/6)*arctan(1/6*sqrt(3)*2^
(5/6)*(2*x - 2^(1/6))) - 1/24*2^(1/6)*log(x^2 + 2^(1/6)*x + 2^(1/3)) + 1/24*2^(1/6)*log(x^2 - 2^(1/6)*x + 2^(1
/3)) - 1/12*2^(1/6)*log(abs(x + 2^(1/6))) + 1/12*2^(1/6)*log(abs(x - 2^(1/6)))

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maple [A]  time = 0.16, size = 111, normalized size = 0.80 \[ -\frac {2^{\frac {1}{6}} \sqrt {3}\, \arctan \left (\frac {2^{\frac {5}{6}} \sqrt {3}\, x}{3}-\frac {\sqrt {3}}{3}\right )}{12}-\frac {2^{\frac {1}{6}} \sqrt {3}\, \arctan \left (\frac {2^{\frac {5}{6}} \sqrt {3}\, x}{3}+\frac {\sqrt {3}}{3}\right )}{12}+\frac {2^{\frac {1}{6}} \ln \left (x -2^{\frac {1}{6}}\right )}{12}-\frac {2^{\frac {1}{6}} \ln \left (x +2^{\frac {1}{6}}\right )}{12}+\frac {2^{\frac {1}{6}} \ln \left (x^{2}-2^{\frac {1}{6}} x +2^{\frac {1}{3}}\right )}{24}-\frac {2^{\frac {1}{6}} \ln \left (x^{2}+2^{\frac {1}{6}} x +2^{\frac {1}{3}}\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6-2),x)

[Out]

1/12*2^(1/6)*ln(x-2^(1/6))-1/24*ln(2^(1/3)+2^(1/6)*x+x^2)*2^(1/6)-1/12*arctan(1/3*3^(1/2)+1/3*2^(5/6)*x*3^(1/2
))*2^(1/6)*3^(1/2)+1/24*ln(2^(1/3)-2^(1/6)*x+x^2)*2^(1/6)-1/12*arctan(-1/3*3^(1/2)+1/3*2^(5/6)*x*3^(1/2))*2^(1
/6)*3^(1/2)-1/12*2^(1/6)*ln(x+2^(1/6))

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maxima [A]  time = 0.95, size = 112, normalized size = 0.81 \[ -\frac {1}{12} \, \sqrt {3} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {5}{6}} {\left (2 \, x + 2^{\frac {1}{6}}\right )}\right ) - \frac {1}{12} \, \sqrt {3} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {5}{6}} {\left (2 \, x - 2^{\frac {1}{6}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {1}{6}} \log \left (x^{2} + 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{24} \cdot 2^{\frac {1}{6}} \log \left (x^{2} - 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) - \frac {1}{12} \cdot 2^{\frac {1}{6}} \log \left (x + 2^{\frac {1}{6}}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \log \left (x - 2^{\frac {1}{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*2^(1/6)*arctan(1/6*sqrt(3)*2^(5/6)*(2*x + 2^(1/6))) - 1/12*sqrt(3)*2^(1/6)*arctan(1/6*sqrt(3)*2^
(5/6)*(2*x - 2^(1/6))) - 1/24*2^(1/6)*log(x^2 + 2^(1/6)*x + 2^(1/3)) + 1/24*2^(1/6)*log(x^2 - 2^(1/6)*x + 2^(1
/3)) - 1/12*2^(1/6)*log(x + 2^(1/6)) + 1/12*2^(1/6)*log(x - 2^(1/6))

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mupad [B]  time = 0.21, size = 140, normalized size = 1.01 \[ -\frac {2^{1/6}\,\mathrm {atanh}\left (\frac {2^{5/6}\,x}{2}\right )}{6}+\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{1/6}\,x\,1{}\mathrm {i}}{2\,\left (-\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}-\frac {2^{1/6}\,\sqrt {3}\,x}{2\,\left (-\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{12}+\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{1/6}\,x\,1{}\mathrm {i}}{2\,\left (\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}+\frac {2^{1/6}\,\sqrt {3}\,x}{2\,\left (\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6 - 2),x)

[Out]

(2^(1/6)*atan((2^(1/6)*x*1i)/(2*((2^(1/3)*3^(1/2)*1i)/2 - 2^(1/3)/2)) - (2^(1/6)*3^(1/2)*x)/(2*((2^(1/3)*3^(1/
2)*1i)/2 - 2^(1/3)/2)))*(3^(1/2)*1i + 1)*1i)/12 - (2^(1/6)*atanh((2^(5/6)*x)/2))/6 + (2^(1/6)*atan((2^(1/6)*x*
1i)/(2*((2^(1/3)*3^(1/2)*1i)/2 + 2^(1/3)/2)) + (2^(1/6)*3^(1/2)*x)/(2*((2^(1/3)*3^(1/2)*1i)/2 + 2^(1/3)/2)))*(
3^(1/2)*1i - 1)*1i)/12

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sympy [A]  time = 0.57, size = 14, normalized size = 0.10 \[ \operatorname {RootSum} {\left (1492992 t^{6} - 1, \left (t \mapsto t \log {\left (- 12 t + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(1492992*_t**6 - 1, Lambda(_t, _t*log(-12*_t + x)))

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