3.1.47 \(\int \frac {1}{-2+x^6} \, dx\) [47]

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

[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 = 0.13, 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 = {216, 648, 632, 210, 642, 212} \begin {gather*} \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}} \end {gather*}

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 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 212

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

Rule 216

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*C
os[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/(r^2 - s^2*x^2), x] + Dis
t[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 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]

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 {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2^{5/6} x\right )}{2\ 2^{5/6}}+\frac {\text {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 = 0.02, size = 122, normalized size = 0.88 \begin {gather*} -\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {-1+2^{5/6} x}{\sqrt {3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {1+2^{5/6} x}{\sqrt {3}}\right )-2 \log \left (2-2^{5/6} x\right )+2 \log \left (2+2^{5/6} x\right )-\log \left (2-2^{5/6} x+2^{2/3} x^2\right )+\log \left (2+2^{5/6} x+2^{2/3} x^2\right )}{12\ 2^{5/6}} \end {gather*}

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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.98, size = 21, normalized size = 0.15 \begin {gather*} \text {RootSum}\left [-1+1492992 \text {\#1}^6\&,\text {Log}\left [x-12 \text {\#1}\right ] \text {\#1}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(x^6-2),x]')

[Out]

RootSum[-1 + 1492992 #1 ^ 6&, Log[x - 12 #1] #1&]

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Maple [A]
time = 0.08, size = 111, normalized size = 0.80

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2\right )}{\sum }\frac {\ln \left (-\textit {\_R} +x \right )}{\textit {\_R}^{5}}\right )}{6}\)	\(22\)
default	\(-\frac {\ln \left (2^{\frac {1}{3}}+2^{\frac {1}{6}} x +x^{2}\right ) 2^{\frac {1}{6}}}{24}-\frac {\arctan \left (\frac {\sqrt {3}}{3}+\frac {2^{\frac {5}{6}} x \sqrt {3}}{3}\right ) 2^{\frac {1}{6}} \sqrt {3}}{12}-\frac {2^{\frac {1}{6}} \ln \left (x +2^{\frac {1}{6}}\right )}{12}+\frac {\ln \left (2^{\frac {1}{3}}-2^{\frac {1}{6}} x +x^{2}\right ) 2^{\frac {1}{6}}}{24}-\frac {\arctan \left (-\frac {\sqrt {3}}{3}+\frac {2^{\frac {5}{6}} x \sqrt {3}}{3}\right ) 2^{\frac {1}{6}} \sqrt {3}}{12}+\frac {2^{\frac {1}{6}} \ln \left (x -2^{\frac {1}{6}}\right )}{12}\)	\(111\)
meijerg	\(\frac {2^{\frac {1}{6}} x \left (\ln \left (1-\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}\right )-\ln \left (1+\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}\right )+\frac {\ln \left (1-\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}+\frac {2^{\frac {2}{3}} \left (x^{6}\right )^{\frac {1}{3}}}{2}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, 2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{4-2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}+\frac {2^{\frac {2}{3}} \left (x^{6}\right )^{\frac {1}{3}}}{2}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, 2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{4+2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{12 \left (x^{6}\right )^{\frac {1}{6}}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))+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 = 0.35, size = 112, normalized size = 0.81 \begin {gather*} -\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 ) \end {gather*}

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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Fricas [A]
time = 0.42, size = 176, normalized size = 1.28 \begin {gather*} \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 (1024 \, x^{2} + 64 \cdot 32^{\frac {5}{6}} x + 512 \cdot 4^{\frac {2}{3}}\right ) + \frac {1}{384} \cdot 32^{\frac {5}{6}} \log \left (1024 \, x^{2} - 64 \cdot 32^{\frac {5}{6}} x + 512 \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 ) \end {gather*}

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(1024*x^2 + 64*32^(5/6)*x + 512*4^(2/3)) + 1/384
*32^(5/6)*log(1024*x^2 - 64*32^(5/6)*x + 512*4^(2/3)) - 1/192*32^(5/6)*log(16*x + 32^(5/6)) + 1/192*32^(5/6)*l
og(16*x - 32^(5/6))

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Sympy [A]
time = 0.30, size = 14, normalized size = 0.10 \begin {gather*} \operatorname {RootSum} {\left (1492992 t^{6} - 1, \left ( t \mapsto t \log {\left (- 12 t + x \right )} \right )\right )} \end {gather*}

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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Giac [A]
time = 0.00, size = 182, normalized size = 1.32 \begin {gather*} \frac {1}{12}\cdot 2^{\frac {1}{6}} \ln \left |x-2^{\frac {1}{6}}\right |-\frac {1}{12}\cdot 2^{\frac {1}{6}} \ln \left |x+2^{\frac {1}{6}}\right |+\frac {1}{24}\cdot 2^{\frac {1}{6}} \ln \left (x^{2}-2^{\frac {1}{6}} x+2^{\frac {1}{6}}\cdot 2^{\frac {1}{6}}\right )-\frac {1}{12}\cdot 2^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {x-\frac {2^{\frac {1}{6}}}{2}}{\frac {1}{2} \sqrt {3}\cdot 2^{\frac {1}{6}}}\right )-\frac {1}{24}\cdot 2^{\frac {1}{6}} \ln \left (x^{2}+2^{\frac {1}{6}} x+2^{\frac {1}{6}}\cdot 2^{\frac {1}{6}}\right )-\frac {1}{12}\cdot 2^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {x+\frac {2^{\frac {1}{6}}}{2}}{\frac {1}{2} \sqrt {3}\cdot 2^{\frac {1}{6}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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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Mupad [B]
time = 0.21, size = 140, normalized size = 1.01 \begin {gather*} -\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} \end {gather*}

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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