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

Optimal. Leaf size=138 \[ -\frac {\log \left (x^2-\sqrt [6]{2} \sqrt {3} x+\sqrt [3]{2}\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (x^2+\sqrt [6]{2} \sqrt {3} x+\sqrt [3]{2}\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\tan ^{-1}\left (\sqrt {3}-2^{5/6} x\right )}{6\ 2^{5/6}}+\frac {\tan ^{-1}\left (2^{5/6} x+\sqrt {3}\right )}{6\ 2^{5/6}} \]

[Out]

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

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Rubi [A]  time = 0.28, 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 = {209, 634, 618, 204, 628, 203} \[ -\frac {\log \left (x^2-\sqrt [6]{2} \sqrt {3} x+\sqrt [3]{2}\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (x^2+\sqrt [6]{2} \sqrt {3} x+\sqrt [3]{2}\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\tan ^{-1}\left (\sqrt {3}-2^{5/6} x\right )}{6\ 2^{5/6}}+\frac {\tan ^{-1}\left (2^{5/6} x+\sqrt {3}\right )}{6\ 2^{5/6}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + x^6)^(-1),x]

[Out]

ArcTan[x/2^(1/6)]/(3*2^(5/6)) - ArcTan[Sqrt[3] - 2^(5/6)*x]/(6*2^(5/6)) + ArcTan[Sqrt[3] + 2^(5/6)*x]/(6*2^(5/
6)) - Log[2^(1/3) - 2^(1/6)*Sqrt[3]*x + x^2]/(4*2^(5/6)*Sqrt[3]) + Log[2^(1/3) + 2^(1/6)*Sqrt[3]*x + x^2]/(4*2
^(5/6)*Sqrt[3])

Rule 203

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

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 209

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x] +
Int[(r + s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[((2*k - 1)*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] && PosQ[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 {\sqrt {3} x}{2}}{\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{3\ 2^{5/6}}+\frac {\int \frac {\sqrt [6]{2}+\frac {\sqrt {3} x}{2}}{\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{3\ 2^{5/6}}+\frac {\int \frac {1}{\sqrt [3]{2}+x^2} \, dx}{3\ 2^{2/3}}\\ &=\frac {\tan ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}+\frac {\int \frac {1}{\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{12\ 2^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{12\ 2^{2/3}}-\frac {\int \frac {-\sqrt [6]{2} \sqrt {3}+2 x}{\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{4\ 2^{5/6} \sqrt {3}}+\frac {\int \frac {\sqrt [6]{2} \sqrt {3}+2 x}{\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{4\ 2^{5/6} \sqrt {3}}\\ &=\frac {\tan ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1-\frac {2^{5/6} x}{\sqrt {3}}\right )}{6\ 2^{5/6} \sqrt {3}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1+\frac {2^{5/6} x}{\sqrt {3}}\right )}{6\ 2^{5/6} \sqrt {3}}\\ &=\frac {\tan ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\tan ^{-1}\left (\sqrt {3}-2^{5/6} x\right )}{6\ 2^{5/6}}+\frac {\tan ^{-1}\left (\sqrt {3}+2^{5/6} x\right )}{6\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 115, normalized size = 0.83 \[ \frac {-\sqrt {3} \log \left (2^{2/3} x^2-2^{5/6} \sqrt {3} x+2\right )+\sqrt {3} \log \left (2^{2/3} x^2+2^{5/6} \sqrt {3} x+2\right )+4 \tan ^{-1}\left (\frac {x}{\sqrt [6]{2}}\right )-2 \tan ^{-1}\left (\sqrt {3}-2^{5/6} x\right )+2 \tan ^{-1}\left (2^{5/6} x+\sqrt {3}\right )}{12\ 2^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x^6)^(-1),x]

[Out]

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

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fricas [A]  time = 0.43, size = 177, normalized size = 1.28 \[ \frac {1}{384} \cdot 32^{\frac {5}{6}} \sqrt {3} \log \left (32^{\frac {5}{6}} \sqrt {3} x + 16 \, x^{2} + 8 \cdot 4^{\frac {2}{3}}\right ) - \frac {1}{384} \cdot 32^{\frac {5}{6}} \sqrt {3} \log \left (-32^{\frac {5}{6}} \sqrt {3} x + 16 \, x^{2} + 8 \cdot 4^{\frac {2}{3}}\right ) - \frac {1}{48} \cdot 32^{\frac {5}{6}} \arctan \left (\frac {1}{4} \cdot 32^{\frac {1}{6}} \sqrt {2} \sqrt {2 \, x^{2} + 4^{\frac {2}{3}}} - \frac {1}{2} \cdot 32^{\frac {1}{6}} x\right ) - \frac {1}{96} \cdot 32^{\frac {5}{6}} \arctan \left (-32^{\frac {1}{6}} x + \frac {1}{4} \cdot 32^{\frac {1}{6}} \sqrt {32^{\frac {5}{6}} \sqrt {3} x + 16 \, x^{2} + 8 \cdot 4^{\frac {2}{3}}} - \sqrt {3}\right ) - \frac {1}{96} \cdot 32^{\frac {5}{6}} \arctan \left (-32^{\frac {1}{6}} x + \frac {1}{4} \cdot 32^{\frac {1}{6}} \sqrt {-32^{\frac {5}{6}} \sqrt {3} x + 16 \, x^{2} + 8 \cdot 4^{\frac {2}{3}}} + \sqrt {3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*32^(5/6)*sqrt(3)*log(32^(5/6)*sqrt(3)*x + 16*x^2 + 8*4^(2/3)) - 1/384*32^(5/6)*sqrt(3)*log(-32^(5/6)*sqr
t(3)*x + 16*x^2 + 8*4^(2/3)) - 1/48*32^(5/6)*arctan(1/4*32^(1/6)*sqrt(2)*sqrt(2*x^2 + 4^(2/3)) - 1/2*32^(1/6)*
x) - 1/96*32^(5/6)*arctan(-32^(1/6)*x + 1/4*32^(1/6)*sqrt(32^(5/6)*sqrt(3)*x + 16*x^2 + 8*4^(2/3)) - sqrt(3))
- 1/96*32^(5/6)*arctan(-32^(1/6)*x + 1/4*32^(1/6)*sqrt(-32^(5/6)*sqrt(3)*x + 16*x^2 + 8*4^(2/3)) + sqrt(3))

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giac [A]  time = 1.25, size = 107, normalized size = 0.78 \[ \frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} + \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) - \frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} - \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x + \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x - \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 95, normalized size = 0.69 \[ \frac {2^{\frac {1}{6}} \arctan \left (\frac {2^{\frac {5}{6}} x}{2}\right )}{6}+\frac {2^{\frac {1}{6}} \arctan \left (2^{\frac {5}{6}} x -\sqrt {3}\right )}{12}+\frac {2^{\frac {1}{6}} \arctan \left (2^{\frac {5}{6}} x +\sqrt {3}\right )}{12}-\frac {2^{\frac {1}{6}} \sqrt {3}\, \ln \left (x^{2}-2^{\frac {1}{6}} \sqrt {3}\, x +2^{\frac {1}{3}}\right )}{24}+\frac {2^{\frac {1}{6}} \sqrt {3}\, \ln \left (x^{2}+2^{\frac {1}{6}} \sqrt {3}\, x +2^{\frac {1}{3}}\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 107, normalized size = 0.78 \[ \frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} + \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) - \frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} - \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x + \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x - \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 135, normalized size = 0.98 \[ \frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{5/6}\,x}{2}\right )}{6}+\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{1/6}\,x}{2\,\left (-\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}+\frac {2^{1/6}\,\sqrt {3}\,x\,1{}\mathrm {i}}{2\,\left (-\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )\,\left (\sqrt {3}-\mathrm {i}\right )\,1{}\mathrm {i}}{12}+\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{1/6}\,x}{2\,\left (\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}-\frac {2^{1/6}\,\sqrt {3}\,x\,1{}\mathrm {i}}{2\,\left (\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )\,\left (\sqrt {3}+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6 + 2),x)

[Out]

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

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sympy [A]  time = 0.28, 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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