3.1369 \(\int \frac{1}{1+x^6} \, dx\)

Optimal. Leaf size=80 \[ -\frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}+\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{1}{6} \tan ^{-1}\left (\sqrt{3}-2 x\right )+\frac{1}{3} \tan ^{-1}(x)+\frac{1}{6} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

[Out]

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

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Rubi [A]  time = 0.170931, antiderivative size = 80, normalized size of antiderivative = 1., 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} \[ -\frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}+\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{1}{6} \tan ^{-1}\left (\sqrt{3}-2 x\right )+\frac{1}{3} \tan ^{-1}(x)+\frac{1}{6} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^6)^(-1),x]

[Out]

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

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 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]

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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.011359, size = 73, normalized size = 0.91 \[ \frac{1}{12} \left (-\sqrt{3} \log \left (x^2-\sqrt{3} x+1\right )+\sqrt{3} \log \left (x^2+\sqrt{3} x+1\right )-2 \tan ^{-1}\left (\sqrt{3}-2 x\right )+4 \tan ^{-1}(x)+2 \tan ^{-1}\left (2 x+\sqrt{3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^6)^(-1),x]

[Out]

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

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Maple [A]  time = 0.011, size = 61, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( x \right ) }{3}}+{\frac{\arctan \left ( 2\,x-\sqrt{3} \right ) }{6}}+{\frac{\arctan \left ( 2\,x+\sqrt{3} \right ) }{6}}-{\frac{\ln \left ( 1+{x}^{2}-x\sqrt{3} \right ) \sqrt{3}}{12}}+{\frac{\ln \left ( 1+{x}^{2}+x\sqrt{3} \right ) \sqrt{3}}{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.60289, size = 81, normalized size = 1.01 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (x^{2} + \sqrt{3} x + 1\right ) - \frac{1}{12} \, \sqrt{3} \log \left (x^{2} - \sqrt{3} x + 1\right ) + \frac{1}{6} \, \arctan \left (2 \, x + \sqrt{3}\right ) + \frac{1}{6} \, \arctan \left (2 \, x - \sqrt{3}\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52775, size = 288, normalized size = 3.6 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (x^{2} + \sqrt{3} x + 1\right ) - \frac{1}{12} \, \sqrt{3} \log \left (x^{2} - \sqrt{3} x + 1\right ) + \frac{1}{3} \, \arctan \left (x\right ) - \frac{1}{3} \, \arctan \left (-2 \, x + \sqrt{3} + 2 \, \sqrt{x^{2} - \sqrt{3} x + 1}\right ) - \frac{1}{3} \, \arctan \left (-2 \, x - \sqrt{3} + 2 \, \sqrt{x^{2} + \sqrt{3} x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.199209, size = 68, normalized size = 0.85 \begin{align*} - \frac{\sqrt{3} \log{\left (x^{2} - \sqrt{3} x + 1 \right )}}{12} + \frac{\sqrt{3} \log{\left (x^{2} + \sqrt{3} x + 1 \right )}}{12} + \frac{\operatorname{atan}{\left (x \right )}}{3} + \frac{\operatorname{atan}{\left (2 x - \sqrt{3} \right )}}{6} + \frac{\operatorname{atan}{\left (2 x + \sqrt{3} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 + sqrt(3)*log(x**2 + sqrt(3)*x + 1)/12 + atan(x)/3 + atan(2*x - sqrt(3))
/6 + atan(2*x + sqrt(3))/6

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Giac [A]  time = 1.19758, size = 81, normalized size = 1.01 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (x^{2} + \sqrt{3} x + 1\right ) - \frac{1}{12} \, \sqrt{3} \log \left (x^{2} - \sqrt{3} x + 1\right ) + \frac{1}{6} \, \arctan \left (2 \, x + \sqrt{3}\right ) + \frac{1}{6} \, \arctan \left (2 \, x - \sqrt{3}\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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