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

Optimal. Leaf size=49 \[ \frac{1}{6} \log \left (x^2+1\right )-\frac{1}{12} \log \left (x^4-x^2+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0369044, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {275, 200, 31, 634, 618, 204, 628} \[ \frac{1}{6} \log \left (x^2+1\right )-\frac{1}{12} \log \left (x^4-x^2+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(1 + x^6),x]

[Out]

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.0140413, size = 78, normalized size = 1.59 \[ \frac{1}{12} \left (2 \log \left (x^2+1\right )-\log \left (x^2-\sqrt{3} x+1\right )-\log \left (x^2+\sqrt{3} x+1\right )-2 \sqrt{3} \tan ^{-1}\left (\sqrt{3}-2 x\right )-2 \sqrt{3} \tan ^{-1}\left (2 x+\sqrt{3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(1 + x^6),x]

[Out]

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

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Maple [A]  time = 0.006, size = 41, normalized size = 0.8 \begin{align*}{\frac{\ln \left ({x}^{2}+1 \right ) }{6}}-{\frac{\ln \left ({x}^{4}-{x}^{2}+1 \right ) }{12}}+{\frac{\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*ln(x^2+1)-1/12*ln(x^4-x^2+1)+1/6*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))

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Maxima [A]  time = 1.41578, size = 54, normalized size = 1.1 \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) - \frac{1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac{1}{6} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/12*log(x^4 - x^2 + 1) + 1/6*log(x^2 + 1)

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Fricas [A]  time = 2.10148, size = 122, normalized size = 2.49 \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) - \frac{1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac{1}{6} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/12*log(x^4 - x^2 + 1) + 1/6*log(x^2 + 1)

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Sympy [A]  time = 0.129665, size = 46, normalized size = 0.94 \begin{align*} \frac{\log{\left (x^{2} + 1 \right )}}{6} - \frac{\log{\left (x^{4} - x^{2} + 1 \right )}}{12} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{2}}{3} - \frac{\sqrt{3}}{3} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**2 + 1)/6 - log(x**4 - x**2 + 1)/12 + sqrt(3)*atan(2*sqrt(3)*x**2/3 - sqrt(3)/3)/6

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Giac [A]  time = 1.07139, size = 54, normalized size = 1.1 \begin{align*} \frac{1}{6} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) - \frac{1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac{1}{6} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/12*log(x^4 - x^2 + 1) + 1/6*log(x^2 + 1)