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

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

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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 {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*}

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Mathematica [A]  time = 0.01, 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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IntegrateAlgebraic [A]  time = 0.03, size = 73, normalized size = 1.49 \[ -\frac {1}{12} \log \left (-x^2+\sqrt {3} x-1\right )+\frac {1}{6} \log \left (x^2+1\right )-\frac {1}{12} \log \left (x^2+\sqrt {3} x+1\right )-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(1 + x^6),x]

[Out]

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

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fricas [A]  time = 1.09, size = 40, normalized size = 0.82 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

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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giac [A]  time = 0.93, size = 40, normalized size = 0.82 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

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)

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maple [A]  time = 0.28, size = 39, normalized size = 0.80

method	result	size
risch	\(\frac {\ln \left (x^{2}+1\right )}{6}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}\)	\(39\)
default	\(-\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\ln \left (x^{2}+1\right )}{6}\)	\(41\)
meijerg	\(\frac {x^{2} \ln \left (1+\left (x^{6}\right )^{\frac {1}{3}}\right )}{6 \left (x^{6}\right )^{\frac {1}{3}}}-\frac {x^{2} \ln \left (1-\left (x^{6}\right )^{\frac {1}{3}}+\left (x^{6}\right )^{\frac {2}{3}}\right )}{12 \left (x^{6}\right )^{\frac {1}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{2-\left (x^{6}\right )^{\frac {1}{3}}}\right )}{6 \left (x^{6}\right )^{\frac {1}{3}}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 40, normalized size = 0.82 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

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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mupad [B]  time = 0.09, size = 52, normalized size = 1.06 \[ \frac {\ln \left (x^2+1\right )}{6}-\ln \left (x^2-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\ln \left (x^2+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^6 + 1),x)

[Out]

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

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sympy [A]  time = 0.15, size = 46, normalized size = 0.94 \[ \frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{6} \]

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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