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

Optimal. Leaf size=33 \[ -\frac {1}{2} \log \left (x^2+x+1\right )+\log (x)-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {705, 29, 634, 618, 204, 628} \begin {gather*} -\frac {1}{2} \log \left (x^2+x+1\right )+\log (x)-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x + x^2)),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 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]

Rule 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \log \left (x^2+x+1\right )+\log (x)-\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x + x^2)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (1+x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x*(1 + x + x^2)),x]

[Out]

IntegrateAlgebraic[1/(x*(1 + x + x^2)), x]

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fricas [A]  time = 0.41, size = 28, normalized size = 0.85 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 29, normalized size = 0.88 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 29, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\ln \relax (x )-\frac {\ln \left (x^{2}+x +1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.01, size = 28, normalized size = 0.85 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 28, normalized size = 0.85 \begin {gather*} \ln \relax (x)-\frac {\ln \left (x^2+x+1\right )}{2}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\left (2\,x+1\right )}{3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x + x^2 + 1)),x)

[Out]

log(x) - log(x + x^2 + 1)/2 - (3^(1/2)*atan((3^(1/2)*(2*x + 1))/3))/3

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sympy [A]  time = 0.14, size = 37, normalized size = 1.12 \begin {gather*} \log {\relax (x )} - \frac {\log {\left (x^{2} + x + 1 \right )}}{2} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(x**2+x+1),x)

[Out]

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

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