3.23.59 \(\int \frac {x^2}{1+x+x^2} \, dx\) [2259]

Optimal. Leaf size=32 \[ x-\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (1+x+x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {717, 648, 632, 210, 642} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^2+x+1\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + x + x^2),x]

[Out]

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

Rule 210

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

Rule 632

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 642

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 648

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 717

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m - 1)/(c*(
m - 1))), x] + Dist[1/c, Int[(d + e*x)^(m - 2)*(Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, 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] && GtQ[m, 1]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 32, normalized size = 1.00 \begin {gather*} x-\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (1+x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + x + x^2),x]

[Out]

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

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Maple [A]
time = 0.82, size = 28, normalized size = 0.88

method	result	size
default	\(x -\frac {\ln \left (x^{2}+x +1\right )}{2}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(28\)
risch	\(x -\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}-\frac {\ln \left (4 x^{2}+4 x +4\right )}{2}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^2+x+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 27, normalized size = 0.84 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.34, size = 27, normalized size = 0.84 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.04, size = 36, normalized size = 1.12 \begin {gather*} 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(x**2/(x**2+x+1),x)

[Out]

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

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Giac [A]
time = 3.31, size = 27, normalized size = 0.84 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.03, size = 29, normalized size = 0.91 \begin {gather*} x-\frac {\ln \left (x^2+x+1\right )}{2}-\frac {\sqrt {3}\,\mathrm {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]

int(x^2/(x + x^2 + 1),x)

[Out]

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

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