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

Optimal. Leaf size=52 \[ -\frac{2 (x+1)}{3 \left (x^2-x+1\right )}+\frac{1}{2} \log \left (x^2-x+1\right )-\frac{11 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

(-2*(1 + x))/(3*(1 - x + x^2)) - (11*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + Log[1 - x + x^2]/2

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Rubi [A]  time = 0.0515259, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {1660, 634, 618, 204, 628} \[ -\frac{2 (x+1)}{3 \left (x^2-x+1\right )}+\frac{1}{2} \log \left (x^2-x+1\right )-\frac{11 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + x))/(3*(1 - x + x^2)) - (11*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + Log[1 - x + x^2]/2

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

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 \left (1+x+x^2\right )}{\left (1-x+x^2\right )^2} \, dx &=-\frac{2 (1+x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \frac{4+3 x}{1-x+x^2} \, dx\\ &=-\frac{2 (1+x)}{3 \left (1-x+x^2\right )}+\frac{1}{2} \int \frac{-1+2 x}{1-x+x^2} \, dx+\frac{11}{6} \int \frac{1}{1-x+x^2} \, dx\\ &=-\frac{2 (1+x)}{3 \left (1-x+x^2\right )}+\frac{1}{2} \log \left (1-x+x^2\right )-\frac{11}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=-\frac{2 (1+x)}{3 \left (1-x+x^2\right )}-\frac{11 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{1}{2} \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0195145, size = 52, normalized size = 1. \[ -\frac{2 (x+1)}{3 \left (x^2-x+1\right )}+\frac{1}{2} \log \left (x^2-x+1\right )+\frac{11 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + x))/(3*(1 - x + x^2)) + (11*ArcTan[(-1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) + Log[1 - x + x^2]/2

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Maple [A]  time = 0.047, size = 45, normalized size = 0.9 \begin{align*}{\frac{1}{{x}^{2}-x+1} \left ( -{\frac{2\,x}{3}}-{\frac{2}{3}} \right ) }+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{2}}+{\frac{11\,\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2/3*x-2/3)/(x^2-x+1)+1/2*ln(x^2-x+1)+11/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))

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Maxima [A]  time = 1.49714, size = 58, normalized size = 1.12 \begin{align*} \frac{11}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{2 \,{\left (x + 1\right )}}{3 \,{\left (x^{2} - x + 1\right )}} + \frac{1}{2} \, \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 2/3*(x + 1)/(x^2 - x + 1) + 1/2*log(x^2 - x + 1)

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Fricas [A]  time = 2.34933, size = 167, normalized size = 3.21 \begin{align*} \frac{22 \, \sqrt{3}{\left (x^{2} - x + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 9 \,{\left (x^{2} - x + 1\right )} \log \left (x^{2} - x + 1\right ) - 12 \, x - 12}{18 \,{\left (x^{2} - x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(22*sqrt(3)*(x^2 - x + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) + 9*(x^2 - x + 1)*log(x^2 - x + 1) - 12*x - 12)/(
x^2 - x + 1)

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Sympy [A]  time = 0.13767, size = 51, normalized size = 0.98 \begin{align*} - \frac{2 x + 2}{3 x^{2} - 3 x + 3} + \frac{\log{\left (x^{2} - x + 1 \right )}}{2} + \frac{11 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*x + 2)/(3*x**2 - 3*x + 3) + log(x**2 - x + 1)/2 + 11*sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/9

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Giac [A]  time = 1.33756, size = 58, normalized size = 1.12 \begin{align*} \frac{11}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{2 \,{\left (x + 1\right )}}{3 \,{\left (x^{2} - x + 1\right )}} + \frac{1}{2} \, \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 2/3*(x + 1)/(x^2 - x + 1) + 1/2*log(x^2 - x + 1)