3.20.27 \(\int \frac {x}{(1+x+x^2)^3} \, dx\)

Optimal. Leaf size=54 \[ -\frac {x+2}{6 \left (x^2+x+1\right )^2}-\frac {2 x+1}{6 \left (x^2+x+1\right )}-\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {638, 614, 618, 204} \begin {gather*} -\frac {x+2}{6 \left (x^2+x+1\right )^2}-\frac {2 x+1}{6 \left (x^2+x+1\right )}-\frac {2 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(1 + x + x^2)^3,x]

[Out]

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

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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 49, normalized size = 0.91 \begin {gather*} \frac {1}{18} \left (-\frac {3 \left (2 x^3+3 x^2+4 x+3\right )}{\left (x^2+x+1\right )^2}-4 \sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(1 + x + x^2)^3,x]

[Out]

((-3*(3 + 4*x + 3*x^2 + 2*x^3))/(1 + x + x^2)^2 - 4*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]])/18

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (1+x+x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x/(1 + x + x^2)^3,x]

[Out]

IntegrateAlgebraic[x/(1 + x + x^2)^3, x]

________________________________________________________________________________________

fricas [A]  time = 0.39, size = 71, normalized size = 1.31 \begin {gather*} -\frac {6 \, x^{3} + 4 \, \sqrt {3} {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 9 \, x^{2} + 12 \, x + 9}{18 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+x+1)^3,x, algorithm="fricas")

[Out]

-1/18*(6*x^3 + 4*sqrt(3)*(x^4 + 2*x^3 + 3*x^2 + 2*x + 1)*arctan(1/3*sqrt(3)*(2*x + 1)) + 9*x^2 + 12*x + 9)/(x^
4 + 2*x^3 + 3*x^2 + 2*x + 1)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 42, normalized size = 0.78 \begin {gather*} -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {2 \, x^{3} + 3 \, x^{2} + 4 \, x + 3}{6 \, {\left (x^{2} + x + 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+x+1)^3,x, algorithm="giac")

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/6*(2*x^3 + 3*x^2 + 4*x + 3)/(x^2 + x + 1)^2

________________________________________________________________________________________

maple [A]  time = 0.05, size = 48, normalized size = 0.89 \begin {gather*} -\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {-x -2}{6 \left (x^{2}+x +1\right )^{2}}-\frac {2 x +1}{6 \left (x^{2}+x +1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2+x+1)^3,x)

[Out]

1/6*(-x-2)/(x^2+x+1)^2-1/6*(2*x+1)/(x^2+x+1)-2/9*3^(1/2)*arctan(1/3*(2*x+1)*3^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.94, size = 54, normalized size = 1.00 \begin {gather*} -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {2 \, x^{3} + 3 \, x^{2} + 4 \, x + 3}{6 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2+x+1)^3,x, algorithm="maxima")

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/6*(2*x^3 + 3*x^2 + 4*x + 3)/(x^4 + 2*x^3 + 3*x^2 + 2*x + 1)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 56, normalized size = 1.04 \begin {gather*} -\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )}{9}-\frac {\frac {x^3}{3}+\frac {x^2}{2}+\frac {2\,x}{3}+\frac {1}{2}}{x^4+2\,x^3+3\,x^2+2\,x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x + x^2 + 1)^3,x)

[Out]

- (2*3^(1/2)*atan((2*3^(1/2)*x)/3 + 3^(1/2)/3))/9 - ((2*x)/3 + x^2/2 + x^3/3 + 1/2)/(2*x + 3*x^2 + 2*x^3 + x^4
+ 1)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 63, normalized size = 1.17 \begin {gather*} \frac {- 2 x^{3} - 3 x^{2} - 4 x - 3}{6 x^{4} + 12 x^{3} + 18 x^{2} + 12 x + 6} - \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**2+x+1)**3,x)

[Out]

(-2*x**3 - 3*x**2 - 4*x - 3)/(6*x**4 + 12*x**3 + 18*x**2 + 12*x + 6) - 2*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/
3)/9

________________________________________________________________________________________