\(\int \frac {x}{(1+x+x^2)^3} \, dx\) [2258]

Optimal result

Integrand size = 10, antiderivative size = 54 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2+x}{6 \left (1+x+x^2\right )^2}-\frac {1+2 x}{6 \left (1+x+x^2\right )}-\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {652, 628, 632, 210} \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {x+2}{6 \left (x^2+x+1\right )^2}-\frac {2 x+1}{6 \left (x^2+x+1\right )} \]

[In]

Int[x/(1 + x + x^2)^3,x]

[Out]

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

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 628

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 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 652

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)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), 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*} \text {integral}& = -\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} \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=\frac {1}{18} \left (-\frac {3 \left (3+4 x+3 x^2+2 x^3\right )}{\left (1+x+x^2\right )^2}-4 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {-\frac {1}{3} x^{3}-\frac {1}{2} x^{2}-\frac {2}{3} x -\frac {1}{2}}{\left (x^{2}+x +1\right )^{2}}-\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\)	\(42\)
default	\(\frac {-2-x}{6 \left (x^{2}+x +1\right )^{2}}-\frac {1+2 x}{6 \left (x^{2}+x +1\right )}-\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\)	\(48\)

[In]

int(x/(x^2+x+1)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.31 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {6 \, x^{3} + 4 \, \sqrt {3} {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 9 \, x^{2} + 12 \, x + 9}{18 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=\frac {- 2 x^{3} - 3 x^{2} - 4 x - 3}{6 x^{4} + 12 x^{3} + 18 x^{2} + 12 x + 6} - \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {2 \, x^{3} + 3 \, x^{2} + 4 \, x + 3}{6 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {2 \, x^{3} + 3 \, x^{2} + 4 \, x + 3}{6 \, {\left (x^{2} + x + 1\right )}^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )}{9}-\frac {\frac {x^3}{3}+\frac {x^2}{2}+\frac {2\,x}{3}+\frac {1}{2}}{x^4+2\,x^3+3\,x^2+2\,x+1} \]

[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)