### 3.2258 $$\int \frac{x}{(1+x+x^2)^3} \, dx$$

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

[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])

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Rubi [A]  time = 0.0205593, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.4, Rules used = {638, 614, 618, 204} $-\frac{x+2}{6 \left (x^2+x+1\right )^2}-\frac{2 x+1}{6 \left (x^2+x+1\right )}-\frac{2 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}}$

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 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]

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 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])

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 = 0.0253443, size = 49, normalized size = 0.91 $\frac{1}{18} \left (-\frac{3 \left (2 x^3+3 x^2+4 x+3\right )}{\left (x^2+x+1\right )^2}-4 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right )$

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

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Maple [A]  time = 0.041, size = 48, normalized size = 0.9 \begin{align*}{\frac{-2-x}{6\, \left ({x}^{2}+x+1 \right ) ^{2}}}-{\frac{1+2\,x}{6\,{x}^{2}+6\,x+6}}-{\frac{2\,\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A]  time = 1.44099, size = 73, normalized size = 1.35 \begin{align*} -\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{align*}

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)

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Fricas [A]  time = 2.87343, size = 189, normalized size = 3.5 \begin{align*} -\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{align*}

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)

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Sympy [A]  time = 0.135304, size = 63, normalized size = 1.17 \begin{align*} - \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{align*}

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

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Giac [A]  time = 1.1185, size = 57, normalized size = 1.06 \begin{align*} -\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{align*}

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