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

Optimal. Leaf size=58 \[ -\frac{1-2 x}{2 \left (x^2-x+1\right )}-\frac{1-x}{2 \left (x^2-x+1\right )^2}-\frac{2 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0209552, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {638, 614, 618, 204} \[ -\frac{1-2 x}{2 \left (x^2-x+1\right )}-\frac{1-x}{2 \left (x^2-x+1\right )^2}-\frac{2 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)/(1 - x + x^2)^3,x]

[Out]

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

Mathematica [A]  time = 0.0278463, size = 49, normalized size = 0.84 \[ \frac{2 x^3-3 x^2+4 x-2}{2 \left (x^2-x+1\right )^2}+\frac{2 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/(1 - x + x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.003, size = 52, normalized size = 0.9 \begin{align*}{\frac{3\,x-3}{6\, \left ({x}^{2}-x+1 \right ) ^{2}}}+{\frac{-1+2\,x}{2\,{x}^{2}-2\,x+2}}+{\frac{2\,\sqrt{3}}{3}\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((1+x)/(x^2-x+1)^3,x)

[Out]

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

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Maxima [A]  time = 1.56341, size = 73, normalized size = 1.26 \begin{align*} \frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{2 \, x^{3} - 3 \, x^{2} + 4 \, x - 2}{2 \,{\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((1+x)/(x^2-x+1)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59001, size = 186, normalized size = 3.21 \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 - 6}{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((1+x)/(x^2-x+1)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.14832, size = 61, normalized size = 1.05 \begin{align*} \frac{2 x^{3} - 3 x^{2} + 4 x - 2}{2 x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 2} + \frac{2 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34664, size = 59, normalized size = 1.02 \begin{align*} \frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{2 \, x^{3} - 3 \, x^{2} + 4 \, x - 2}{2 \,{\left (x^{2} - x + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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