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

Optimal. Leaf size=56 \[ -\frac{2 (1-2 x)}{3 \left (x^2-x+1\right )}-\frac{1}{2} \log \left (x^2-x+1\right )+\log (x)-\frac{11 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0894687, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1646, 800, 634, 618, 204, 628} \[ -\frac{2 (1-2 x)}{3 \left (x^2-x+1\right )}-\frac{1}{2} \log \left (x^2-x+1\right )+\log (x)-\frac{11 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1646

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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{1+x+x^2}{x \left (1-x+x^2\right )^2} \, dx &=-\frac{2 (1-2 x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \frac{3+4 x}{x \left (1-x+x^2\right )} \, dx\\ &=-\frac{2 (1-2 x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \left (\frac{3}{x}+\frac{7-3 x}{1-x+x^2}\right ) \, dx\\ &=-\frac{2 (1-2 x)}{3 \left (1-x+x^2\right )}+\log (x)+\frac{1}{3} \int \frac{7-3 x}{1-x+x^2} \, dx\\ &=-\frac{2 (1-2 x)}{3 \left (1-x+x^2\right )}+\log (x)-\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 (1-2 x)}{3 \left (1-x+x^2\right )}+\log (x)-\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 (1-2 x)}{3 \left (1-x+x^2\right )}-\frac{11 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+\log (x)-\frac{1}{2} \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0249741, size = 56, normalized size = 1. \[ \frac{2 (2 x-1)}{3 \left (x^2-x+1\right )}-\frac{1}{2} \log \left (x^2-x+1\right )+\log (x)+\frac{11 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.048, size = 48, normalized size = 0.9 \begin{align*} \ln \left ( x \right ) -{\frac{1}{{x}^{2}-x+1} \left ( -{\frac{4\,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^2+x+1)/x/(x^2-x+1)^2,x)

[Out]

ln(x)-(-4/3*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))

________________________________________________________________________________________

Maxima [A]  time = 1.53454, size = 63, 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 (2 \, x - 1\right )}}{3 \,{\left (x^{2} - x + 1\right )}} - \frac{1}{2} \, \log \left (x^{2} - x + 1\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.62718, size = 203, normalized size = 3.62 \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 ) + 18 \,{\left (x^{2} - x + 1\right )} \log \left (x\right ) + 24 \, x - 12}{18 \,{\left (x^{2} - x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x+1)/x/(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) + 18*(x^2 - x
+ 1)*log(x) + 24*x - 12)/(x^2 - x + 1)

________________________________________________________________________________________

Sympy [A]  time = 0.178549, size = 54, normalized size = 0.96 \begin{align*} \frac{4 x - 2}{3 x^{2} - 3 x + 3} + \log{\left (x \right )} - \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**2+x+1)/x/(x**2-x+1)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.23469, size = 65, normalized size = 1.16 \begin{align*} \frac{11}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{2 \,{\left (2 \, x - 1\right )}}{3 \,{\left (x^{2} - x + 1\right )}} - \frac{1}{2} \, \log \left (x^{2} - x + 1\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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