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

Optimal. Leaf size=61 \[ \frac {2 (x+1)}{3 \left (x^2-x+1\right )}-\frac {3}{2} \log \left (x^2-x+1\right )-\frac {1}{x}+3 \log (x)-\frac {7 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

-1/x+2/3*(1+x)/(x^2-x+1)+3*ln(x)-3/2*ln(x^2-x+1)-7/9*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1646, 1628, 634, 618, 204, 628} \[ \frac {2 (x+1)}{3 \left (x^2-x+1\right )}-\frac {3}{2} \log \left (x^2-x+1\right )-\frac {1}{x}+3 \log (x)-\frac {7 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + (2*(1 + x))/(3*(1 - x + x^2)) - (7*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 3*Log[x] - (3*Log[1 - x
+ x^2])/2

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 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 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]

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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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]

Rubi steps

\begin {align*} \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx &=\frac {2 (1+x)}{3 \left (1-x+x^2\right )}+\frac {1}{3} \int \frac {3+6 x+2 x^2}{x^2 \left (1-x+x^2\right )} \, dx\\ &=\frac {2 (1+x)}{3 \left (1-x+x^2\right )}+\frac {1}{3} \int \left (\frac {3}{x^2}+\frac {9}{x}+\frac {8-9 x}{1-x+x^2}\right ) \, dx\\ &=-\frac {1}{x}+\frac {2 (1+x)}{3 \left (1-x+x^2\right )}+3 \log (x)+\frac {1}{3} \int \frac {8-9 x}{1-x+x^2} \, dx\\ &=-\frac {1}{x}+\frac {2 (1+x)}{3 \left (1-x+x^2\right )}+3 \log (x)+\frac {7}{6} \int \frac {1}{1-x+x^2} \, dx-\frac {3}{2} \int \frac {-1+2 x}{1-x+x^2} \, dx\\ &=-\frac {1}{x}+\frac {2 (1+x)}{3 \left (1-x+x^2\right )}+3 \log (x)-\frac {3}{2} \log \left (1-x+x^2\right )-\frac {7}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=-\frac {1}{x}+\frac {2 (1+x)}{3 \left (1-x+x^2\right )}-\frac {7 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+3 \log (x)-\frac {3}{2} \log \left (1-x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 1.00 \[ \frac {2 (x+1)}{3 \left (x^2-x+1\right )}-\frac {3}{2} \log \left (x^2-x+1\right )-\frac {1}{x}+3 \log (x)+\frac {7 \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + (2*(1 + x))/(3*(1 - x + x^2)) + (7*ArcTan[(-1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 3*Log[x] - (3*Log[1 - x
+ x^2])/2

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fricas [A]  time = 1.20, size = 85, normalized size = 1.39 \[ \frac {14 \, \sqrt {3} {\left (x^{3} - x^{2} + x\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 6 \, x^{2} - 27 \, {\left (x^{3} - x^{2} + x\right )} \log \left (x^{2} - x + 1\right ) + 54 \, {\left (x^{3} - x^{2} + x\right )} \log \relax (x) + 30 \, x - 18}{18 \, {\left (x^{3} - x^{2} + x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(14*sqrt(3)*(x^3 - x^2 + x)*arctan(1/3*sqrt(3)*(2*x - 1)) - 6*x^2 - 27*(x^3 - x^2 + x)*log(x^2 - x + 1) +
54*(x^3 - x^2 + x)*log(x) + 30*x - 18)/(x^3 - x^2 + x)

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giac [A]  time = 0.15, size = 55, normalized size = 0.90 \[ \frac {7}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x^{2} - 5 \, x + 3}{3 \, {\left (x^{3} - x^{2} + x\right )}} - \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) + 3 \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/3*(x^2 - 5*x + 3)/(x^3 - x^2 + x) - 3/2*log(x^2 - x + 1) + 3*log
(abs(x))

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maple [A]  time = 0.01, size = 55, normalized size = 0.90 \[ \frac {7 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+3 \ln \relax (x )-\frac {3 \ln \left (x^{2}-x +1\right )}{2}-\frac {1}{x}-\frac {-\frac {2 x}{3}-\frac {2}{3}}{x^{2}-x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-2/3*x-2/3)/(x^2-x+1)-3/2*ln(x^2-x+1)+7/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-1/x+3*ln(x)

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maxima [A]  time = 0.96, size = 54, normalized size = 0.89 \[ \frac {7}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x^{2} - 5 \, x + 3}{3 \, {\left (x^{3} - x^{2} + x\right )}} - \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) + 3 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/3*(x^2 - 5*x + 3)/(x^3 - x^2 + x) - 3/2*log(x^2 - x + 1) + 3*log
(x)

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mupad [B]  time = 4.13, size = 68, normalized size = 1.11 \[ 3\,\ln \relax (x)-\frac {\frac {x^2}{3}-\frac {5\,x}{3}+1}{x^3-x^2+x}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,7{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {3}{2}+\frac {\sqrt {3}\,7{}\mathrm {i}}{18}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^2 + 1)/(x^2*(x^2 - x + 1)^2),x)

[Out]

3*log(x) - (x^2/3 - (5*x)/3 + 1)/(x - x^2 + x^3) - log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*7i)/18 + 3/2) + log
(x + (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*7i)/18 - 3/2)

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sympy [A]  time = 0.20, size = 65, normalized size = 1.07 \[ \frac {- x^{2} + 5 x - 3}{3 x^{3} - 3 x^{2} + 3 x} + 3 \log {\relax (x )} - \frac {3 \log {\left (x^{2} - x + 1 \right )}}{2} + \frac {7 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x**2 + 5*x - 3)/(3*x**3 - 3*x**2 + 3*x) + 3*log(x) - 3*log(x**2 - x + 1)/2 + 7*sqrt(3)*atan(2*sqrt(3)*x/3 -
sqrt(3)/3)/9

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