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

Optimal. Leaf size=68 \[ \frac {2 (2-x)}{3 \left (x^2-x+1\right )}-\frac {1}{2 x^2}-2 \log \left (x^2-x+1\right )-\frac {3}{x}+4 \log (x)+\frac {10 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

-1/2/x^2-3/x+2/3*(2-x)/(x^2-x+1)+4*ln(x)-2*ln(x^2-x+1)+10/9*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 68, 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} \[ \frac {2 (2-x)}{3 \left (x^2-x+1\right )}-\frac {1}{2 x^2}-2 \log \left (x^2-x+1\right )-\frac {3}{x}+4 \log (x)+\frac {10 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x + x^2)/(x^3*(1 - x + x^2)^2),x]

[Out]

-1/(2*x^2) - 3/x + (2*(2 - x))/(3*(1 - x + x^2)) + (10*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 4*Log[x] - 2*L
og[1 - x + x^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^3 \left (1-x+x^2\right )^2} \, dx &=\frac {2 (2-x)}{3 \left (1-x+x^2\right )}+\frac {1}{3} \int \frac {3+6 x+6 x^2-2 x^3}{x^3 \left (1-x+x^2\right )} \, dx\\ &=\frac {2 (2-x)}{3 \left (1-x+x^2\right )}+\frac {1}{3} \int \left (\frac {3}{x^3}+\frac {9}{x^2}+\frac {12}{x}+\frac {1-12 x}{1-x+x^2}\right ) \, dx\\ &=-\frac {1}{2 x^2}-\frac {3}{x}+\frac {2 (2-x)}{3 \left (1-x+x^2\right )}+4 \log (x)+\frac {1}{3} \int \frac {1-12 x}{1-x+x^2} \, dx\\ &=-\frac {1}{2 x^2}-\frac {3}{x}+\frac {2 (2-x)}{3 \left (1-x+x^2\right )}+4 \log (x)-\frac {5}{3} \int \frac {1}{1-x+x^2} \, dx-2 \int \frac {-1+2 x}{1-x+x^2} \, dx\\ &=-\frac {1}{2 x^2}-\frac {3}{x}+\frac {2 (2-x)}{3 \left (1-x+x^2\right )}+4 \log (x)-2 \log \left (1-x+x^2\right )+\frac {10}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=-\frac {1}{2 x^2}-\frac {3}{x}+\frac {2 (2-x)}{3 \left (1-x+x^2\right )}+\frac {10 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+4 \log (x)-2 \log \left (1-x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.97 \[ -\frac {2 (x-2)}{3 \left (x^2-x+1\right )}-\frac {1}{2 x^2}-2 \log \left (x^2-x+1\right )-\frac {3}{x}+4 \log (x)-\frac {10 \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + x^2)/(x^3*(1 - x + x^2)^2),x]

[Out]

-1/2*1/x^2 - 3/x - (2*(-2 + x))/(3*(1 - x + x^2)) - (10*ArcTan[(-1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) + 4*Log[x] - 2
*Log[1 - x + x^2]

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fricas [A]  time = 0.92, size = 98, normalized size = 1.44 \[ -\frac {66 \, x^{3} + 20 \, \sqrt {3} {\left (x^{4} - x^{3} + x^{2}\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 69 \, x^{2} + 36 \, {\left (x^{4} - x^{3} + x^{2}\right )} \log \left (x^{2} - x + 1\right ) - 72 \, {\left (x^{4} - x^{3} + x^{2}\right )} \log \relax (x) + 45 \, x + 9}{18 \, {\left (x^{4} - x^{3} + x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(66*x^3 + 20*sqrt(3)*(x^4 - x^3 + x^2)*arctan(1/3*sqrt(3)*(2*x - 1)) - 69*x^2 + 36*(x^4 - x^3 + x^2)*log
(x^2 - x + 1) - 72*(x^4 - x^3 + x^2)*log(x) + 45*x + 9)/(x^4 - x^3 + x^2)

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giac [A]  time = 0.16, size = 63, normalized size = 0.93 \[ -\frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {22 \, x^{3} - 23 \, x^{2} + 15 \, x + 3}{6 \, {\left (x^{2} - x + 1\right )} x^{2}} - 2 \, \log \left (x^{2} - x + 1\right ) + 4 \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*(22*x^3 - 23*x^2 + 15*x + 3)/((x^2 - x + 1)*x^2) - 2*log(x^2
- x + 1) + 4*log(abs(x))

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maple [A]  time = 0.01, size = 60, normalized size = 0.88 \[ -\frac {10 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+4 \ln \relax (x )-2 \ln \left (x^{2}-x +1\right )-\frac {3}{x}-\frac {1}{2 x^{2}}-\frac {\frac {2 x}{3}-\frac {4}{3}}{x^{2}-x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2/3*x-4/3)/(x^2-x+1)-2*ln(x^2-x+1)-10/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-1/2/x^2-3/x+4*ln(x)

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maxima [A]  time = 0.95, size = 63, normalized size = 0.93 \[ -\frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {22 \, x^{3} - 23 \, x^{2} + 15 \, x + 3}{6 \, {\left (x^{4} - x^{3} + x^{2}\right )}} - 2 \, \log \left (x^{2} - x + 1\right ) + 4 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*(22*x^3 - 23*x^2 + 15*x + 3)/(x^4 - x^3 + x^2) - 2*log(x^2 -
x + 1) + 4*log(x)

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mupad [B]  time = 0.10, size = 75, normalized size = 1.10 \[ 4\,\ln \relax (x)+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-2+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\frac {\frac {11\,x^3}{3}-\frac {23\,x^2}{6}+\frac {5\,x}{2}+\frac {1}{2}}{x^4-x^3+x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(x) + log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*5i)/9 - 2) - log(x + (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*5i)/9
+ 2) - ((5*x)/2 - (23*x^2)/6 + (11*x^3)/3 + 1/2)/(x^2 - x^3 + x^4)

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sympy [A]  time = 0.22, size = 71, normalized size = 1.04 \[ 4 \log {\relax (x )} - 2 \log {\left (x^{2} - x + 1 \right )} - \frac {10 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{9} + \frac {- 22 x^{3} + 23 x^{2} - 15 x - 3}{6 x^{4} - 6 x^{3} + 6 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(x) - 2*log(x**2 - x + 1) - 10*sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/9 + (-22*x**3 + 23*x**2 - 15*x - 3
)/(6*x**4 - 6*x**3 + 6*x**2)

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