### 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*(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]

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Rubi [A]  time = 0.109582, antiderivative size = 68, 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, 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 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 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 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^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*}

Mathematica [A]  time = 0.029881, 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*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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Maple [A]  time = 0.048, size = 60, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{x}^{2}}}-3\,{x}^{-1}+4\,\ln \left ( x \right ) -{\frac{1}{{x}^{2}-x+1} \left ({\frac{2\,x}{3}}-{\frac{4}{3}} \right ) }-2\,\ln \left ({x}^{2}-x+1 \right ) -{\frac{10\,\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^3/(x^2-x+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.50708, size = 85, normalized size = 1.25 \begin{align*} -\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 \left (x\right ) \end{align*}

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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Fricas [A]  time = 1.74637, size = 250, normalized size = 3.68 \begin{align*} -\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 \left (x\right ) + 45 \, x + 9}{18 \,{\left (x^{4} - x^{3} + x^{2}\right )}} \end{align*}

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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Sympy [A]  time = 0.187243, size = 71, normalized size = 1.04 \begin{align*} 4 \log{\left (x \right )} - 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}} \end{align*}

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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Giac [A]  time = 1.26833, size = 85, normalized size = 1.25 \begin{align*} -\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 ) \end{align*}

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))