### 3.160 $$\int \frac{x^3 (1+x+x^2)}{(1-x+x^2)^2} \, dx$$

Optimal. Leaf size=62 $\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (x^2-x+1\right )}+2 \log \left (x^2-x+1\right )+3 x+\frac{10 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}$

[Out]

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

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Rubi [A]  time = 0.0743509, antiderivative size = 62, 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 = {1660, 1657, 634, 618, 204, 628} $\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (x^2-x+1\right )}+2 \log \left (x^2-x+1\right )+3 x+\frac{10 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1660

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

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, 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{x^3 \left (1+x+x^2\right )}{\left (1-x+x^2\right )^2} \, dx &=\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \frac{-2+6 x+6 x^2+3 x^3}{1-x+x^2} \, dx\\ &=\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+\frac{1}{3} \int \left (9+3 x-\frac{11-12 x}{1-x+x^2}\right ) \, dx\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}-\frac{1}{3} \int \frac{11-12 x}{1-x+x^2} \, dx\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}-\frac{5}{3} \int \frac{1}{1-x+x^2} \, dx+2 \int \frac{-1+2 x}{1-x+x^2} \, dx\\ &=3 x+\frac{x^2}{2}+\frac{2 (2-x)}{3 \left (1-x+x^2\right )}+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 )\\ &=3 x+\frac{x^2}{2}+\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}}+2 \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0332674, size = 60, normalized size = 0.97 $\frac{x^2}{2}-\frac{2 (x-2)}{3 \left (x^2-x+1\right )}+2 \log \left (x^2-x+1\right )+3 x-\frac{10 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 53, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2}}+3\,x+{\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^3*(x^2+x+1)/(x^2-x+1)^2,x)

[Out]

1/2*x^2+3*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 = 1.43281, size = 69, normalized size = 1.11 \begin{align*} \frac{1}{2} \, x^{2} - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 3 \, x - \frac{2 \,{\left (x - 2\right )}}{3 \,{\left (x^{2} - x + 1\right )}} + 2 \, \log \left (x^{2} - x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3*x - 2/3*(x - 2)/(x^2 - x + 1) + 2*log(x^2 - x + 1)

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Fricas [A]  time = 2.34005, size = 204, normalized size = 3.29 \begin{align*} \frac{9 \, x^{4} + 45 \, x^{3} - 20 \, \sqrt{3}{\left (x^{2} - x + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - 45 \, x^{2} + 36 \,{\left (x^{2} - x + 1\right )} \log \left (x^{2} - x + 1\right ) + 42 \, x + 24}{18 \,{\left (x^{2} - x + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(9*x^4 + 45*x^3 - 20*sqrt(3)*(x^2 - x + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) - 45*x^2 + 36*(x^2 - x + 1)*log(
x^2 - x + 1) + 42*x + 24)/(x^2 - x + 1)

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Sympy [A]  time = 0.143439, size = 60, normalized size = 0.97 \begin{align*} \frac{x^{2}}{2} + 3 x - \frac{2 x - 4}{3 x^{2} - 3 x + 3} + 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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19867, size = 69, normalized size = 1.11 \begin{align*} \frac{1}{2} \, x^{2} - \frac{10}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 3 \, x - \frac{2 \,{\left (x - 2\right )}}{3 \,{\left (x^{2} - x + 1\right )}} + 2 \, \log \left (x^{2} - x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - 10/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3*x - 2/3*(x - 2)/(x^2 - x + 1) + 2*log(x^2 - x + 1)