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

Optimal. Leaf size=44 \[ \frac{1}{6} \log \left (x^2-x+1\right )-x+\frac{2}{3} \log (x+1)-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

-x - ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] + (2*Log[1 + x])/3 + Log[1 - x + x^2]/6

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Rubi [A]  time = 0.0598778, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {1887, 1874, 31, 634, 618, 204, 628} \[ \frac{1}{6} \log \left (x^2-x+1\right )-x+\frac{2}{3} \log (x+1)-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[((1 - x)*x^2)/(1 + x^3),x]

[Out]

-x - ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] + (2*Log[1 + x])/3 + Log[1 - x + x^2]/6

Rule 1887

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rule 1874

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2], q = (a/b)^(1/3)}, Dist[(q*(A - B*q + C*q^2))/(3*a), Int[1/(q + x), x], x] + Dist[q/(3*a), Int[(q*(2*A + B
*q - C*q^2) - (A - B*q - 2*C*q^2)*x)/(q^2 - q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A - B*q + C*q^2
, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2] && GtQ[a/b, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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

Mathematica [A]  time = 0.0086687, size = 53, normalized size = 1.2 \[ -\frac{1}{6} \log \left (x^2-x+1\right )+\frac{1}{3} \log \left (x^3+1\right )-x+\frac{1}{3} \log (x+1)+\frac{\tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - x)*x^2)/(1 + x^3),x]

[Out]

-x + ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] + Log[1 + x]/3 - Log[1 - x + x^2]/6 + Log[1 + x^3]/3

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Maple [A]  time = 0.003, size = 38, normalized size = 0.9 \begin{align*} -x+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{6}}+{\frac{\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{2\,\ln \left ( 1+x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x+1/6*ln(x^2-x+1)+1/3*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+2/3*ln(1+x)

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Maxima [A]  time = 1.41157, size = 50, normalized size = 1.14 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - x + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac{2}{3} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - x + 1/6*log(x^2 - x + 1) + 2/3*log(x + 1)

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Fricas [A]  time = 1.49866, size = 117, normalized size = 2.66 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - x + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac{2}{3} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - x + 1/6*log(x^2 - x + 1) + 2/3*log(x + 1)

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Sympy [A]  time = 0.123329, size = 44, normalized size = 1. \begin{align*} - x + \frac{2 \log{\left (x + 1 \right )}}{3} + \frac{\log{\left (x^{2} - x + 1 \right )}}{6} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(x + 1)/3 + log(x**2 - x + 1)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/3

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Giac [A]  time = 1.06137, size = 51, normalized size = 1.16 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - x + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac{2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - x + 1/6*log(x^2 - x + 1) + 2/3*log(abs(x + 1))