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

Optimal. Leaf size=49 \[ \frac{1}{6} \log \left (x^2-x+1\right )-\frac{1}{x}-\log (x)+\frac{2}{3} \log (x+1)+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.049683, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {1834, 634, 618, 204, 628} \[ \frac{1}{6} \log \left (x^2-x+1\right )-\frac{1}{x}-\log (x)+\frac{2}{3} \log (x+1)+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)/(x^2*(1 + x^3)),x]

[Out]

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

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 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 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 \left (1+x^3\right )} \, dx &=\int \left (\frac{1}{x^2}-\frac{1}{x}+\frac{2}{3 (1+x)}+\frac{-2+x}{3 \left (1-x+x^2\right )}\right ) \, dx\\ &=-\frac{1}{x}-\log (x)+\frac{2}{3} \log (1+x)+\frac{1}{3} \int \frac{-2+x}{1-x+x^2} \, dx\\ &=-\frac{1}{x}-\log (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\\ &=-\frac{1}{x}-\log (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 )\\ &=-\frac{1}{x}-\frac{\tan ^{-1}\left (\frac{-1+2 x}{\sqrt{3}}\right )}{\sqrt{3}}-\log (x)+\frac{2}{3} \log (1+x)+\frac{1}{6} \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0153023, size = 60, normalized size = 1.22 \[ -\frac{1}{6} \log \left (x^2-x+1\right )+\frac{1}{3} \log \left (x^3+1\right )-\frac{1}{x}-\log (x)+\frac{1}{3} \log (x+1)-\frac{\tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)/(x^2*(1 + x^3)),x]

[Out]

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

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Maple [A]  time = 0.008, size = 44, normalized size = 0.9 \begin{align*}{\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 ) }-{x}^{-1}-\ln \left ( x \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]

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

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Maxima [A]  time = 1.47203, size = 58, normalized size = 1.18 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{x} + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac{2}{3} \, \log \left (x + 1\right ) - \log \left (x\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)) - 1/x + 1/6*log(x^2 - x + 1) + 2/3*log(x + 1) - log(x)

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Fricas [A]  time = 1.53385, size = 144, normalized size = 2.94 \begin{align*} -\frac{2 \, \sqrt{3} x \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - x \log \left (x^{2} - x + 1\right ) - 4 \, x \log \left (x + 1\right ) + 6 \, x \log \left (x\right ) + 6}{6 \, x} \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]

-1/6*(2*sqrt(3)*x*arctan(1/3*sqrt(3)*(2*x - 1)) - x*log(x^2 - x + 1) - 4*x*log(x + 1) + 6*x*log(x) + 6)/x

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Sympy [A]  time = 0.194682, size = 49, normalized size = 1. \begin{align*} - \log{\left (x \right )} + \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} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.06451, size = 61, normalized size = 1.24 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{x} + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac{2}{3} \, \log \left ({\left | x + 1 \right |}\right ) - \log \left ({\left | x \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)) - 1/x + 1/6*log(x^2 - x + 1) + 2/3*log(abs(x + 1)) - log(abs(x))