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

Optimal. Leaf size=32 \[ -\frac{1}{2 x^2}+\frac{1}{3} \log \left (x^2-x+1\right )+\frac{1}{x}-\frac{2}{3} \log (x+1) \]

[Out]

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

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Rubi [A]  time = 0.0326286, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1834, 628} \[ -\frac{1}{2 x^2}+\frac{1}{3} \log \left (x^2-x+1\right )+\frac{1}{x}-\frac{2}{3} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)/(x^3*(1 + x^3)),x]

[Out]

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

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

Mathematica [A]  time = 0.003987, size = 32, normalized size = 1. \[ -\frac{1}{2 x^2}+\frac{1}{3} \log \left (x^2-x+1\right )+\frac{1}{x}-\frac{2}{3} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)/(x^3*(1 + x^3)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 27, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,{x}^{2}}}+{x}^{-1}-{\frac{2\,\ln \left ( 1+x \right ) }{3}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.40062, size = 38, normalized size = 1.19 \begin{align*} \frac{2 \, x - 1}{2 \, x^{2}} + \frac{1}{3} \, \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^3/(x^3+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42257, size = 85, normalized size = 2.66 \begin{align*} \frac{2 \, x^{2} \log \left (x^{2} - x + 1\right ) - 4 \, x^{2} \log \left (x + 1\right ) + 6 \, x - 3}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.109497, size = 27, normalized size = 0.84 \begin{align*} - \frac{2 \log{\left (x + 1 \right )}}{3} + \frac{\log{\left (x^{2} - x + 1 \right )}}{3} + \frac{2 x - 1}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(x + 1)/3 + log(x**2 - x + 1)/3 + (2*x - 1)/(2*x**2)

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Giac [A]  time = 1.04714, size = 39, normalized size = 1.22 \begin{align*} \frac{2 \, x - 1}{2 \, x^{2}} + \frac{1}{3} \, \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^3/(x^3+1),x, algorithm="giac")

[Out]

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