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

Optimal. Leaf size=22 \[ \frac{2}{3} \log (x+1)-\frac{1}{3} \log \left (x^2-x+1\right ) \]

[Out]

(2*Log[1 + x])/3 - Log[1 - x + x^2]/3

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Rubi [A]  time = 0.0126704, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1860, 31, 628} \[ \frac{2}{3} \log (x+1)-\frac{1}{3} \log \left (x^2-x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)/(1 + x^3),x]

[Out]

(2*Log[1 + x])/3 - Log[1 - x + x^2]/3

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 31

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

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

Mathematica [A]  time = 0.0036034, size = 22, normalized size = 1. \[ \frac{2}{3} \log (x+1)-\frac{1}{3} \log \left (x^2-x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)/(1 + x^3),x]

[Out]

(2*Log[1 + x])/3 - Log[1 - x + x^2]/3

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Maple [A]  time = 0.004, size = 19, normalized size = 0.9 \begin{align*}{\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+1),x)

[Out]

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

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Maxima [A]  time = 1.47228, size = 24, normalized size = 1.09 \begin{align*} -\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+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.00901, size = 54, normalized size = 2.45 \begin{align*} -\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+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.091979, size = 17, normalized size = 0.77 \begin{align*} \frac{2 \log{\left (x + 1 \right )}}{3} - \frac{\log{\left (x^{2} - x + 1 \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x + 1)/3 - log(x**2 - x + 1)/3

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Giac [A]  time = 1.11873, size = 26, normalized size = 1.18 \begin{align*} -\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+1),x, algorithm="giac")

[Out]

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