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

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

[Out]

(2*Log[1 - x])/3 - Log[1 + x + x^2]/3

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Rubi [A]  time = 0.0106187, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1861, 31, 628} \[ \frac{2}{3} \log (1-x)-\frac{1}{3} \log \left (x^2+x+1\right ) \]

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 1861

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 3]], s = Denominato
r[Rt[-(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] && Ne
gQ[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 = 0.0026044, size = 22, normalized size = 1. \[ \frac{2}{3} \log (1-x)-\frac{1}{3} \log \left (x^2+x+1\right ) \]

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 = 17, normalized size = 0.8 \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 = 1.42956, size = 22, normalized size = 1. \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 = 0.973768, 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 = 0.086469, 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 = 1.11627, size = 23, normalized size = 1.05 \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))