3.120 \(\int \frac{1+x+4 x^2}{-1+x^3} \, dx\)

Optimal. Leaf size=16 \[ \log \left (x^2+x+1\right )+2 \log (1-x) \]

[Out]

2*Log[1 - x] + Log[1 + x + x^2]

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Rubi [A]  time = 0.0167353, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1875, 31, 628} \[ \log \left (x^2+x+1\right )+2 \log (1-x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x + 4*x^2)/(-1 + x^3),x]

[Out]

2*Log[1 - x] + Log[1 + x + x^2]

Rule 1875

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] && LtQ[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 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+4 x^2}{-1+x^3} \, dx &=-\left (\frac{1}{3} \int \frac{-3-6 x}{1+x+x^2} \, dx\right )-2 \int \frac{1}{1-x} \, dx\\ &=2 \log (1-x)+\log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0037856, size = 16, normalized size = 1. \[ \log \left (x^2+x+1\right )+2 \log (1-x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + 4*x^2)/(-1 + x^3),x]

[Out]

2*Log[1 - x] + Log[1 + x + x^2]

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Maple [A]  time = 0.003, size = 15, normalized size = 0.9 \begin{align*} 2\,\ln \left ( -1+x \right ) +\ln \left ({x}^{2}+x+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+x+1)/(x^3-1),x)

[Out]

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

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Maxima [A]  time = 1.40583, size = 19, normalized size = 1.19 \begin{align*} \log \left (x^{2} + x + 1\right ) + 2 \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+x+1)/(x^3-1),x, algorithm="maxima")

[Out]

log(x^2 + x + 1) + 2*log(x - 1)

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Fricas [A]  time = 0.747851, size = 45, normalized size = 2.81 \begin{align*} \log \left (x^{2} + x + 1\right ) + 2 \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+x+1)/(x^3-1),x, algorithm="fricas")

[Out]

log(x^2 + x + 1) + 2*log(x - 1)

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Sympy [A]  time = 0.085175, size = 14, normalized size = 0.88 \begin{align*} 2 \log{\left (x - 1 \right )} + \log{\left (x^{2} + x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+x+1)/(x**3-1),x)

[Out]

2*log(x - 1) + log(x**2 + x + 1)

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Giac [A]  time = 1.08864, size = 20, normalized size = 1.25 \begin{align*} \log \left (x^{2} + x + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+x+1)/(x^3-1),x, algorithm="giac")

[Out]

log(x^2 + x + 1) + 2*log(abs(x - 1))