3.96 \(\int \frac{-1+x+x^2}{-6 x+x^2+x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac{1}{2} \log (2-x)+\frac{\log (x)}{6}+\frac{1}{3} \log (x+3) \]

[Out]

Log[2 - x]/2 + Log[x]/6 + Log[3 + x]/3

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Rubi [A]  time = 0.0365624, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {1594, 1628} \[ \frac{1}{2} \log (2-x)+\frac{\log (x)}{6}+\frac{1}{3} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x + x^2)/(-6*x + x^2 + x^3),x]

[Out]

Log[2 - x]/2 + Log[x]/6 + Log[3 + x]/3

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.005858, size = 25, normalized size = 1. \[ \frac{1}{2} \log (2-x)+\frac{\log (x)}{6}+\frac{1}{3} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x + x^2)/(-6*x + x^2 + x^3),x]

[Out]

Log[2 - x]/2 + Log[x]/6 + Log[3 + x]/3

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Maple [A]  time = 0.008, size = 18, normalized size = 0.7 \begin{align*}{\frac{\ln \left ( x \right ) }{6}}+{\frac{\ln \left ( 3+x \right ) }{3}}+{\frac{\ln \left ( -2+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+x-1)/(x^3+x^2-6*x),x)

[Out]

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

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Maxima [A]  time = 0.931829, size = 23, normalized size = 0.92 \begin{align*} \frac{1}{3} \, \log \left (x + 3\right ) + \frac{1}{2} \, \log \left (x - 2\right ) + \frac{1}{6} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x-1)/(x^3+x^2-6*x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.88495, size = 62, normalized size = 2.48 \begin{align*} \frac{1}{3} \, \log \left (x + 3\right ) + \frac{1}{2} \, \log \left (x - 2\right ) + \frac{1}{6} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x-1)/(x^3+x^2-6*x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.117081, size = 17, normalized size = 0.68 \begin{align*} \frac{\log{\left (x \right )}}{6} + \frac{\log{\left (x - 2 \right )}}{2} + \frac{\log{\left (x + 3 \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+x-1)/(x**3+x**2-6*x),x)

[Out]

log(x)/6 + log(x - 2)/2 + log(x + 3)/3

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Giac [A]  time = 1.06005, size = 27, normalized size = 1.08 \begin{align*} \frac{1}{3} \, \log \left ({\left | x + 3 \right |}\right ) + \frac{1}{2} \, \log \left ({\left | x - 2 \right |}\right ) + \frac{1}{6} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x-1)/(x^3+x^2-6*x),x, algorithm="giac")

[Out]

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