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

Optimal. Leaf size=23 \[ -\log (1-x)+\frac {\log (x)}{2}+\frac {3}{2} \log (x+2) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1594, 1628} \[ -\log (1-x)+\frac {\log (x)}{2}+\frac {3}{2} \log (x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ -\log (1-x)+\frac {\log (x)}{2}+\frac {3}{2} \log (x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 17, normalized size = 0.74 \[ \frac {3}{2} \, \log \left (x + 2\right ) - \log \left (x - 1\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 20, normalized size = 0.87 \[ \frac {3}{2} \, \log \left ({\left | x + 2 \right |}\right ) - \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 18, normalized size = 0.78 \[ \frac {\ln \relax (x )}{2}-\ln \left (x -1\right )+\frac {3 \ln \left (x +2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.40, size = 17, normalized size = 0.74 \[ \frac {3}{2} \, \log \left (x + 2\right ) - \log \left (x - 1\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.19, size = 17, normalized size = 0.74 \[ \frac {3\,\ln \left (x+2\right )}{2}-\ln \left (x-1\right )+\frac {\ln \relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.14, size = 17, normalized size = 0.74 \[ \frac {\log {\relax (x )}}{2} - \log {\left (x - 1 \right )} + \frac {3 \log {\left (x + 2 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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