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

Optimal. Leaf size=23 \[ \log \left (\frac {\log ^2\left (e^{2 x}\right )}{x \left (2+x-x^2\right )^2}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1594, 1628} \begin {gather*} -2 \log (2-x)+\log (x)-2 \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*Log[2 - x] + Log[x] - 2*Log[1 + x]

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

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.61 \begin {gather*} \log (x)-2 \log \left (2+x-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - 2*Log[2 + x - x^2]

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fricas [A]  time = 1.22, size = 14, normalized size = 0.61 \begin {gather*} -2 \, \log \left (x^{2} - x - 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(x^2 - x - 2) + log(x)

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giac [A]  time = 0.12, size = 18, normalized size = 0.78 \begin {gather*} -2 \, \log \left ({\left | x + 1 \right |}\right ) - 2 \, \log \left ({\left | x - 2 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 15, normalized size = 0.65

method	result	size
risch	\(\ln \relax (x )-2 \ln \left (x^{2}-x -2\right )\)	\(15\)
default	\(\ln \relax (x )-2 \ln \left (x -2\right )-2 \ln \left (x +1\right )\)	\(16\)
norman	\(\ln \relax (x )-2 \ln \left (x -2\right )-2 \ln \left (x +1\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2+x-2)/(x^3-x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 15, normalized size = 0.65 \begin {gather*} -2 \, \log \left (x + 1\right ) - 2 \, \log \left (x - 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(x + 1) - 2*log(x - 2) + log(x)

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mupad [B]  time = 0.07, size = 14, normalized size = 0.61 \begin {gather*} \ln \relax (x)-2\,\ln \left (x^2-x-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 2*log(x^2 - x - 2)

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sympy [A]  time = 0.11, size = 12, normalized size = 0.52 \begin {gather*} \log {\relax (x )} - 2 \log {\left (x^{2} - x - 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - 2*log(x**2 - x - 2)

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