3.85.100 $\int \frac{-2+x-3x^2}{-2x-x^2+x^3}dx$

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

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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} $$\begin{array}{c}-2\log (2-x)+\log (x)-2\log (x+1)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-2+x-3x^2}{x(-2-x+x^2)}dx\\ & =\int (-\frac{2}{-2+x}+\frac{1}{x}-\frac{2}{1+x})dx\\ & =-2\log (2-x)+\log (x)-2\log (1+x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.61 $$\begin{array}{c}\log (x)-2\log (2+x-x^2)\end{array}$$

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 $$\begin{array}{c}-2\log (x^2-x-2)+\log (x)\end{array}$$

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 $$\begin{array}{c}-2\log \left(|x+1|\right)-2\log \left(|x-2|\right)+\log \left(\left|x\right|\right)\end{array}$$

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	$\ln (x)-2\ln (x^2-x-2)$	$15$
default	$\ln (x)-2\ln (x-2)-2\ln (x+1)$	$16$
norman	$\ln (x)-2\ln (x-2)-2\ln (x+1)$	$16$

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 $$\begin{array}{c}-2\log (x+1)-2\log (x-2)+\log (x)\end{array}$$

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 $$\begin{array}{c}\ln (x)-2\ln (x^2-x-2)\end{array}$$

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 $$\begin{array}{c}\log (x)-2\log (x^2-x-2)\end{array}$$

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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