3.118 \(\int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {x^2}{2}-x-\log (1-x)+3 \log (x)+\log (x+2) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 27, 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} \[ \frac {x^2}{2}-x-\log (1-x)+3 \log (x)+\log (x+2) \]

Antiderivative was successfully verified.

[In]

Int[(-6 + 2*x + x^4)/(-2*x + x^2 + x^3),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {x^2}{2}-x-\log (1-x)+3 \log (x)+\log (x+2) \]

Antiderivative was successfully verified.

[In]

Integrate[(-6 + 2*x + x^4)/(-2*x + x^2 + x^3),x]

[Out]

-x + x^2/2 - Log[1 - x] + 3*Log[x] + Log[2 + x]

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fricas [A]  time = 0.38, size = 23, normalized size = 0.85 \[ \frac {1}{2} \, x^{2} - x + \log \left (x + 2\right ) - \log \left (x - 1\right ) + 3 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+2*x-6)/(x^3+x^2-2*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.01, size = 26, normalized size = 0.96 \[ \frac {1}{2} \, x^{2} - x + \log \left ({\left | x + 2 \right |}\right ) - \log \left ({\left | x - 1 \right |}\right ) + 3 \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+2*x-6)/(x^3+x^2-2*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 24, normalized size = 0.89 \[ \frac {x^{2}}{2}-x +3 \ln \relax (x )-\ln \left (x -1\right )+\ln \left (x +2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+2*x-6)/(x^3+x^2-2*x),x)

[Out]

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

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maxima [A]  time = 0.50, size = 23, normalized size = 0.85 \[ \frac {1}{2} \, x^{2} - x + \log \left (x + 2\right ) - \log \left (x - 1\right ) + 3 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+2*x-6)/(x^3+x^2-2*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 30, normalized size = 1.11 \[ 3\,\ln \relax (x)-x+\frac {x^2}{2}+\mathrm {atan}\left (\frac {192{}\mathrm {i}}{7\,\left (28\,x-40\right )}+\frac {9}{7}{}\mathrm {i}\right )\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + x^4 - 6)/(x^2 - 2*x + x^3),x)

[Out]

atan(192i/(7*(28*x - 40)) + 9i/7)*2i - x + 3*log(x) + x^2/2

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sympy [A]  time = 0.13, size = 20, normalized size = 0.74 \[ \frac {x^{2}}{2} - x + 3 \log {\relax (x )} - \log {\left (x - 1 \right )} + \log {\left (x + 2 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+2*x-6)/(x**3+x**2-2*x),x)

[Out]

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

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