3.127 \(\int \frac {x^2}{-6+x+x^2} \, dx\)

Optimal. Leaf size=20 \[ x+\frac {4}{5} \log (2-x)-\frac {9}{5} \log (x+3) \]

[Out]

x+4/5*ln(2-x)-9/5*ln(3+x)

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {703, 632, 31} \[ x+\frac {4}{5} \log (2-x)-\frac {9}{5} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(-6 + x + x^2),x]

[Out]

x + (4*Log[2 - x])/5 - (9*Log[3 + x])/5

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(a + b*x + c*x^2),
x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*
e, 0] && GtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {x^2}{-6+x+x^2} \, dx &=x+\int \frac {6-x}{-6+x+x^2} \, dx\\ &=x+\frac {4}{5} \int \frac {1}{-2+x} \, dx-\frac {9}{5} \int \frac {1}{3+x} \, dx\\ &=x+\frac {4}{5} \log (2-x)-\frac {9}{5} \log (3+x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \[ x+\frac {4}{5} \log (2-x)-\frac {9}{5} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(-6 + x + x^2),x]

[Out]

x + (4*Log[2 - x])/5 - (9*Log[3 + x])/5

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fricas [A]  time = 0.38, size = 14, normalized size = 0.70 \[ x - \frac {9}{5} \, \log \left (x + 3\right ) + \frac {4}{5} \, \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 9/5*log(x + 3) + 4/5*log(x - 2)

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giac [A]  time = 0.01, size = 16, normalized size = 0.80 \[ x - \frac {9}{5} \, \log \left ({\left | x + 3 \right |}\right ) + \frac {4}{5} \, \log \left ({\left | x - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 9/5*log(abs(x + 3)) + 4/5*log(abs(x - 2))

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maple [A]  time = 0.01, size = 15, normalized size = 0.75 \[ x +\frac {4 \ln \left (x -2\right )}{5}-\frac {9 \ln \left (x +3\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^2+x-6),x)

[Out]

x-9/5*ln(x+3)+4/5*ln(x-2)

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maxima [A]  time = 0.51, size = 14, normalized size = 0.70 \[ x - \frac {9}{5} \, \log \left (x + 3\right ) + \frac {4}{5} \, \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 9/5*log(x + 3) + 4/5*log(x - 2)

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mupad [B]  time = 0.04, size = 14, normalized size = 0.70 \[ x+\frac {4\,\ln \left (x-2\right )}{5}-\frac {9\,\ln \left (x+3\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x + x^2 - 6),x)

[Out]

x + (4*log(x - 2))/5 - (9*log(x + 3))/5

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sympy [A]  time = 0.11, size = 17, normalized size = 0.85 \[ x + \frac {4 \log {\left (x - 2 \right )}}{5} - \frac {9 \log {\left (x + 3 \right )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4*log(x - 2)/5 - 9*log(x + 3)/5

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