3.54.64 \(\int \frac {-6 e+8 x-24 x^2+18 x^3}{4-12 x+9 x^2} \, dx\)

Optimal. Leaf size=30 \[ x^2+4 \left (-6-\frac {e}{-1+\left (-5+\frac {5-x}{x}\right ) x}+\log (3)\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 0.47, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 1850} \begin {gather*} x^2-\frac {2 e}{2-3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*E + 8*x - 24*x^2 + 18*x^3)/(4 - 12*x + 9*x^2),x]

[Out]

(-2*E)/(2 - 3*x) + x^2

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6 e+8 x-24 x^2+18 x^3}{(-2+3 x)^2} \, dx\\ &=\int \left (2 x-\frac {6 e}{(-2+3 x)^2}\right ) \, dx\\ &=-\frac {2 e}{2-3 x}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.93 \begin {gather*} \frac {18 e+(2-3 x)^2 (2+3 x)}{9 (-2+3 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*E + 8*x - 24*x^2 + 18*x^3)/(4 - 12*x + 9*x^2),x]

[Out]

(18*E + (2 - 3*x)^2*(2 + 3*x))/(9*(-2 + 3*x))

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fricas [A]  time = 1.12, size = 23, normalized size = 0.77 \begin {gather*} \frac {3 \, x^{3} - 2 \, x^{2} + 2 \, e}{3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(1)+18*x^3-24*x^2+8*x)/(9*x^2-12*x+4),x, algorithm="fricas")

[Out]

(3*x^3 - 2*x^2 + 2*e)/(3*x - 2)

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giac [A]  time = 0.17, size = 15, normalized size = 0.50 \begin {gather*} x^{2} + \frac {2 \, e}{3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(1)+18*x^3-24*x^2+8*x)/(9*x^2-12*x+4),x, algorithm="giac")

[Out]

x^2 + 2*e/(3*x - 2)

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maple [A]  time = 0.30, size = 14, normalized size = 0.47

method	result	size
risch	\(x^{2}+\frac {2 \,{\mathrm e}}{3 \left (x -\frac {2}{3}\right )}\)	\(14\)
default	\(x^{2}+\frac {2 \,{\mathrm e}}{3 x -2}\)	\(16\)
gosper	\(\frac {3 x^{3}-2 x^{2}+2 \,{\mathrm e}}{3 x -2}\)	\(24\)
norman	\(\frac {3 x \,{\mathrm e}-2 x^{2}+3 x^{3}}{3 x -2}\)	\(25\)
meijerg	\(-\frac {3 \,{\mathrm e} x}{2 \left (1-\frac {3 x}{2}\right )}+\frac {x \left (-\frac {9}{2} x^{2}-9 x +12\right )}{3-\frac {9 x}{2}}-\frac {8 x \left (-\frac {9 x}{2}+6\right )}{9 \left (1-\frac {3 x}{2}\right )}+\frac {4 x}{3 \left (1-\frac {3 x}{2}\right )}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*exp(1)+18*x^3-24*x^2+8*x)/(9*x^2-12*x+4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 15, normalized size = 0.50 \begin {gather*} x^{2} + \frac {2 \, e}{3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(1)+18*x^3-24*x^2+8*x)/(9*x^2-12*x+4),x, algorithm="maxima")

[Out]

x^2 + 2*e/(3*x - 2)

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mupad [B]  time = 0.05, size = 15, normalized size = 0.50 \begin {gather*} \frac {2\,\mathrm {e}}{3\,x-2}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - 6*exp(1) - 24*x^2 + 18*x^3)/(9*x^2 - 12*x + 4),x)

[Out]

(2*exp(1))/(3*x - 2) + x^2

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sympy [A]  time = 0.09, size = 12, normalized size = 0.40 \begin {gather*} x^{2} + \frac {2 e}{3 x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(1)+18*x**3-24*x**2+8*x)/(9*x**2-12*x+4),x)

[Out]

x**2 + 2*E/(3*x - 2)

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