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

Optimal. Leaf size=17 \[ 5+\log \left (\left (-\frac {2}{3}+e+\frac {x}{2}\right ) x^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 631} \begin {gather*} \log (2 (2-3 e)-3 x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[2*(2 - 3*E) - 3*x] + 2*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 631

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8+12 e+9 x}{(-4+6 e) x+3 x^2} \, dx\\ &=\int \left (\frac {2}{x}+\frac {3}{-4+6 e+3 x}\right ) \, dx\\ &=\log (2 (2-3 e)-3 x)+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.82 \begin {gather*} \log (4-6 e-3 x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4 - 6*E - 3*x] + 2*Log[x]

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fricas [A]  time = 1.12, size = 15, normalized size = 0.88 \begin {gather*} \log \left (3 \, x + 6 \, e - 4\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*x + 6*e - 4) + 2*log(x)

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giac [A]  time = 0.24, size = 17, normalized size = 1.00 \begin {gather*} \log \left ({\left | 3 \, x + 6 \, e - 4 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(3*x + 6*e - 4)) + 2*log(abs(x))

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maple [A]  time = 0.46, size = 16, normalized size = 0.94

method	result	size
default	\(2 \ln \relax (x )+\ln \left (6 \,{\mathrm e}+3 x -4\right )\)	\(16\)
norman	\(2 \ln \relax (x )+\ln \left (6 \,{\mathrm e}+3 x -4\right )\)	\(16\)
risch	\(2 \ln \relax (x )+\ln \left (6 \,{\mathrm e}+3 x -4\right )\)	\(16\)
meijerg	\(\frac {18 \,{\mathrm e} \left (2 \,{\mathrm e}-\frac {4}{3}\right ) \left (\ln \relax (x )+\ln \relax (3)-\ln \relax (2)-\ln \left (3 \,{\mathrm e}-2\right )-\ln \left (1+\frac {3 x}{2 \left (3 \,{\mathrm e}-2\right )}\right )\right )}{\left (3 \,{\mathrm e}-2\right ) \left (6 \,{\mathrm e}-4\right )}+\frac {9 \left (2 \,{\mathrm e}-\frac {4}{3}\right ) \ln \left (1+\frac {3 x}{2 \left (3 \,{\mathrm e}-2\right )}\right )}{6 \,{\mathrm e}-4}-\frac {12 \left (2 \,{\mathrm e}-\frac {4}{3}\right ) \left (\ln \relax (x )+\ln \relax (3)-\ln \relax (2)-\ln \left (3 \,{\mathrm e}-2\right )-\ln \left (1+\frac {3 x}{2 \left (3 \,{\mathrm e}-2\right )}\right )\right )}{\left (6 \,{\mathrm e}-4\right ) \left (3 \,{\mathrm e}-2\right )}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*ln(x)+ln(6*exp(1)+3*x-4)

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maxima [A]  time = 0.45, size = 15, normalized size = 0.88 \begin {gather*} \log \left (3 \, x + 6 \, e - 4\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*x + 6*e - 4) + 2*log(x)

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mupad [B]  time = 1.83, size = 13, normalized size = 0.76 \begin {gather*} \ln \left (x+2\,\mathrm {e}-\frac {4}{3}\right )+2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 2*exp(1) - 4/3) + 2*log(x)

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sympy [A]  time = 0.47, size = 15, normalized size = 0.88 \begin {gather*} 2 \log {\relax (x )} + \log {\left (x - \frac {4}{3} + 2 e \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x) + log(x - 4/3 + 2*E)

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