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

Optimal. Leaf size=13 \[ \frac {\log (3)}{3+e^x-2 x} \]

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Rubi [A]  time = 0.10, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6688, 12, 6686} \begin {gather*} \frac {\log (3)}{-2 x+e^x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[3]/(3 + E^x - 2*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} \frac {\log (3)}{3+e^x-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[3]/(3 + E^x - 2*x)

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fricas [A]  time = 1.42, size = 15, normalized size = 1.15 \begin {gather*} -\frac {\log \relax (3)}{2 \, x - e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(3)/(2*x - e^x - 3)

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giac [A]  time = 0.17, size = 15, normalized size = 1.15 \begin {gather*} -\frac {\log \relax (3)}{2 \, x - e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(3)/(2*x - e^x - 3)

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maple [A]  time = 0.07, size = 16, normalized size = 1.23




method result size



norman \(-\frac {\ln \relax (3)}{2 x -3-{\mathrm e}^{x}}\) \(16\)
risch \(-\frac {\ln \relax (3)}{2 x -3-{\mathrm e}^{x}}\) \(16\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(3)/(2*x-3-exp(x))

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maxima [A]  time = 0.52, size = 15, normalized size = 1.15 \begin {gather*} -\frac {\log \relax (3)}{2 \, x - e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(3)/(2*x - e^x - 3)

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mupad [B]  time = 2.28, size = 12, normalized size = 0.92 \begin {gather*} \frac {\ln \relax (3)}{{\mathrm {e}}^x-2\,x+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3)/(exp(x) - 2*x + 3)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.77 \begin {gather*} \frac {\log {\relax (3 )}}{- 2 x + e^{x} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3)/(-2*x + exp(x) + 3)

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