3.48.76 \(\int \frac {8-4 e^{2+x}}{e^{4+x}-2 e^2 x} \, dx\)

Optimal. Leaf size=15 \[ -\frac {4 \log \left (e^{2+x}-2 x\right )}{e^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6684} \begin {gather*} -\frac {4 \log \left (e^{x+2}-2 x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Log[E^(2 + x) - 2*x])/E^2

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {4 \log \left (e^{2+x}-2 x\right )}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 15, normalized size = 1.00 \begin {gather*} -\frac {4 \log \left (e^{2+x}-2 x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Log[E^(2 + x) - 2*x])/E^2

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fricas [A]  time = 0.70, size = 15, normalized size = 1.00 \begin {gather*} -4 \, e^{\left (-2\right )} \log \left (-2 \, x e^{2} + e^{\left (x + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2+x)+8)/(exp(2)*exp(2+x)-2*exp(2)*x),x, algorithm="fricas")

[Out]

-4*e^(-2)*log(-2*x*e^2 + e^(x + 4))

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giac [A]  time = 0.14, size = 15, normalized size = 1.00 \begin {gather*} -4 \, e^{\left (-2\right )} \log \left (2 \, x - e^{\left (x + 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2+x)+8)/(exp(2)*exp(2+x)-2*exp(2)*x),x, algorithm="giac")

[Out]

-4*e^(-2)*log(2*x - e^(x + 2))

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maple [A]  time = 0.14, size = 18, normalized size = 1.20

method	result	size
norman	\(-4 \,{\mathrm e}^{-2} \ln \left (2 x -{\mathrm e}^{2+x}\right )\)	\(18\)
risch	\(8 \,{\mathrm e}^{-2}-4 \ln \left ({\mathrm e}^{2+x}-2 x \right ) {\mathrm e}^{-2}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(2+x)+8)/(exp(2)*exp(2+x)-2*exp(2)*x),x,method=_RETURNVERBOSE)

[Out]

-4/exp(2)*ln(2*x-exp(2+x))

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maxima [A]  time = 0.36, size = 17, normalized size = 1.13 \begin {gather*} -4 \, e^{\left (-2\right )} \log \left (2 \, x e^{2} - e^{\left (x + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2+x)+8)/(exp(2)*exp(2+x)-2*exp(2)*x),x, algorithm="maxima")

[Out]

-4*e^(-2)*log(2*x*e^2 - e^(x + 4))

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mupad [B]  time = 3.46, size = 15, normalized size = 1.00 \begin {gather*} -4\,\ln \left (2\,x-{\mathrm {e}}^{x+2}\right )\,{\mathrm {e}}^{-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(x + 2) - 8)/(exp(x + 2)*exp(2) - 2*x*exp(2)),x)

[Out]

-4*log(2*x - exp(x + 2))*exp(-2)

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sympy [A]  time = 0.12, size = 15, normalized size = 1.00 \begin {gather*} - \frac {4 \log {\left (- 2 x + e^{x + 2} \right )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2+x)+8)/(exp(2)*exp(2+x)-2*exp(2)*x),x)

[Out]

-4*exp(-2)*log(-2*x + exp(x + 2))

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