3.10.51 \(\int \frac {e^{-2 x} (-e^2+e^2 (-2 e-2 x) \log (e+x))}{(2 e+2 x) \log ^2(e+x)} \, dx\)

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

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Rubi [A]  time = 0.13, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2288} \begin {gather*} \frac {e^{2-2 x}}{2 \log (x+e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 - 2*x)/(2*Log[E + x])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{2-2 x}}{2 \log (e+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 - 2*x)/(2*Log[E + x])

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fricas [A]  time = 0.96, size = 15, normalized size = 0.88 \begin {gather*} \frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)-2*x)*exp(2)*log(x+exp(1))-exp(2))/(2*exp(1)+2*x)/exp(x)^2/log(x+exp(1))^2,x, algorithm="
fricas")

[Out]

1/2*e^(-2*x + 2)/log(x + e)

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giac [A]  time = 0.34, size = 15, normalized size = 0.88 \begin {gather*} \frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)-2*x)*exp(2)*log(x+exp(1))-exp(2))/(2*exp(1)+2*x)/exp(x)^2/log(x+exp(1))^2,x, algorithm="
giac")

[Out]

1/2*e^(-2*x + 2)/log(x + e)

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maple [A]  time = 0.15, size = 16, normalized size = 0.94

method	result	size
norman	\(\frac {{\mathrm e}^{2} {\mathrm e}^{-2 x}}{2 \ln \left (x +{\mathrm e}\right )}\)	\(16\)
risch	\(\frac {{\mathrm e}^{-2 x +2}}{2 \ln \left (x +{\mathrm e}\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(1)-2*x)*exp(2)*ln(x+exp(1))-exp(2))/(2*exp(1)+2*x)/exp(x)^2/ln(x+exp(1))^2,x,method=_RETURNVERBOS
E)

[Out]

1/2*exp(2)/exp(x)^2/ln(x+exp(1))

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maxima [A]  time = 0.43, size = 15, normalized size = 0.88 \begin {gather*} \frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)-2*x)*exp(2)*log(x+exp(1))-exp(2))/(2*exp(1)+2*x)/exp(x)^2/log(x+exp(1))^2,x, algorithm="
maxima")

[Out]

1/2*e^(-2*x + 2)/log(x + e)

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mupad [B]  time = 0.92, size = 15, normalized size = 0.88 \begin {gather*} \frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2}{2\,\ln \left (x+\mathrm {e}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*(exp(2) + exp(2)*log(x + exp(1))*(2*x + 2*exp(1))))/(log(x + exp(1))^2*(2*x + 2*exp(1))),x)

[Out]

(exp(-2*x)*exp(2))/(2*log(x + exp(1)))

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sympy [A]  time = 0.30, size = 15, normalized size = 0.88 \begin {gather*} \frac {e^{2} e^{- 2 x}}{2 \log {\left (x + e \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)-2*x)*exp(2)*ln(x+exp(1))-exp(2))/(2*exp(1)+2*x)/exp(x)**2/ln(x+exp(1))**2,x)

[Out]

exp(2)*exp(-2*x)/(2*log(x + E))

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