\(\int \frac {e^{-2 x} (-e^2+e^2 (-2 e-2 x) \log (e+x))}{(2 e+2 x) \log ^2(e+x)} \, dx\) [951]

Optimal result

Integrand size = 42, antiderivative size = 17 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2-2 x}}{2 \log (e+x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2326} \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2-2 x}}{2 \log (x+e)} \]

[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 2326

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2-2 x}}{2 \log (e+x)} \]

[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])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{2} e^{- 2 x}}{2 \log {\left (x + e \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {e^{\left (-2 \, x + 2\right )}}{2 \, \log \left (x + e\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.50 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 x} \left (-e^2+e^2 (-2 e-2 x) \log (e+x)\right )}{(2 e+2 x) \log ^2(e+x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2}{2\,\ln \left (x+\mathrm {e}\right )} \]

[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)))