\(\int \frac {20 \log (4 e^5)}{e^{10}} \, dx\) [5046]

Optimal result

Integrand size = 11, antiderivative size = 12 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=\frac {20 x \log \left (4 e^5\right )}{e^{10}} \]

[Out]

20/exp(10)*x*ln(4*exp(5))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {8} \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=\frac {20 x (5+\log (4))}{e^{10}} \]

[In]

Int[(20*Log[4*E^5])/E^10,x]

[Out]

(20*x*(5 + Log[4]))/E^10

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \frac {20 x (5+\log (4))}{e^{10}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=\frac {20 x \log \left (4 e^5\right )}{e^{10}} \]

[In]

Integrate[(20*Log[4*E^5])/E^10,x]

[Out]

(20*x*Log[4*E^5])/E^10

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
default	\(20 \,{\mathrm e}^{-10} x \ln \left (4 \,{\mathrm e}^{5}\right )\)	\(13\)
parallelrisch	\(20 \,{\mathrm e}^{-10} x \ln \left (4 \,{\mathrm e}^{5}\right )\)	\(13\)
norman	\(20 \left (2 \ln \left (2\right )+5\right ) {\mathrm e}^{-10} x\)	\(14\)
risch	\(40 \,{\mathrm e}^{-10} x \ln \left (2\right )+100 \,{\mathrm e}^{-10} x\)	\(14\)

[In]

int(20*ln(4*exp(5))/exp(10),x,method=_RETURNVERBOSE)

[Out]

20/exp(10)*x*ln(4*exp(5))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=20 \, {\left (2 \, x \log \left (2\right ) + 5 \, x\right )} e^{\left (-10\right )} \]

[In]

integrate(20*log(4*exp(5))/exp(10),x, algorithm="fricas")

[Out]

20*(2*x*log(2) + 5*x)*e^(-10)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=\frac {20 x \log {\left (4 e^{5} \right )}}{e^{10}} \]

[In]

integrate(20*ln(4*exp(5))/exp(10),x)

[Out]

20*x*exp(-10)*log(4*exp(5))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=20 \, x e^{\left (-10\right )} \log \left (4 \, e^{5}\right ) \]

[In]

integrate(20*log(4*exp(5))/exp(10),x, algorithm="maxima")

[Out]

20*x*e^(-10)*log(4*e^5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=20 \, x e^{\left (-10\right )} \log \left (4 \, e^{5}\right ) \]

[In]

integrate(20*log(4*exp(5))/exp(10),x, algorithm="giac")

[Out]

20*x*e^(-10)*log(4*e^5)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {20 \log \left (4 e^5\right )}{e^{10}} \, dx=20\,x\,\ln \left (4\,{\mathrm {e}}^5\right )\,{\mathrm {e}}^{-10} \]

[In]

int(20*log(4*exp(5))*exp(-10),x)

[Out]

20*x*log(4*exp(5))*exp(-10)