\(\int (-1-x+\log (\frac {5 e^{-x}}{2 x})) \, dx\) [6705]

Optimal result

Integrand size = 18, antiderivative size = 15 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=x \log \left (\frac {5 e^{-x}}{2 x}\right ) \]

[Out]

x*ln(5/2/exp(x)/x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2628} \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=x \log \left (\frac {5 e^{-x}}{2 x}\right ) \]

[In]

Int[-1 - x + Log[5/(2*E^x*x)],x]

[Out]

x*Log[5/(2*E^x*x)]

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = -x-\frac {x^2}{2}+\int \log \left (\frac {5 e^{-x}}{2 x}\right ) \, dx \\ & = -x-\frac {x^2}{2}+x \log \left (\frac {5 e^{-x}}{2 x}\right )-\int (-1-x) \, dx \\ & = x \log \left (\frac {5 e^{-x}}{2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=x \log \left (\frac {5 e^{-x}}{2 x}\right ) \]

[In]

Integrate[-1 - x + Log[5/(2*E^x*x)],x]

[Out]

x*Log[5/(2*E^x*x)]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
default	\(x \ln \left (\frac {5 \,{\mathrm e}^{-x}}{2 x}\right )\)	\(13\)
norman	\(x \ln \left (\frac {5 \,{\mathrm e}^{-x}}{2 x}\right )\)	\(13\)
parallelrisch	\(x \ln \left (\frac {5 \,{\mathrm e}^{-x}}{2 x}\right )\)	\(13\)
parts	\(x \ln \left (\frac {5 \,{\mathrm e}^{-x}}{2 x}\right )\)	\(13\)
risch	\(-x \ln \left ({\mathrm e}^{x}\right )-x \ln \left (x \right )+\frac {i x \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )}{2}+\frac {i x \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}-\frac {i x \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}-\frac {i x \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{x}\right )^{3}}{2}+x \ln \left (5\right )-x \ln \left (2\right )\)	\(122\)

[In]

int(ln(5/2/exp(x)/x)-x-1,x,method=_RETURNVERBOSE)

[Out]

x*ln(5/2/exp(x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=x \log \left (\frac {5 \, e^{\left (-x\right )}}{2 \, x}\right ) \]

[In]

integrate(log(5/2/exp(x)/x)-x-1,x, algorithm="fricas")

[Out]

x*log(5/2*e^(-x)/x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=x \log {\left (\frac {5 e^{- x}}{2 x} \right )} \]

[In]

integrate(ln(5/2/exp(x)/x)-x-1,x)

[Out]

x*log(5*exp(-x)/(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=x \log \left (\frac {5 \, e^{\left (-x\right )}}{2 \, x}\right ) \]

[In]

integrate(log(5/2/exp(x)/x)-x-1,x, algorithm="maxima")

[Out]

x*log(5/2*e^(-x)/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=-x^{2} + x \log \left (5\right ) - x \log \left (2\right ) - x \log \left (x\right ) \]

[In]

integrate(log(5/2/exp(x)/x)-x-1,x, algorithm="giac")

[Out]

-x^2 + x*log(5) - x*log(2) - x*log(x)

Mupad [B] (verification not implemented)

Time = 11.66 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (-1-x+\log \left (\frac {5 e^{-x}}{2 x}\right )\right ) \, dx=-x\,\left (x-\ln \left (\frac {5}{2\,x}\right )\right ) \]

[In]

int(log((5*exp(-x))/(2*x)) - x - 1,x)

[Out]

-x*(x - log(5/(2*x)))