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

Optimal. Leaf size=15 \[ x \log \left (\frac {5 e^{-x}}{2 x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2548} \begin {gather*} x \log \left (\frac {5 e^{-x}}{2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 - x + Log[5/(2*E^x*x)],x]

[Out]

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

Rule 2548

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} x \log \left (\frac {5 e^{-x}}{2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 12, normalized size = 0.80 \begin {gather*} x \log \left (\frac {5 \, e^{\left (-x\right )}}{2 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 20, normalized size = 1.33 \begin {gather*} -x^{2} + x \log \relax (5) - x \log \relax (2) - x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 13, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 12, normalized size = 0.80 \begin {gather*} x \log \left (\frac {5 \, e^{\left (-x\right )}}{2 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.10, size = 13, normalized size = 0.87 \begin {gather*} -x\,\left (x-\ln \left (\frac {5}{2\,x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((5*exp(-x))/(2*x)) - x - 1,x)

[Out]

-x*(x - log(5/(2*x)))

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sympy [A]  time = 0.13, size = 10, normalized size = 0.67 \begin {gather*} x \log {\left (\frac {5 e^{- x}}{2 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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