3.4.23 \(\int (2+4 x+2 \log (\frac {1}{4} e^{-8+2 x} x)) \, dx\)

Optimal. Leaf size=18 \[ -1+2 x \log \left (\frac {1}{4} e^{-8+2 x} x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2548} \begin {gather*} 2 x \log \left (\frac {1}{4} e^{2 x-8} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2 + 4*x + 2*Log[(E^(-8 + 2*x)*x)/4],x]

[Out]

2*x*Log[(E^(-8 + 2*x)*x)/4]

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} &=2 x+2 x^2+2 \int \log \left (\frac {1}{4} e^{-8+2 x} x\right ) \, dx\\ &=2 x+2 x^2+2 x \log \left (\frac {1}{4} e^{-8+2 x} x\right )-2 \int (1+2 x) \, dx\\ &=2 x \log \left (\frac {1}{4} e^{-8+2 x} x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.89 \begin {gather*} 2 x \log \left (\frac {1}{4} e^{-8+2 x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2 + 4*x + 2*Log[(E^(-8 + 2*x)*x)/4],x]

[Out]

2*x*Log[(E^(-8 + 2*x)*x)/4]

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fricas [A]  time = 0.56, size = 13, normalized size = 0.72 \begin {gather*} 2 \, x \log \left (\frac {1}{4} \, x e^{\left (2 \, x - 8\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(1/4*x*exp(x-4)^2)+4*x+2,x, algorithm="fricas")

[Out]

2*x*log(1/4*x*e^(2*x - 8))

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giac [A]  time = 0.24, size = 19, normalized size = 1.06 \begin {gather*} 4 \, x^{2} - 4 \, x \log \relax (2) + 2 \, x \log \relax (x) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(1/4*x*exp(x-4)^2)+4*x+2,x, algorithm="giac")

[Out]

4*x^2 - 4*x*log(2) + 2*x*log(x) - 16*x

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maple [A]  time = 0.10, size = 14, normalized size = 0.78

method	result	size
norman	\(2 x \ln \left (\frac {x \,{\mathrm e}^{2 x -8}}{4}\right )\)	\(14\)
default	\(2 x \ln \left (\frac {x \,{\mathrm e}^{2 x -8}}{4}\right )+40\)	\(16\)
risch	\(4 x \ln \left ({\mathrm e}^{x -4}\right )+2 x \ln \relax (x )-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x -8}\right )^{3} x -i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x -8}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{2 x -8}\right ) x +i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x -8}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{2 x -8}\right )^{2} x -i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{2 x -8}\right )^{3} x +i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{2 x -8}\right )^{2} x +2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x -4}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x -8}\right )^{2} x -i \pi \mathrm {csgn}\left (i {\mathrm e}^{x -4}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x -8}\right ) x -4 x \ln \relax (2)\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*ln(1/4*x*exp(x-4)^2)+4*x+2,x,method=_RETURNVERBOSE)

[Out]

2*x*ln(1/4*x*exp(x-4)^2)

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maxima [A]  time = 0.45, size = 13, normalized size = 0.72 \begin {gather*} 2 \, x \log \left (\frac {1}{4} \, x e^{\left (2 \, x - 8\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(1/4*x*exp(x-4)^2)+4*x+2,x, algorithm="maxima")

[Out]

2*x*log(1/4*x*e^(2*x - 8))

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mupad [B]  time = 0.37, size = 12, normalized size = 0.67 \begin {gather*} 2\,x\,\left (2\,x+\ln \left (\frac {x}{4}\right )-8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*x + 2*log((x*exp(2*x - 8))/4) + 2,x)

[Out]

2*x*(2*x + log(x/4) - 8)

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sympy [A]  time = 0.14, size = 14, normalized size = 0.78 \begin {gather*} 2 x \log {\left (\frac {x e^{2 x - 8}}{4} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*ln(1/4*x*exp(x-4)**2)+4*x+2,x)

[Out]

2*x*log(x*exp(2*x - 8)/4)

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