3.85.57 \(\int \frac {e^3 (-2-2 x)-2 \log (\frac {e^2}{2})}{e^3} \, dx\)

Optimal. Leaf size=21 \[ 8-\left (1+x+\frac {\log \left (\frac {e^2}{2}\right )}{e^3}\right )^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12} \begin {gather*} -(x+1)^2-\frac {2 x (2-\log (2))}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(-2 - 2*x) - 2*Log[E^2/2])/E^3,x]

[Out]

-(1 + x)^2 - (2*x*(2 - Log[2]))/E^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (e^3 (-2-2 x)-2 \log \left (\frac {e^2}{2}\right )\right ) \, dx}{e^3}\\ &=-(1+x)^2-\frac {2 x (2-\log (2))}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.24 \begin {gather*} \frac {-4 x-2 e^3 x-e^3 x^2+x \log (4)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(-2 - 2*x) - 2*Log[E^2/2])/E^3,x]

[Out]

(-4*x - 2*E^3*x - E^3*x^2 + x*Log[4])/E^3

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fricas [A]  time = 0.56, size = 23, normalized size = 1.10 \begin {gather*} -{\left ({\left (x^{2} + 2 \, x\right )} e^{3} - 2 \, x \log \relax (2) + 4 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(1/2*exp(2))+(-2*x-2)*exp(3))/exp(3),x, algorithm="fricas")

[Out]

-((x^2 + 2*x)*e^3 - 2*x*log(2) + 4*x)*e^(-3)

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giac [A]  time = 0.16, size = 23, normalized size = 1.10 \begin {gather*} -{\left ({\left (x^{2} + 2 \, x\right )} e^{3} + 2 \, x \log \left (\frac {1}{2} \, e^{2}\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(1/2*exp(2))+(-2*x-2)*exp(3))/exp(3),x, algorithm="giac")

[Out]

-((x^2 + 2*x)*e^3 + 2*x*log(1/2*e^2))*e^(-3)

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maple [A]  time = 0.03, size = 22, normalized size = 1.05

method	result	size
norman	\(-x^{2}-2 \left ({\mathrm e}^{3}-\ln \relax (2)+2\right ) {\mathrm e}^{-3} x\)	\(22\)
risch	\(-x^{2}-2 x +2 \,{\mathrm e}^{-3} \ln \relax (2) x -4 x \,{\mathrm e}^{-3}\)	\(22\)
gosper	\(-x \left (x \,{\mathrm e}^{3}+2 \,{\mathrm e}^{3}+2 \ln \left (\frac {{\mathrm e}^{2}}{2}\right )\right ) {\mathrm e}^{-3}\)	\(24\)
default	\({\mathrm e}^{-3} \left (-2 \ln \left (\frac {{\mathrm e}^{2}}{2}\right ) x +{\mathrm e}^{3} \left (-x^{2}-2 x \right )\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(1/2*exp(2))+(-2*x-2)*exp(3))/exp(3),x,method=_RETURNVERBOSE)

[Out]

-x^2-2*(exp(3)-ln(2)+2)/exp(3)*x

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maxima [A]  time = 0.36, size = 23, normalized size = 1.10 \begin {gather*} -{\left ({\left (x^{2} + 2 \, x\right )} e^{3} + 2 \, x \log \left (\frac {1}{2} \, e^{2}\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(1/2*exp(2))+(-2*x-2)*exp(3))/exp(3),x, algorithm="maxima")

[Out]

-((x^2 + 2*x)*e^3 + 2*x*log(1/2*e^2))*e^(-3)

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mupad [B]  time = 5.52, size = 20, normalized size = 0.95 \begin {gather*} -\frac {{\mathrm {e}}^{-6}\,{\left (\ln \left (\frac {{\mathrm {e}}^4}{4}\right )+{\mathrm {e}}^3\,\left (2\,x+2\right )\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-3)*(2*log(exp(2)/2) + exp(3)*(2*x + 2)),x)

[Out]

-(exp(-6)*(log(exp(4)/4) + exp(3)*(2*x + 2))^2)/4

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sympy [A]  time = 0.06, size = 19, normalized size = 0.90 \begin {gather*} - x^{2} + \frac {x \left (- 2 e^{3} - 4 + 2 \log {\relax (2 )}\right )}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(1/2*exp(2))+(-2*x-2)*exp(3))/exp(3),x)

[Out]

-x**2 + x*(-2*exp(3) - 4 + 2*log(2))*exp(-3)

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