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

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

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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} $$\begin{array}{c}-(x+1{)}^2-\frac{2x(2-\log (2))}{e^3}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int (e^3(-2-2x)-2\log \left(\frac{e^2}{2}\right))dx}{e^3}\\ & =-(1+x{)}^2-\frac{2x(2-\log (2))}{e^3}\end{array}\end{array}$$

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

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 $$\begin{array}{c}-((x^2+2x)e^3-2x\log (2)+4x)e^{(-3)}\end{array}$$

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 $$\begin{array}{c}-((x^2+2x)e^3+2x\log \left(\frac{1}{2}{e}^2\right))e^{(-3)}\end{array}$$

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	$-x^2-2({\mathrm{e}}^3-\ln (2)+2){\mathrm{e}}^{-3}x$	$22$
risch	$-x^2-2x+2{\mathrm{e}}^{-3}\ln (2)x-4x{\mathrm{e}}^{-3}$	$22$
gosper	$-x(x{\mathrm{e}}^3+2{\mathrm{e}}^3+2\ln \left(\frac{{\mathrm{e}}^2}{2}\right)){\mathrm{e}}^{-3}$	$24$
default	${\mathrm{e}}^{-3}(-2\ln \left(\frac{{\mathrm{e}}^2}{2}\right)x+{\mathrm{e}}^3(-x^2-2x))$	$27$

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 $$\begin{array}{c}-((x^2+2x)e^3+2x\log \left(\frac{1}{2}{e}^2\right))e^{(-3)}\end{array}$$

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 $$\begin{array}{c}-\frac{{\mathrm{e}}^{-6}{(\ln \left(\frac{{\mathrm{e}}^4}{4}\right)+{\mathrm{e}}^3(2x+2))}^2}{4}\end{array}$$

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 $$\begin{array}{c}-x^2+\frac{x(-2e^3-4+2\log (2))}{e^3}\end{array}$$

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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