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

Optimal. Leaf size=24 \[ e^2-2 x+\frac {-1-3 x+x (-x+\log (2))}{e^3} \]

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Rubi [A]  time = 0.00, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {9} \begin {gather*} -\frac {\left (2 x+2 e^3+3-\log (2)\right )^2}{4 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(3 + 2*E^3 + 2*x - Log[2])^2/E^3

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {\left (3+2 e^3+2 x-\log (2)\right )^2}{4 e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 0.96 \begin {gather*} \frac {-3 x-2 e^3 x-x^2+x \log (2)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x - 2*E^3*x - x^2 + x*Log[2])/E^3

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fricas [A]  time = 0.92, size = 21, normalized size = 0.88 \begin {gather*} -{\left (x^{2} + 2 \, x e^{3} - x \log \relax (2) + 3 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 21, normalized size = 0.88 \begin {gather*} -{\left (x^{2} + 2 \, x e^{3} - x \log \relax (2) + 3 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 19, normalized size = 0.79

method	result	size
gosper	\(-x \left (x +2 \,{\mathrm e}^{3}-\ln \relax (2)+3\right ) {\mathrm e}^{-3}\)	\(19\)
risch	\({\mathrm e}^{-3} \ln \relax (2) x -2 x -x^{2} {\mathrm e}^{-3}-3 x \,{\mathrm e}^{-3}\)	\(23\)
default	\({\mathrm e}^{-3} \left (x \ln \relax (2)-2 x \,{\mathrm e}^{3}-x^{2}-3 x \right )\)	\(24\)
norman	\(-x^{2} {\mathrm e}^{-3}-{\mathrm e}^{-3} \left (2 \,{\mathrm e}^{3}-\ln \relax (2)+3\right ) x\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 21, normalized size = 0.88 \begin {gather*} -{\left (x^{2} + 2 \, x e^{3} - x \log \relax (2) + 3 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.35, size = 19, normalized size = 0.79 \begin {gather*} -\frac {{\mathrm {e}}^{-3}\,{\left (2\,x+2\,{\mathrm {e}}^3-\ln \relax (2)+3\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.05, size = 20, normalized size = 0.83 \begin {gather*} - \frac {x^{2}}{e^{3}} + \frac {x \left (- 2 e^{3} - 3 + \log {\relax (2 )}\right )}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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