\(\int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx\) [7957]

Optimal result

Integrand size = 16, antiderivative size = 24 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=e^2-2 x+\frac {-1-3 x+x (-x+\log (2))}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {9} \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=-\frac {\left (2 x+2 e^3+3-\log (2)\right )^2}{4 e^3} \]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=\frac {-3 x-2 e^3 x-x^2+x \log (2)}{e^3} \]

[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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
gosper	\(-x \left (x +2 \,{\mathrm e}^{3}-\ln \left (2\right )+3\right ) {\mathrm e}^{-3}\)	\(19\)
parallelrisch	\({\mathrm e}^{-3} \left (-x^{2}+\left (-2 \,{\mathrm e}^{3}+\ln \left (2\right )-3\right ) x \right )\)	\(22\)
risch	\({\mathrm e}^{-3} x \ln \left (2\right )-2 x -x^{2} {\mathrm e}^{-3}-3 x \,{\mathrm e}^{-3}\)	\(23\)
default	\({\mathrm e}^{-3} \left (x \ln \left (2\right )-2 x \,{\mathrm e}^{3}-x^{2}-3 x \right )\)	\(24\)
norman	\(-x^{2} {\mathrm e}^{-3}-{\mathrm e}^{-3} \left (2 \,{\mathrm e}^{3}-\ln \left (2\right )+3\right ) x\)	\(28\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=-{\left (x^{2} + 2 \, x e^{3} - x \log \left (2\right ) + 3 \, x\right )} e^{\left (-3\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=- \frac {x^{2}}{e^{3}} + \frac {x \left (- 2 e^{3} - 3 + \log {\left (2 \right )}\right )}{e^{3}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=-{\left (x^{2} + 2 \, x e^{3} - x \log \left (2\right ) + 3 \, x\right )} e^{\left (-3\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=-{\left (x^{2} + 2 \, x e^{3} - x \log \left (2\right ) + 3 \, x\right )} e^{\left (-3\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.80 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-3-2 e^3-2 x+\log (2)}{e^3} \, dx=-\frac {{\mathrm {e}}^{-3}\,{\left (2\,x+2\,{\mathrm {e}}^3-\ln \left (2\right )+3\right )}^2}{4} \]

[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