\(\int \frac {-72+4 e^5}{e^3} \, dx\) [7582]

Optimal result

Integrand size = 11, antiderivative size = 12 \[ \int \frac {-72+4 e^5}{e^3} \, dx=4 \left (-\frac {18}{e^3}+e^2\right ) x \]

[Out]

4*(exp(2)-18/exp(3))*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {8} \[ \int \frac {-72+4 e^5}{e^3} \, dx=-\frac {4 \left (18-e^5\right ) x}{e^3} \]

[In]

Int[(-72 + 4*E^5)/E^3,x]

[Out]

(-4*(18 - E^5)*x)/E^3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \left (18-e^5\right ) x}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {-72+4 e^5}{e^3} \, dx=-\frac {72 x}{e^3}+4 e^2 x \]

[In]

Integrate[(-72 + 4*E^5)/E^3,x]

[Out]

(-72*x)/E^3 + 4*E^2*x

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17

method	result	size
risch	\(4 \,{\mathrm e}^{-3} x \,{\mathrm e}^{5}-72 \,{\mathrm e}^{-3} x\)	\(14\)
default	\(4 \left ({\mathrm e}^{2} {\mathrm e}^{3}-18\right ) {\mathrm e}^{-3} x\)	\(15\)
norman	\(4 \left ({\mathrm e}^{2} {\mathrm e}^{3}-18\right ) {\mathrm e}^{-3} x\)	\(15\)
parallelrisch	\(\left (4 \,{\mathrm e}^{2} {\mathrm e}^{3}-72\right ) {\mathrm e}^{-3} x\)	\(15\)

[In]

int((4*exp(2)*exp(3)-72)/exp(3),x,method=_RETURNVERBOSE)

[Out]

4*exp(-3)*x*exp(5)-72*exp(-3)*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {-72+4 e^5}{e^3} \, dx=4 \, {\left (x e^{5} - 18 \, x\right )} e^{\left (-3\right )} \]

[In]

integrate((4*exp(2)*exp(3)-72)/exp(3),x, algorithm="fricas")

[Out]

4*(x*e^5 - 18*x)*e^(-3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {-72+4 e^5}{e^3} \, dx=\frac {x \left (-72 + 4 e^{5}\right )}{e^{3}} \]

[In]

integrate((4*exp(2)*exp(3)-72)/exp(3),x)

[Out]

x*(-72 + 4*exp(5))*exp(-3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {-72+4 e^5}{e^3} \, dx=4 \, x {\left (e^{5} - 18\right )} e^{\left (-3\right )} \]

[In]

integrate((4*exp(2)*exp(3)-72)/exp(3),x, algorithm="maxima")

[Out]

4*x*(e^5 - 18)*e^(-3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {-72+4 e^5}{e^3} \, dx=4 \, x {\left (e^{5} - 18\right )} e^{\left (-3\right )} \]

[In]

integrate((4*exp(2)*exp(3)-72)/exp(3),x, algorithm="giac")

[Out]

4*x*(e^5 - 18)*e^(-3)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {-72+4 e^5}{e^3} \, dx=x\,{\mathrm {e}}^{-3}\,\left (4\,{\mathrm {e}}^5-72\right ) \]

[In]

int(exp(-3)*(4*exp(5) - 72),x)

[Out]

x*exp(-3)*(4*exp(5) - 72)