\(\int 4 e^{\frac {1}{18} (-e^3-e^5)} \, dx\) [3203]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=-2+4 e^{\frac {1}{18} \left (-e^3-e^5\right )} x \]

[Out]

4*x/exp(1/18*exp(5)+1/18*exp(3))-2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {8} \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=4 e^{-\frac {1}{18} e^3 \left (1+e^2\right )} x \]

[In]

Int[4*E^((-E^3 - E^5)/18),x]

[Out]

(4*x)/E^((E^3*(1 + E^2))/18)

Rule 8

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

Rubi steps \begin{align*} \text {integral}& = 4 e^{-\frac {1}{18} e^3 \left (1+e^2\right )} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=4 e^{\frac {1}{18} \left (-e^3-e^5\right )} x \]

[In]

Integrate[4*E^((-E^3 - E^5)/18),x]

[Out]

4*E^((-E^3 - E^5)/18)*x

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64

method	result	size
norman	\(4 \,{\mathrm e}^{-\frac {{\mathrm e}^{5}}{18}} {\mathrm e}^{-\frac {{\mathrm e}^{3}}{18}} x\)	\(14\)
risch	\(4 x \,{\mathrm e}^{-\frac {{\mathrm e}^{5}}{18}-\frac {{\mathrm e}^{3}}{18}}\)	\(14\)
default	\(4 x \,{\mathrm e}^{-\frac {{\mathrm e}^{5}}{18}-\frac {{\mathrm e}^{3}}{18}}\)	\(16\)
parallelrisch	\(4 x \,{\mathrm e}^{-\frac {{\mathrm e}^{5}}{18}-\frac {{\mathrm e}^{3}}{18}}\)	\(16\)

[In]

int(4/exp(1/18*exp(5)+1/18*exp(3)),x,method=_RETURNVERBOSE)

[Out]

4/exp(exp(5))^(1/18)/exp(exp(3))^(1/18)*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=4 \, x e^{\left (-\frac {1}{18} \, e^{5} - \frac {1}{18} \, e^{3}\right )} \]

[In]

integrate(4/exp(1/18*exp(5)+1/18*exp(3)),x, algorithm="fricas")

[Out]

4*x*e^(-1/18*e^5 - 1/18*e^3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=\frac {4 x}{e^{\frac {e^{3}}{18} + \frac {e^{5}}{18}}} \]

[In]

integrate(4/exp(1/18*exp(5)+1/18*exp(3)),x)

[Out]

4*x*exp(-exp(5)/18 - exp(3)/18)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=4 \, x e^{\left (-\frac {1}{18} \, e^{5} - \frac {1}{18} \, e^{3}\right )} \]

[In]

integrate(4/exp(1/18*exp(5)+1/18*exp(3)),x, algorithm="maxima")

[Out]

4*x*e^(-1/18*e^5 - 1/18*e^3)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=4 \, x e^{\left (-\frac {1}{18} \, e^{5} - \frac {1}{18} \, e^{3}\right )} \]

[In]

integrate(4/exp(1/18*exp(5)+1/18*exp(3)),x, algorithm="giac")

[Out]

4*x*e^(-1/18*e^5 - 1/18*e^3)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} \, dx=4\,x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^3}{18}-\frac {{\mathrm {e}}^5}{18}} \]

[In]

int(4*exp(- exp(3)/18 - exp(5)/18),x)

[Out]

4*x*exp(- exp(3)/18 - exp(5)/18)