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

Optimal. Leaf size=22 \[ -2+4 e^{\frac {1}{18} \left (-e^3-e^5\right )} x \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {8} \begin {gather*} 4 e^{-\frac {1}{18} e^3 \left (1+e^2\right )} x \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]  time = 0.00, size = 20, normalized size = 0.91 \begin {gather*} 4 e^{\frac {1}{18} \left (-e^3-e^5\right )} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[4*E^((-E^3 - E^5)/18),x]

[Out]

4*E^((-E^3 - E^5)/18)*x

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fricas [A]  time = 0.52, size = 13, normalized size = 0.59 \begin {gather*} 4 \, x e^{\left (-\frac {1}{18} \, e^{5} - \frac {1}{18} \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.26, size = 13, normalized size = 0.59 \begin {gather*} 4 \, x e^{\left (-\frac {1}{18} \, e^{5} - \frac {1}{18} \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 14, normalized size = 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	\(4 x \,{\mathrm e}^{-\frac {{\mathrm e}^{5}}{18}-\frac {{\mathrm e}^{3}}{18}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.60, size = 13, normalized size = 0.59 \begin {gather*} 4 \, x e^{\left (-\frac {1}{18} \, e^{5} - \frac {1}{18} \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.00, size = 13, normalized size = 0.59 \begin {gather*} 4\,x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^3}{18}-\frac {{\mathrm {e}}^5}{18}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*exp(- exp(3)/18 - exp(5)/18),x)

[Out]

4*x*exp(- exp(3)/18 - exp(5)/18)

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sympy [A]  time = 0.05, size = 14, normalized size = 0.64 \begin {gather*} \frac {4 x}{e^{\frac {e^{3}}{18} + \frac {e^{5}}{18}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x*exp(-exp(5)/18 - exp(3)/18)

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