3.65.57 \(\int -\frac {1}{2} e^{5+\frac {1}{2} (2-e^5 x)} \, dx\)

Optimal. Leaf size=12 \[ e^{1-\frac {e^5 x}{2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2203} \begin {gather*} e^{\frac {1}{2} \left (2-e^5 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1/2*E^(5 + (2 - E^5*x)/2),x]

[Out]

E^((2 - E^5*x)/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2203

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\frac {1}{2} \int e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx\right )\\ &=e^{\frac {1}{2} \left (2-e^5 x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} e^{1-\frac {e^5 x}{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1/2*E^(5 + (2 - E^5*x)/2),x]

[Out]

E^(1 - (E^5*x)/2)

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fricas [A]  time = 0.65, size = 8, normalized size = 0.67 \begin {gather*} e^{\left (-\frac {1}{2} \, x e^{5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*exp(5)*exp(-1/2*x*exp(5)+1),x, algorithm="fricas")

[Out]

e^(-1/2*x*e^5 + 1)

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giac [A]  time = 0.13, size = 8, normalized size = 0.67 \begin {gather*} e^{\left (-\frac {1}{2} \, x e^{5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*exp(5)*exp(-1/2*x*exp(5)+1),x, algorithm="giac")

[Out]

e^(-1/2*x*e^5 + 1)

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maple [A]  time = 0.04, size = 9, normalized size = 0.75




method result size



gosper \({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\) \(9\)
derivativedivides \({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\) \(9\)
default \({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\) \(9\)
norman \({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\) \(9\)
risch \({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\) \(9\)
meijerg \(-{\mathrm e} \left (1-{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}}\right )\) \(15\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/2*exp(5)*exp(-1/2*x*exp(5)+1),x,method=_RETURNVERBOSE)

[Out]

exp(-1/2*x*exp(5)+1)

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maxima [A]  time = 0.37, size = 8, normalized size = 0.67 \begin {gather*} e^{\left (-\frac {1}{2} \, x e^{5} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*exp(5)*exp(-1/2*x*exp(5)+1),x, algorithm="maxima")

[Out]

e^(-1/2*x*e^5 + 1)

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mupad [B]  time = 0.04, size = 9, normalized size = 0.75 \begin {gather*} \mathrm {e}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1 - (x*exp(5))/2)*exp(5))/2,x)

[Out]

exp(1)*exp(-(x*exp(5))/2)

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sympy [A]  time = 0.09, size = 8, normalized size = 0.67 \begin {gather*} e^{- \frac {x e^{5}}{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/2*exp(5)*exp(-1/2*x*exp(5)+1),x)

[Out]

exp(-x*exp(5)/2 + 1)

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