3.12.20 \(\int \frac {25}{4} e^{\frac {1}{4} (-36-100 e^3+25 x)} \, dx\)

Optimal. Leaf size=15 \[ e^{-9-25 \left (e^3-\frac {x}{4}\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.27, 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, 2194} \begin {gather*} e^{\frac {1}{4} \left (25 x-4 \left (9+25 e^3\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*E^((-36 - 100*E^3 + 25*x)/4))/4,x]

[Out]

E^((-4*(9 + 25*E^3) + 25*x)/4)

Rule 12

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

Rule 2194

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {25}{4} \int e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx\\ &=e^{\frac {1}{4} \left (-4 \left (9+25 e^3\right )+25 x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.93 \begin {gather*} e^{-9-25 e^3+\frac {25 x}{4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*E^((-36 - 100*E^3 + 25*x)/4))/4,x]

[Out]

E^(-9 - 25*E^3 + (25*x)/4)

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fricas [A]  time = 0.71, size = 10, normalized size = 0.67 \begin {gather*} e^{\left (\frac {25}{4} \, x - 25 \, e^{3} - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*exp(-25*exp(3)+25/4*x-9),x, algorithm="fricas")

[Out]

e^(25/4*x - 25*e^3 - 9)

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giac [A]  time = 0.35, size = 10, normalized size = 0.67 \begin {gather*} e^{\left (\frac {25}{4} \, x - 25 \, e^{3} - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*exp(-25*exp(3)+25/4*x-9),x, algorithm="giac")

[Out]

e^(25/4*x - 25*e^3 - 9)

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maple [A]  time = 0.02, size = 11, normalized size = 0.73

method	result	size
gosper	\({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\)	\(11\)
derivativedivides	\({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\)	\(11\)
default	\({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\)	\(11\)
norman	\({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\)	\(11\)
risch	\({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\)	\(11\)
meijerg	\(-{\mathrm e}^{-25 \,{\mathrm e}^{3}-9} \left (1-{\mathrm e}^{\frac {25 x}{4}}\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(25/4*exp(-25*exp(3)+25/4*x-9),x,method=_RETURNVERBOSE)

[Out]

exp(-25*exp(3)+25/4*x-9)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.67 \begin {gather*} e^{\left (\frac {25}{4} \, x - 25 \, e^{3} - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*exp(-25*exp(3)+25/4*x-9),x, algorithm="maxima")

[Out]

e^(25/4*x - 25*e^3 - 9)

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mupad [B]  time = 0.77, size = 12, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{-25\,{\mathrm {e}}^3}\,{\mathrm {e}}^{\frac {25\,x}{4}}\,{\mathrm {e}}^{-9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp((25*x)/4 - 25*exp(3) - 9))/4,x)

[Out]

exp(-25*exp(3))*exp((25*x)/4)*exp(-9)

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sympy [A]  time = 0.08, size = 12, normalized size = 0.80 \begin {gather*} e^{\frac {25 x}{4} - 25 e^{3} - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*exp(-25*exp(3)+25/4*x-9),x)

[Out]

exp(25*x/4 - 25*exp(3) - 9)

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