\(\int \frac {25}{4} e^{\frac {1}{4} (-36-100 e^3+25 x)} \, dx\) [1120]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{-9-25 \left (e^3-\frac {x}{4}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2225} \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{\frac {1}{4} \left (25 x-4 \left (9+25 e^3\right )\right )} \]

[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 2225

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{-9-25 e^3+\frac {25 x}{4}} \]

[In]

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

[Out]

E^(-9 - 25*E^3 + (25*x)/4)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 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 \({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\) \(11\)
parallelrisch \({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\) \(11\)
parts \({\mathrm e}^{-25 \,{\mathrm e}^{3}+\frac {25 x}{4}-9}\) \(11\)
meijerg \(-{\mathrm e}^{-25 \,{\mathrm e}^{3}-9} \left (1-{\mathrm e}^{\frac {25 x}{4}}\right )\) \(18\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{\left (\frac {25}{4} \, x - 25 \, e^{3} - 9\right )} \]

[In]

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

[Out]

e^(25/4*x - 25*e^3 - 9)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{\frac {25 x}{4} - 25 e^{3} - 9} \]

[In]

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

[Out]

exp(25*x/4 - 25*exp(3) - 9)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{\left (\frac {25}{4} \, x - 25 \, e^{3} - 9\right )} \]

[In]

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

[Out]

e^(25/4*x - 25*e^3 - 9)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx=e^{\left (\frac {25}{4} \, x - 25 \, e^{3} - 9\right )} \]

[In]

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

[Out]

e^(25/4*x - 25*e^3 - 9)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {25}{4} e^{\frac {1}{4} \left (-36-100 e^3+25 x\right )} \, dx={\mathrm {e}}^{-25\,{\mathrm {e}}^3}\,{\mathrm {e}}^{\frac {25\,x}{4}}\,{\mathrm {e}}^{-9} \]

[In]

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

[Out]

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