\(\int -15 e^{\frac {1}{5} (198-75 x)} \, dx\) [8111]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{9+15 \left (\frac {51}{25}-x\right )} \]

[Out]

exp(-15*x+198/5)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2225} \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{\frac {3}{5} (66-25 x)} \]

[In]

Int[-15*E^((198 - 75*x)/5),x]

[Out]

E^((3*(66 - 25*x))/5)

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}& = -\left (15 \int e^{\frac {1}{5} (198-75 x)} \, dx\right ) \\ & = e^{\frac {3}{5} (66-25 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{\frac {198}{5}-15 x} \]

[In]

Integrate[-15*E^((198 - 75*x)/5),x]

[Out]

E^(198/5 - 15*x)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54

method	result	size
gosper	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
derivativedivides	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
default	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
norman	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
risch	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
parallelrisch	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
parts	\({\mathrm e}^{-15 x +\frac {198}{5}}\)	\(7\)
meijerg	\(-{\mathrm e}^{\frac {198}{5}} \left (1-{\mathrm e}^{-15 x}\right )\)	\(13\)

[In]

int(-15*exp(-15*x+198/5),x,method=_RETURNVERBOSE)

[Out]

exp(-15*x+198/5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.46 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{\left (-15 \, x + \frac {198}{5}\right )} \]

[In]

integrate(-15*exp(-15*x+198/5),x, algorithm="fricas")

[Out]

e^(-15*x + 198/5)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{\frac {198}{5} - 15 x} \]

[In]

integrate(-15*exp(-15*x+198/5),x)

[Out]

exp(198/5 - 15*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.46 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{\left (-15 \, x + \frac {198}{5}\right )} \]

[In]

integrate(-15*exp(-15*x+198/5),x, algorithm="maxima")

[Out]

e^(-15*x + 198/5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.46 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx=e^{\left (-15 \, x + \frac {198}{5}\right )} \]

[In]

integrate(-15*exp(-15*x+198/5),x, algorithm="giac")

[Out]

e^(-15*x + 198/5)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int -15 e^{\frac {1}{5} (198-75 x)} \, dx={\mathrm {e}}^{-15\,x}\,{\mathrm {e}}^{198/5} \]

[In]

int(-15*exp(198/5 - 15*x),x)

[Out]

exp(-15*x)*exp(198/5)