\(\int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} (1-e^{35}) \, dx\) [7580]

Optimal result

Integrand size = 27, antiderivative size = 12 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=e^{-5-x+\frac {x}{e^{35}}} \]

[Out]

exp(exp(-35)*x-5-x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2259, 2234} \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=e^{-\frac {5 e^{35}-\left (1-e^{35}\right ) x}{e^{35}}} \]

[In]

Int[E^(-35 + (E^35*(-5 - x) + x)/E^35)*(1 - E^35),x]

[Out]

E^(-((5*E^35 - (1 - E^35)*x)/E^35))

Rule 12

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

Rule 2234

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]

Rule 2259

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \left (1-e^{35}\right ) \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \, dx \\ & = \left (1-e^{35}\right ) \int e^{-35+\frac {-5 e^{35}+\left (1-e^{35}\right ) x}{e^{35}}} \, dx \\ & = e^{-\frac {5 e^{35}-\left (1-e^{35}\right ) x}{e^{35}}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(25\) vs. \(2(12)=24\).

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 2.08 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=-\frac {e^{-40+\left (-1+\frac {1}{e^{35}}\right ) x} \left (-1+e^{35}\right )}{-1+\frac {1}{e^{35}}} \]

[In]

Integrate[E^(-35 + (E^35*(-5 - x) + x)/E^35)*(1 - E^35),x]

[Out]

-((E^(-40 + (-1 + E^(-35))*x)*(-1 + E^35))/(-1 + E^(-35)))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\({\mathrm e}^{\left (\left (-x -5\right ) {\mathrm e}^{35}+x \right ) {\mathrm e}^{-35}}\)	\(17\)
default	\({\mathrm e}^{\left (\left (-x -5\right ) {\mathrm e}^{35}+x \right ) {\mathrm e}^{-35}}\)	\(17\)
norman	\({\mathrm e}^{\left (\left (-x -5\right ) {\mathrm e}^{35}+x \right ) {\mathrm e}^{-35}}\)	\(17\)
gosper	\({\mathrm e}^{-\left ({\mathrm e}^{35} x +5 \,{\mathrm e}^{35}-x \right ) {\mathrm e}^{-35}}\)	\(20\)
parallelrisch	\(-\frac {\left (-{\mathrm e}^{35}+1\right ) {\mathrm e}^{\left (\left (-x -5\right ) {\mathrm e}^{35}+x \right ) {\mathrm e}^{-35}}}{{\mathrm e}^{35}-1}\)	\(31\)
meijerg	\(-\frac {{\mathrm e}^{30} \left (1-{\mathrm e}^{-x \left ({\mathrm e}^{35}-1\right ) {\mathrm e}^{-35}}\right )}{{\mathrm e}^{35}-1}+\frac {{\mathrm e}^{-5} \left (1-{\mathrm e}^{-x \left ({\mathrm e}^{35}-1\right ) {\mathrm e}^{-35}}\right )}{{\mathrm e}^{35}-1}\)	\(49\)
risch	\(\frac {{\mathrm e}^{-\left ({\mathrm e}^{35} x +5 \,{\mathrm e}^{35}-x \right ) {\mathrm e}^{-35}} {\mathrm e}^{35}}{{\mathrm e}^{35}-1}-\frac {{\mathrm e}^{-\left ({\mathrm e}^{35} x +5 \,{\mathrm e}^{35}-x \right ) {\mathrm e}^{-35}}}{{\mathrm e}^{35}-1}\)	\(53\)
parts	\(-\frac {{\mathrm e}^{\left (\left (-x -5\right ) {\mathrm e}^{35}+x \right ) {\mathrm e}^{-35}} {\mathrm e}^{35}}{-{\mathrm e}^{35}+1}+\frac {{\mathrm e}^{\left (\left (-x -5\right ) {\mathrm e}^{35}+x \right ) {\mathrm e}^{-35}}}{-{\mathrm e}^{35}+1}\)	\(55\)

[In]

int((-exp(35)+1)*exp(((-x-5)*exp(35)+x)/exp(35))/exp(35),x,method=_RETURNVERBOSE)

[Out]

exp(((-x-5)*exp(35)+x)/exp(35))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=e^{\left (-{\left ({\left (x + 40\right )} e^{35} - x\right )} e^{\left (-35\right )} + 35\right )} \]

[In]

integrate((-exp(35)+1)*exp(((-x-5)*exp(35)+x)/exp(35))/exp(35),x, algorithm="fricas")

[Out]

e^(-((x + 40)*e^35 - x)*e^(-35) + 35)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=e^{\frac {x + \left (- x - 5\right ) e^{35}}{e^{35}}} \]

[In]

integrate((-exp(35)+1)*exp(((-x-5)*exp(35)+x)/exp(35))/exp(35),x)

[Out]

exp((x + (-x - 5)*exp(35))*exp(-35))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=e^{\left (-{\left ({\left (x + 5\right )} e^{35} - x\right )} e^{\left (-35\right )}\right )} \]

[In]

integrate((-exp(35)+1)*exp(((-x-5)*exp(35)+x)/exp(35))/exp(35),x, algorithm="maxima")

[Out]

e^(-((x + 5)*e^35 - x)*e^(-35))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx=e^{\left (-{\left ({\left (x + 5\right )} e^{35} - x\right )} e^{\left (-35\right )}\right )} \]

[In]

integrate((-exp(35)+1)*exp(((-x-5)*exp(35)+x)/exp(35))/exp(35),x, algorithm="giac")

[Out]

e^(-((x + 5)*e^35 - x)*e^(-35))

Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int e^{-35+\frac {e^{35} (-5-x)+x}{e^{35}}} \left (1-e^{35}\right ) \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-35}} \]

[In]

int(-exp(-35)*exp(exp(-35)*(x - exp(35)*(x + 5)))*(exp(35) - 1),x)

[Out]

exp(-x)*exp(-5)*exp(x*exp(-35))