\(\int 25 e^{6+e^8+e^{1-x}} (1-e^{1-x} x) \, dx\) [7437]

Optimal result

Integrand size = 28, antiderivative size = 17 \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=25 e^{6+e^8+e^{1-x}} x \]

[Out]

x*exp(exp(1-x)+2*ln(5)+exp(4)^2+6)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 2326} \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=25 e^{e^{1-x}+6+e^8} x \]

[In]

Int[25*E^(6 + E^8 + E^(1 - x))*(1 - E^(1 - x)*x),x]

[Out]

25*E^(6 + E^8 + E^(1 - x))*x

Rule 12

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

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 25 \int e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx \\ & = 25 e^{6+e^8+e^{1-x}} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=25 e^{6+e^8+e^{1-x}} x \]

[In]

Integrate[25*E^(6 + E^8 + E^(1 - x))*(1 - E^(1 - x)*x),x]

[Out]

25*E^(6 + E^8 + E^(1 - x))*x

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
risch	\(25 x \,{\mathrm e}^{{\mathrm e}^{1-x}+{\mathrm e}^{8}+6}\)	\(15\)
norman	\(x \,{\mathrm e}^{{\mathrm e}^{1-x}+2 \ln \left (5\right )+{\mathrm e}^{8}+6}\)	\(20\)
parallelrisch	\(x \,{\mathrm e}^{{\mathrm e}^{1-x}+2 \ln \left (5\right )+{\mathrm e}^{8}+6}\)	\(20\)

[In]

int((-x*exp(1-x)+1)*exp(exp(1-x)+2*ln(5)+exp(4)^2+6),x,method=_RETURNVERBOSE)

[Out]

25*x*exp(exp(1-x)+exp(8)+6)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=x e^{\left (e^{8} + e^{\left (-x + 1\right )} + 2 \, \log \left (5\right ) + 6\right )} \]

[In]

integrate((-x*exp(1-x)+1)*exp(exp(1-x)+2*log(5)+exp(4)^2+6),x, algorithm="fricas")

[Out]

x*e^(e^8 + e^(-x + 1) + 2*log(5) + 6)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=25 x e^{e^{1 - x} + 6 + e^{8}} \]

[In]

integrate((-x*exp(1-x)+1)*exp(exp(1-x)+2*ln(5)+exp(4)**2+6),x)

[Out]

25*x*exp(exp(1 - x) + 6 + exp(8))

Maxima [F]

\[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=\int { -{\left (x e^{\left (-x + 1\right )} - 1\right )} e^{\left (e^{8} + e^{\left (-x + 1\right )} + 2 \, \log \left (5\right ) + 6\right )} \,d x } \]

[In]

integrate((-x*exp(1-x)+1)*exp(exp(1-x)+2*log(5)+exp(4)^2+6),x, algorithm="maxima")

[Out]

25*(x*e^(e^8 + 7) + e^(e^8 + 7))*e^(-x + e^(-x + 1)) - 25*Ei(e^(-x + 1))*e^(e^8 + 6) + 25*integrate((x*e^(e^8
+ 8) + e^(e^8 + 8))*e^(-2*x + e^(-x + 1)), x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=25 \, x e^{\left (e^{8} + e^{\left (-x + 1\right )} + 6\right )} \]

[In]

integrate((-x*exp(1-x)+1)*exp(exp(1-x)+2*log(5)+exp(4)^2+6),x, algorithm="giac")

[Out]

25*x*e^(e^8 + e^(-x + 1) + 6)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int 25 e^{6+e^8+e^{1-x}} \left (1-e^{1-x} x\right ) \, dx=25\,x\,{\mathrm {e}}^6\,{\mathrm {e}}^{{\mathrm {e}}^{-x}\,\mathrm {e}}\,{\mathrm {e}}^{{\mathrm {e}}^8} \]

[In]

int(-exp(exp(8) + 2*log(5) + exp(1 - x) + 6)*(x*exp(1 - x) - 1),x)

[Out]

25*x*exp(6)*exp(exp(-x)*exp(1))*exp(exp(8))