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

Optimal. Leaf size=17 \[ 25 e^{6+e^8+e^{1-x}} x \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 2288} \begin {gather*} 25 e^{e^{1-x}+6+e^8} x \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 17, normalized size = 1.00 \begin {gather*} 25 e^{6+e^8+e^{1-x}} x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.84, size = 17, normalized size = 1.00 \begin {gather*} x e^{\left (e^{8} + e^{\left (-x + 1\right )} + 2 \, \log \relax (5) + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(e^8 + e^(-x + 1) + 2*log(5) + 6)

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giac [A]  time = 0.22, size = 14, normalized size = 0.82 \begin {gather*} 25 \, x e^{\left (e^{8} + e^{\left (-x + 1\right )} + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*x*e^(e^8 + e^(-x + 1) + 6)

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maple [A]  time = 0.06, size = 15, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[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)+6+exp(8))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 25 \, {\left (x e^{\left (e^{8} + 7\right )} + e^{\left (e^{8} + 7\right )}\right )} e^{\left (-x + e^{\left (-x + 1\right )}\right )} - 25 \, {\rm Ei}\left (e^{\left (-x + 1\right )}\right ) e^{\left (e^{8} + 6\right )} + 25 \, \int {\left (x e^{\left (e^{8} + 8\right )} + e^{\left (e^{8} + 8\right )}\right )} e^{\left (-2 \, x + e^{\left (-x + 1\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(-x+1)+1)*exp(exp(-x+1)+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)

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mupad [B]  time = 0.08, size = 16, normalized size = 0.94 \begin {gather*} 25\,x\,{\mathrm {e}}^6\,{\mathrm {e}}^{{\mathrm {e}}^{-x}\,\mathrm {e}}\,{\mathrm {e}}^{{\mathrm {e}}^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.14, size = 14, normalized size = 0.82 \begin {gather*} 25 x e^{e^{1 - x} + 6 + e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*x*exp(exp(1 - x) + 6 + exp(8))

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