3.72.59 \(\int \frac {1}{2} e^{3+x-e^{16+x} x-x^2} (1+e^{16+x} (-1-x)-2 x) \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{2} \left (-5-e^{e^4}+e^{3+x-x \left (e^{16+x}+x\right )}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6706} \begin {gather*} \frac {1}{2} e^{-x^2-e^{x+16} x+x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 + x - E^(16 + x)*x - x^2)*(1 + E^(16 + x)*(-1 - x) - 2*x))/2,x]

[Out]

E^(3 + x - E^(16 + x)*x - x^2)/2

Rule 12

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int e^{3+x-e^{16+x} x-x^2} \left (1+e^{16+x} (-1-x)-2 x\right ) \, dx\\ &=\frac {1}{2} e^{3+x-e^{16+x} x-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 22, normalized size = 0.79 \begin {gather*} \frac {1}{2} e^{3+x-e^{16+x} x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + x - E^(16 + x)*x - x^2)*(1 + E^(16 + x)*(-1 - x) - 2*x))/2,x]

[Out]

E^(3 + x - E^(16 + x)*x - x^2)/2

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fricas [A]  time = 0.60, size = 18, normalized size = 0.64 \begin {gather*} \frac {1}{2} \, e^{\left (-x^{2} - x e^{\left (x + 16\right )} + x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-x-1)*exp(x+16)+1-2*x)*exp(-x*exp(x+16)-x^2+x+3),x, algorithm="fricas")

[Out]

1/2*e^(-x^2 - x*e^(x + 16) + x + 3)

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giac [A]  time = 0.17, size = 18, normalized size = 0.64 \begin {gather*} \frac {1}{2} \, e^{\left (-x^{2} - x e^{\left (x + 16\right )} + x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-x-1)*exp(x+16)+1-2*x)*exp(-x*exp(x+16)-x^2+x+3),x, algorithm="giac")

[Out]

1/2*e^(-x^2 - x*e^(x + 16) + x + 3)

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maple [A]  time = 0.06, size = 19, normalized size = 0.68

method	result	size
norman	\(\frac {{\mathrm e}^{-x \,{\mathrm e}^{x +16}-x^{2}+x +3}}{2}\)	\(19\)
risch	\(\frac {{\mathrm e}^{-x \,{\mathrm e}^{x +16}-x^{2}+x +3}}{2}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-x-1)*exp(x+16)+1-2*x)*exp(-x*exp(x+16)-x^2+x+3),x,method=_RETURNVERBOSE)

[Out]

1/2*exp(-x*exp(x+16)-x^2+x+3)

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maxima [A]  time = 0.42, size = 18, normalized size = 0.64 \begin {gather*} \frac {1}{2} \, e^{\left (-x^{2} - x e^{\left (x + 16\right )} + x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-x-1)*exp(x+16)+1-2*x)*exp(-x*exp(x+16)-x^2+x+3),x, algorithm="maxima")

[Out]

1/2*e^(-x^2 - x*e^(x + 16) + x + 3)

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mupad [B]  time = 0.07, size = 20, normalized size = 0.71 \begin {gather*} \frac {{\mathrm {e}}^3\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{16}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - x*exp(x + 16) - x^2 + 3)*(2*x + exp(x + 16)*(x + 1) - 1))/2,x)

[Out]

(exp(3)*exp(-x*exp(16)*exp(x))*exp(-x^2)*exp(x))/2

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sympy [A]  time = 0.22, size = 15, normalized size = 0.54 \begin {gather*} \frac {e^{- x^{2} - x e^{x + 16} + x + 3}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-x-1)*exp(x+16)+1-2*x)*exp(-x*exp(x+16)-x**2+x+3),x)

[Out]

exp(-x**2 - x*exp(x + 16) + x + 3)/2

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