3.73.35 \(\int \frac {1}{16} e^{e^{\frac {1}{16} (80+18 x-e^5 x-x^2)}-x} (-16+e^{\frac {1}{16} (80+18 x-e^5 x-x^2)} (18-e^5-2 x)) \, dx\)

Optimal. Leaf size=26 \[ e^{e^{5+x+\frac {1}{16} \left (2-e^5-x\right ) x}-x} \]

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Rubi [A]  time = 0.28, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {12, 6706} \begin {gather*} e^{e^{\frac {1}{16} \left (-x^2-e^5 x+18 x+80\right )}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^((80 + 18*x - E^5*x - x^2)/16) - x)*(-16 + E^((80 + 18*x - E^5*x - x^2)/16)*(18 - E^5 - 2*x)))/16,x]

[Out]

E^(E^((80 + 18*x - E^5*x - x^2)/16) - 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 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}{16} \int e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx\\ &=e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 21, normalized size = 0.81 \begin {gather*} e^{e^{5-\frac {1}{16} x \left (-18+e^5+x\right )}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^((80 + 18*x - E^5*x - x^2)/16) - x)*(-16 + E^((80 + 18*x - E^5*x - x^2)/16)*(18 - E^5 - 2*x)))
/16,x]

[Out]

E^(E^(5 - (x*(-18 + E^5 + x))/16) - x)

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fricas [A]  time = 0.67, size = 21, normalized size = 0.81 \begin {gather*} e^{\left (-x + e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-exp(5)-2*x+18)*exp(-1/16*x*exp(5)-1/16*x^2+9/8*x+5)-16)*exp(exp(-1/16*x*exp(5)-1/16*x^2+9/8*
x+5)-x),x, algorithm="fricas")

[Out]

e^(-x + e^(-1/16*x^2 - 1/16*x*e^5 + 9/8*x + 5))

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giac [A]  time = 0.20, size = 21, normalized size = 0.81 \begin {gather*} e^{\left (-x + e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-exp(5)-2*x+18)*exp(-1/16*x*exp(5)-1/16*x^2+9/8*x+5)-16)*exp(exp(-1/16*x*exp(5)-1/16*x^2+9/8*
x+5)-x),x, algorithm="giac")

[Out]

e^(-x + e^(-1/16*x^2 - 1/16*x*e^5 + 9/8*x + 5))

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maple [A]  time = 0.08, size = 22, normalized size = 0.85

method	result	size
norman	\({\mathrm e}^{{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{16}-\frac {x^{2}}{16}+\frac {9 x}{8}+5}-x}\)	\(22\)
risch	\({\mathrm e}^{{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{16}-\frac {x^{2}}{16}+\frac {9 x}{8}+5}-x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*((-exp(5)-2*x+18)*exp(-1/16*x*exp(5)-1/16*x^2+9/8*x+5)-16)*exp(exp(-1/16*x*exp(5)-1/16*x^2+9/8*x+5)-x
),x,method=_RETURNVERBOSE)

[Out]

exp(exp(-1/16*x*exp(5)-1/16*x^2+9/8*x+5)-x)

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maxima [A]  time = 0.49, size = 21, normalized size = 0.81 \begin {gather*} e^{\left (-x + e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-exp(5)-2*x+18)*exp(-1/16*x*exp(5)-1/16*x^2+9/8*x+5)-16)*exp(exp(-1/16*x*exp(5)-1/16*x^2+9/8*
x+5)-x),x, algorithm="maxima")

[Out]

e^(-x + e^(-1/16*x^2 - 1/16*x*e^5 + 9/8*x + 5))

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mupad [B]  time = 0.27, size = 25, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {9\,x}{8}}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {x^2}{16}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{16}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp((9*x)/8 - (x*exp(5))/16 - x^2/16 + 5) - x)*(exp((9*x)/8 - (x*exp(5))/16 - x^2/16 + 5)*(2*x + exp
(5) - 18) + 16))/16,x)

[Out]

exp(-x)*exp(exp((9*x)/8)*exp(5)*exp(-x^2/16)*exp(-(x*exp(5))/16))

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sympy [A]  time = 0.65, size = 22, normalized size = 0.85 \begin {gather*} e^{- x + e^{- \frac {x^{2}}{16} - \frac {x e^{5}}{16} + \frac {9 x}{8} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-exp(5)-2*x+18)*exp(-1/16*x*exp(5)-1/16*x**2+9/8*x+5)-16)*exp(exp(-1/16*x*exp(5)-1/16*x**2+9/
8*x+5)-x),x)

[Out]

exp(-x + exp(-x**2/16 - x*exp(5)/16 + 9*x/8 + 5))

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