3.27.54 \(\int e^{\frac {1-e^{\frac {5}{2}+2 x} x-e^{5/2} x^2}{e^{5/2}}} (e^{2 x} (-1-2 x)-2 x) \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {1}{e^{5/2}}-e^{2 x} x-x^2} \]

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Rubi [A]  time = 0.18, antiderivative size = 32, normalized size of antiderivative = 1.52, number of steps used = 1, number of rules used = 1, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6706} \begin {gather*} e^{\frac {-e^{5/2} x^2-e^{2 x+\frac {5}{2}} x+1}{e^{5/2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((1 - E^(5/2 + 2*x)*x - E^(5/2)*x^2)/E^(5/2))*(E^(2*x)*(-1 - 2*x) - 2*x),x]

[Out]

E^((1 - E^(5/2 + 2*x)*x - E^(5/2)*x^2)/E^(5/2))

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} &=e^{\frac {1-e^{\frac {5}{2}+2 x} x-e^{5/2} x^2}{e^{5/2}}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 21, normalized size = 1.00 \begin {gather*} e^{\frac {1}{e^{5/2}}-e^{2 x} x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^((1 - E^(5/2 + 2*x)*x - E^(5/2)*x^2)/E^(5/2))*(E^(2*x)*(-1 - 2*x) - 2*x),x]

[Out]

E^(E^(-5/2) - E^(2*x)*x - x^2)

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fricas [A]  time = 0.65, size = 21, normalized size = 1.00 \begin {gather*} e^{\left (-{\left (x^{2} e^{\frac {5}{2}} + x e^{\left (2 \, x + \frac {5}{2}\right )} - 1\right )} e^{\left (-\frac {5}{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 16, normalized size = 0.76 \begin {gather*} e^{\left (-x^{2} - x e^{\left (2 \, x\right )} + e^{\left (-\frac {5}{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^2 - x*e^(2*x) + e^(-5/2))

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maple [A]  time = 0.06, size = 22, normalized size = 1.05

method	result	size
risch	\({\mathrm e}^{-\left (x^{2} {\mathrm e}^{\frac {5}{2}}+x \,{\mathrm e}^{\frac {5}{2}+2 x}-1\right ) {\mathrm e}^{-\frac {5}{2}}}\)	\(22\)
norman	\({\mathrm e}^{\left (-x \,{\mathrm e}^{\frac {5}{2}} {\mathrm e}^{2 x}-x^{2} {\mathrm e}^{\frac {5}{2}}+1\right ) {\mathrm e}^{-\frac {5}{2}}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.92, size = 16, normalized size = 0.76 \begin {gather*} e^{\left (-x^{2} - x e^{\left (2 \, x\right )} + e^{\left (-\frac {5}{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-x^2 - x*e^(2*x) + e^(-5/2))

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mupad [B]  time = 0.07, size = 18, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{-x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-exp(-5/2)*(x^2*exp(5/2) + x*exp(2*x)*exp(5/2) - 1))*(2*x + exp(2*x)*(2*x + 1)),x)

[Out]

exp(-x*exp(2*x))*exp(-x^2)*exp(exp(-5/2))

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sympy [A]  time = 0.17, size = 27, normalized size = 1.29 \begin {gather*} e^{\frac {- x^{2} e^{\frac {5}{2}} - x e^{\frac {5}{2}} e^{2 x} + 1}{e^{\frac {5}{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*exp(2*x)-2*x)*exp((-x*exp(5/2)*exp(2*x)-x**2*exp(5/2)+1)/exp(5/2)),x)

[Out]

exp((-x**2*exp(5/2) - x*exp(5/2)*exp(2*x) + 1)*exp(-5/2))

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