3.65.76 \(\int e^{3+16 x-8 e^2 x+e^4 x-x^2} (16-8 e^2+e^4-2 x) \, dx\)

Optimal. Leaf size=19 \[ 4+e^{3+\left (\left (-4+e^2\right )^2-x\right ) x} \]

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {2244, 2236} \begin {gather*} e^{-x^2+\left (4-e^2\right )^2 x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3 + 16*x - 8*E^2*x + E^4*x - x^2)*(16 - 8*E^2 + E^4 - 2*x),x]

[Out]

E^(3 + (4 - E^2)^2*x - x^2)

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{3+\left (4-e^2\right )^2 x-x^2} \left (\left (-4+e^2\right )^2-2 x\right ) \, dx\\ &=e^{3+\left (4-e^2\right )^2 x-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 0.95 \begin {gather*} e^{3+\left (-4+e^2\right )^2 x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3 + 16*x - 8*E^2*x + E^4*x - x^2)*(16 - 8*E^2 + E^4 - 2*x),x]

[Out]

E^(3 + (-4 + E^2)^2*x - x^2)

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fricas [A]  time = 0.54, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (-x^{2} + x e^{4} - 8 \, x e^{2} + 16 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)^2-8*exp(2)+16-2*x)*exp(x*exp(2)^2-8*exp(2)*x-x^2+16*x+3),x, algorithm="fricas")

[Out]

e^(-x^2 + x*e^4 - 8*x*e^2 + 16*x + 3)

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giac [A]  time = 0.17, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (-x^{2} + x e^{4} - 8 \, x e^{2} + 16 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)^2-8*exp(2)+16-2*x)*exp(x*exp(2)^2-8*exp(2)*x-x^2+16*x+3),x, algorithm="giac")

[Out]

e^(-x^2 + x*e^4 - 8*x*e^2 + 16*x + 3)

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maple [A]  time = 0.06, size = 21, normalized size = 1.11

method	result	size
risch	\({\mathrm e}^{x \,{\mathrm e}^{4}-8 \,{\mathrm e}^{2} x -x^{2}+16 x +3}\)	\(21\)
gosper	\({\mathrm e}^{x \,{\mathrm e}^{4}-8 \,{\mathrm e}^{2} x -x^{2}+16 x +3}\)	\(23\)
derivativedivides	\({\mathrm e}^{x \,{\mathrm e}^{4}-8 \,{\mathrm e}^{2} x -x^{2}+16 x +3}\)	\(23\)
default	\({\mathrm e}^{x \,{\mathrm e}^{4}-8 \,{\mathrm e}^{2} x -x^{2}+16 x +3}\)	\(23\)
norman	\({\mathrm e}^{x \,{\mathrm e}^{4}-8 \,{\mathrm e}^{2} x -x^{2}+16 x +3}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)^2-8*exp(2)+16-2*x)*exp(x*exp(2)^2-8*exp(2)*x-x^2+16*x+3),x,method=_RETURNVERBOSE)

[Out]

exp(x*exp(4)-8*exp(2)*x-x^2+16*x+3)

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maxima [A]  time = 0.38, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (-x^{2} + x e^{4} - 8 \, x e^{2} + 16 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)^2-8*exp(2)+16-2*x)*exp(x*exp(2)^2-8*exp(2)*x-x^2+16*x+3),x, algorithm="maxima")

[Out]

e^(-x^2 + x*e^4 - 8*x*e^2 + 16*x + 3)

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mupad [B]  time = 0.19, size = 24, normalized size = 1.26 \begin {gather*} {\mathrm {e}}^{16\,x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{x\,{\mathrm {e}}^4}\,{\mathrm {e}}^{-8\,x\,{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(16*x - 8*x*exp(2) + x*exp(4) - x^2 + 3)*(2*x + 8*exp(2) - exp(4) - 16),x)

[Out]

exp(16*x)*exp(3)*exp(-x^2)*exp(x*exp(4))*exp(-8*x*exp(2))

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sympy [A]  time = 0.12, size = 20, normalized size = 1.05 \begin {gather*} e^{- x^{2} - 8 x e^{2} + 16 x + x e^{4} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)**2-8*exp(2)+16-2*x)*exp(x*exp(2)**2-8*exp(2)*x-x**2+16*x+3),x)

[Out]

exp(-x**2 - 8*x*exp(2) + 16*x + x*exp(4) + 3)

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