3.30.50 \(\int e^{65-16 x+x^2} (-16+2 x) \, dx\)

Optimal. Leaf size=11 \[ e^{1+(8-x)^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 0.91, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2236} \begin {gather*} e^{x^2-16 x+65} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(65 - 16*x + x^2)*(-16 + 2*x),x]

[Out]

E^(65 - 16*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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{65-16 x+x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 10, normalized size = 0.91 \begin {gather*} e^{65-16 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(65 - 16*x + x^2)*(-16 + 2*x),x]

[Out]

E^(65 - 16*x + x^2)

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fricas [A]  time = 0.62, size = 9, normalized size = 0.82 \begin {gather*} e^{\left (x^{2} - 16 \, x + 65\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-16)*exp(x^2-16*x+65),x, algorithm="fricas")

[Out]

e^(x^2 - 16*x + 65)

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giac [A]  time = 0.21, size = 9, normalized size = 0.82 \begin {gather*} e^{\left (x^{2} - 16 \, x + 65\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-16)*exp(x^2-16*x+65),x, algorithm="giac")

[Out]

e^(x^2 - 16*x + 65)

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maple [A]  time = 0.04, size = 10, normalized size = 0.91

method	result	size
gosper	\({\mathrm e}^{x^{2}-16 x +65}\)	\(10\)
derivativedivides	\({\mathrm e}^{x^{2}-16 x +65}\)	\(10\)
default	\({\mathrm e}^{x^{2}-16 x +65}\)	\(10\)
norman	\({\mathrm e}^{x^{2}-16 x +65}\)	\(10\)
risch	\({\mathrm e}^{x^{2}-16 x +65}\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x-16)*exp(x^2-16*x+65),x,method=_RETURNVERBOSE)

[Out]

exp(x^2-16*x+65)

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maxima [A]  time = 0.43, size = 9, normalized size = 0.82 \begin {gather*} e^{\left (x^{2} - 16 \, x + 65\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-16)*exp(x^2-16*x+65),x, algorithm="maxima")

[Out]

e^(x^2 - 16*x + 65)

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mupad [B]  time = 0.07, size = 11, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-16\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{65} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2 - 16*x + 65)*(2*x - 16),x)

[Out]

exp(-16*x)*exp(x^2)*exp(65)

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sympy [A]  time = 0.08, size = 8, normalized size = 0.73 \begin {gather*} e^{x^{2} - 16 x + 65} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-16)*exp(x**2-16*x+65),x)

[Out]

exp(x**2 - 16*x + 65)

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