3.8.41 \(\int (-5+5 e^{-2+x}+e^{-59-16 x-x^2} (-960-120 x)) \, dx\)

Optimal. Leaf size=24 \[ 5 \left (e^{-2+x}+12 e^{5-(8+x)^2}-x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2194, 2236} \begin {gather*} 60 e^{-x^2-16 x-59}-5 x+5 e^{x-2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-5 + 5*E^(-2 + x) + E^(-59 - 16*x - x^2)*(-960 - 120*x),x]

[Out]

5*E^(-2 + x) + 60*E^(-59 - 16*x - x^2) - 5*x

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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} &=-5 x+5 \int e^{-2+x} \, dx+\int e^{-59-16 x-x^2} (-960-120 x) \, dx\\ &=5 e^{-2+x}+60 e^{-59-16 x-x^2}-5 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 25, normalized size = 1.04 \begin {gather*} 5 \left (e^{-2+x}+12 e^{-59-16 x-x^2}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-5 + 5*E^(-2 + x) + E^(-59 - 16*x - x^2)*(-960 - 120*x),x]

[Out]

5*(E^(-2 + x) + 12*E^(-59 - 16*x - x^2) - x)

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fricas [A]  time = 0.73, size = 23, normalized size = 0.96 \begin {gather*} -5 \, x + 60 \, e^{\left (-x^{2} - 16 \, x - 59\right )} + 5 \, e^{\left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5*exp(x-2)+(-120*x-960)*exp(-x^2-16*x-59)-5,x, algorithm="fricas")

[Out]

-5*x + 60*e^(-x^2 - 16*x - 59) + 5*e^(x - 2)

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giac [A]  time = 0.47, size = 23, normalized size = 0.96 \begin {gather*} -5 \, x + 60 \, e^{\left (-x^{2} - 16 \, x - 59\right )} + 5 \, e^{\left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5*exp(x-2)+(-120*x-960)*exp(-x^2-16*x-59)-5,x, algorithm="giac")

[Out]

-5*x + 60*e^(-x^2 - 16*x - 59) + 5*e^(x - 2)

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maple [A]  time = 0.03, size = 24, normalized size = 1.00

method	result	size
default	\(-5 x +60 \,{\mathrm e}^{-x^{2}-16 x -59}+5 \,{\mathrm e}^{x -2}\)	\(24\)
norman	\(-5 x +60 \,{\mathrm e}^{-x^{2}-16 x -59}+5 \,{\mathrm e}^{x -2}\)	\(24\)
risch	\(-5 x +60 \,{\mathrm e}^{-x^{2}-16 x -59}+5 \,{\mathrm e}^{x -2}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*exp(x-2)+(-120*x-960)*exp(-x^2-16*x-59)-5,x,method=_RETURNVERBOSE)

[Out]

-5*x+60*exp(-x^2-16*x-59)+5*exp(x-2)

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maxima [A]  time = 0.51, size = 23, normalized size = 0.96 \begin {gather*} -5 \, x + 60 \, e^{\left (-x^{2} - 16 \, x - 59\right )} + 5 \, e^{\left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5*exp(x-2)+(-120*x-960)*exp(-x^2-16*x-59)-5,x, algorithm="maxima")

[Out]

-5*x + 60*e^(-x^2 - 16*x - 59) + 5*e^(x - 2)

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mupad [B]  time = 0.10, size = 24, normalized size = 1.00 \begin {gather*} 5\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x-5\,x+60\,{\mathrm {e}}^{-16\,x}\,{\mathrm {e}}^{-59}\,{\mathrm {e}}^{-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*exp(x - 2) - exp(- 16*x - x^2 - 59)*(120*x + 960) - 5,x)

[Out]

5*exp(-2)*exp(x) - 5*x + 60*exp(-16*x)*exp(-59)*exp(-x^2)

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sympy [A]  time = 0.14, size = 22, normalized size = 0.92 \begin {gather*} - 5 x + 5 e^{x - 2} + 60 e^{- x^{2} - 16 x - 59} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5*exp(x-2)+(-120*x-960)*exp(-x**2-16*x-59)-5,x)

[Out]

-5*x + 5*exp(x - 2) + 60*exp(-x**2 - 16*x - 59)

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