3.62.5 \(\int e^{-x} (-16+48 x-2 e^x x-16 x^2+e^3 (-24+72 x-24 x^2)) \, dx\)

Optimal. Leaf size=25 \[ x \left (-x+4 e^{-x} \left (2+3 e^3\right ) (-2+2 x)\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 46, normalized size of antiderivative = 1.84, number of steps used = 16, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6742, 2194, 2176, 2196} \begin {gather*} 24 e^{3-x} x^2+16 e^{-x} x^2-x^2-24 e^{3-x} x-16 e^{-x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 48*x - 2*E^x*x - 16*x^2 + E^3*(-24 + 72*x - 24*x^2))/E^x,x]

[Out]

-24*E^(3 - x)*x - (16*x)/E^x - x^2 + 24*E^(3 - x)*x^2 + (16*x^2)/E^x

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-16 e^{-x}-2 x+48 e^{-x} x-16 e^{-x} x^2-24 e^{3-x} \left (1-3 x+x^2\right )\right ) \, dx\\ &=-x^2-16 \int e^{-x} \, dx-16 \int e^{-x} x^2 \, dx-24 \int e^{3-x} \left (1-3 x+x^2\right ) \, dx+48 \int e^{-x} x \, dx\\ &=16 e^{-x}-48 e^{-x} x-x^2+16 e^{-x} x^2-24 \int \left (e^{3-x}-3 e^{3-x} x+e^{3-x} x^2\right ) \, dx-32 \int e^{-x} x \, dx+48 \int e^{-x} \, dx\\ &=-32 e^{-x}-16 e^{-x} x-x^2+16 e^{-x} x^2-24 \int e^{3-x} \, dx-24 \int e^{3-x} x^2 \, dx-32 \int e^{-x} \, dx+72 \int e^{3-x} x \, dx\\ &=24 e^{3-x}-72 e^{3-x} x-16 e^{-x} x-x^2+24 e^{3-x} x^2+16 e^{-x} x^2-48 \int e^{3-x} x \, dx+72 \int e^{3-x} \, dx\\ &=-48 e^{3-x}-24 e^{3-x} x-16 e^{-x} x-x^2+24 e^{3-x} x^2+16 e^{-x} x^2-48 \int e^{3-x} \, dx\\ &=-24 e^{3-x} x-16 e^{-x} x-x^2+24 e^{3-x} x^2+16 e^{-x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 27, normalized size = 1.08 \begin {gather*} -e^{-x} x \left (-16 (-1+x)-24 e^3 (-1+x)+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 48*x - 2*E^x*x - 16*x^2 + E^3*(-24 + 72*x - 24*x^2))/E^x,x]

[Out]

-((x*(-16*(-1 + x) - 24*E^3*(-1 + x) + E^x*x))/E^x)

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fricas [A]  time = 0.58, size = 32, normalized size = 1.28 \begin {gather*} -{\left (x^{2} e^{x} - 16 \, x^{2} - 24 \, {\left (x^{2} - x\right )} e^{3} + 16 \, x\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)*x+(-24*x^2+72*x-24)*exp(3)-16*x^2+48*x-16)/exp(x),x, algorithm="fricas")

[Out]

-(x^2*e^x - 16*x^2 - 24*(x^2 - x)*e^3 + 16*x)*e^(-x)

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giac [A]  time = 0.15, size = 34, normalized size = 1.36 \begin {gather*} -x^{2} + 16 \, {\left (x^{2} - x\right )} e^{\left (-x\right )} + 24 \, {\left (x^{2} - x\right )} e^{\left (-x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)*x+(-24*x^2+72*x-24)*exp(3)-16*x^2+48*x-16)/exp(x),x, algorithm="giac")

[Out]

-x^2 + 16*(x^2 - x)*e^(-x) + 24*(x^2 - x)*e^(-x + 3)

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maple [A]  time = 0.04, size = 32, normalized size = 1.28

method	result	size
norman	\(\left (\left (-24 \,{\mathrm e}^{3}-16\right ) x +\left (24 \,{\mathrm e}^{3}+16\right ) x^{2}-{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}\)	\(32\)
risch	\(-x^{2}+\left (24 x^{2} {\mathrm e}^{3}-24 x \,{\mathrm e}^{3}+16 x^{2}-16 x \right ) {\mathrm e}^{-x}\)	\(33\)
default	\(-x^{2}-16 x \,{\mathrm e}^{-x}+16 x^{2} {\mathrm e}^{-x}+24 \,{\mathrm e}^{-x} {\mathrm e}^{3}+72 \,{\mathrm e}^{3} \left (-x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\right )-24 \,{\mathrm e}^{3} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(x)*x+(-24*x^2+72*x-24)*exp(3)-16*x^2+48*x-16)/exp(x),x,method=_RETURNVERBOSE)

[Out]

((-24*exp(3)-16)*x+(24*exp(3)+16)*x^2-exp(x)*x^2)/exp(x)

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maxima [B]  time = 0.37, size = 78, normalized size = 3.12 \begin {gather*} -x^{2} + 24 \, {\left (x^{2} e^{3} + 2 \, x e^{3} + 2 \, e^{3}\right )} e^{\left (-x\right )} + 16 \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} - 72 \, {\left (x e^{3} + e^{3}\right )} e^{\left (-x\right )} - 48 \, {\left (x + 1\right )} e^{\left (-x\right )} + 16 \, e^{\left (-x\right )} + 24 \, e^{\left (-x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)*x+(-24*x^2+72*x-24)*exp(3)-16*x^2+48*x-16)/exp(x),x, algorithm="maxima")

[Out]

-x^2 + 24*(x^2*e^3 + 2*x*e^3 + 2*e^3)*e^(-x) + 16*(x^2 + 2*x + 2)*e^(-x) - 72*(x*e^3 + e^3)*e^(-x) - 48*(x + 1
)*e^(-x) + 16*e^(-x) + 24*e^(-x + 3)

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mupad [B]  time = 4.36, size = 33, normalized size = 1.32 \begin {gather*} x^2\,{\mathrm {e}}^{-x}\,\left (24\,{\mathrm {e}}^3+16\right )-x\,{\mathrm {e}}^{-x}\,\left (24\,{\mathrm {e}}^3+16\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x)*(exp(3)*(24*x^2 - 72*x + 24) - 48*x + 2*x*exp(x) + 16*x^2 + 16),x)

[Out]

x^2*exp(-x)*(24*exp(3) + 16) - x*exp(-x)*(24*exp(3) + 16) - x^2

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sympy [A]  time = 0.13, size = 29, normalized size = 1.16 \begin {gather*} - x^{2} + \left (16 x^{2} + 24 x^{2} e^{3} - 24 x e^{3} - 16 x\right ) e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)*x+(-24*x**2+72*x-24)*exp(3)-16*x**2+48*x-16)/exp(x),x)

[Out]

-x**2 + (16*x**2 + 24*x**2*exp(3) - 24*x*exp(3) - 16*x)*exp(-x)

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