3.37.11 \(\int e^{-4-x} (-e^{4+x}-30 x^2+10 x^3) \, dx\)

Optimal. Leaf size=21 \[ 83-x-2 \left (5+5 e^{-4-x} x^3\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 11, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6688, 2196, 2176, 2194} \begin {gather*} -10 e^{-x-4} x^3-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-4 - x)*(-E^(4 + x) - 30*x^2 + 10*x^3),x]

[Out]

-x - 10*E^(-4 - x)*x^3

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+10 e^{-4-x} (-3+x) x^2\right ) \, dx\\ &=-x+10 \int e^{-4-x} (-3+x) x^2 \, dx\\ &=-x+10 \int \left (-3 e^{-4-x} x^2+e^{-4-x} x^3\right ) \, dx\\ &=-x+10 \int e^{-4-x} x^3 \, dx-30 \int e^{-4-x} x^2 \, dx\\ &=-x+30 e^{-4-x} x^2-10 e^{-4-x} x^3+30 \int e^{-4-x} x^2 \, dx-60 \int e^{-4-x} x \, dx\\ &=-x+60 e^{-4-x} x-10 e^{-4-x} x^3-60 \int e^{-4-x} \, dx+60 \int e^{-4-x} x \, dx\\ &=60 e^{-4-x}-x-10 e^{-4-x} x^3+60 \int e^{-4-x} \, dx\\ &=-x-10 e^{-4-x} x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.76 \begin {gather*} -x-10 e^{-4-x} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-4 - x)*(-E^(4 + x) - 30*x^2 + 10*x^3),x]

[Out]

-x - 10*E^(-4 - x)*x^3

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fricas [A]  time = 0.68, size = 20, normalized size = 0.95 \begin {gather*} -{\left (10 \, x^{3} + x e^{\left (x + 4\right )}\right )} e^{\left (-x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(4+x)+10*x^3-30*x^2)/exp(4+x),x, algorithm="fricas")

[Out]

-(10*x^3 + x*e^(x + 4))*e^(-x - 4)

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giac [A]  time = 0.19, size = 18, normalized size = 0.86 \begin {gather*} -{\left (10 \, x^{3} e^{\left (-x\right )} + x e^{4}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(4+x)+10*x^3-30*x^2)/exp(4+x),x, algorithm="giac")

[Out]

-(10*x^3*e^(-x) + x*e^4)*e^(-4)

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maple [A]  time = 0.06, size = 16, normalized size = 0.76

method	result	size
risch	\(-x -10 \,{\mathrm e}^{-x -4} x^{3}\)	\(16\)
norman	\(\left (-10 x^{3}-x \,{\mathrm e}^{4+x}\right ) {\mathrm e}^{-x -4}\)	\(21\)
derivativedivides	\(-4-x +640 \,{\mathrm e}^{-x -4}-480 \,{\mathrm e}^{-x -4} \left (4+x \right )+120 \,{\mathrm e}^{-x -4} \left (4+x \right )^{2}-10 \,{\mathrm e}^{-x -4} \left (4+x \right )^{3}\)	\(51\)
default	\(-4-x +640 \,{\mathrm e}^{-x -4}-480 \,{\mathrm e}^{-x -4} \left (4+x \right )+120 \,{\mathrm e}^{-x -4} \left (4+x \right )^{2}-10 \,{\mathrm e}^{-x -4} \left (4+x \right )^{3}\)	\(51\)
meijerg	\(-\frac {{\mathrm e}^{-x +x \,{\mathrm e}^{-4}+4} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{-4} \left (1-{\mathrm e}^{4}\right )}\right )}{1-{\mathrm e}^{4}}+10 \,{\mathrm e}^{12-x +x \,{\mathrm e}^{-4}} \left (6-\frac {\left (4 x^{3} {\mathrm e}^{-12}+12 x^{2} {\mathrm e}^{-8}+24 x \,{\mathrm e}^{-4}+24\right ) {\mathrm e}^{-x \,{\mathrm e}^{-4}}}{4}\right )-30 \,{\mathrm e}^{8-x +x \,{\mathrm e}^{-4}} \left (2-\frac {\left (3 x^{2} {\mathrm e}^{-8}+6 x \,{\mathrm e}^{-4}+6\right ) {\mathrm e}^{-x \,{\mathrm e}^{-4}}}{3}\right )\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(4+x)+10*x^3-30*x^2)/exp(4+x),x,method=_RETURNVERBOSE)

[Out]

-x-10*exp(-x-4)*x^3

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maxima [B]  time = 0.39, size = 41, normalized size = 1.95 \begin {gather*} -10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right )} e^{\left (-x - 4\right )} + 30 \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x - 4\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(4+x)+10*x^3-30*x^2)/exp(4+x),x, algorithm="maxima")

[Out]

-10*(x^3 + 3*x^2 + 6*x + 6)*e^(-x - 4) + 30*(x^2 + 2*x + 2)*e^(-x - 4) - x

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mupad [B]  time = 2.14, size = 15, normalized size = 0.71 \begin {gather*} -x-10\,x^3\,{\mathrm {e}}^{-x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(- x - 4)*(exp(x + 4) + 30*x^2 - 10*x^3),x)

[Out]

- x - 10*x^3*exp(- x - 4)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.67 \begin {gather*} - 10 x^{3} e^{- x - 4} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(4+x)+10*x**3-30*x**2)/exp(4+x),x)

[Out]

-10*x**3*exp(-x - 4) - x

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