3.26.97 \(\int e^{-4 x} (-32 x^3-4 e^{4 x} x^3+52 x^4-16 x^5) \, dx\)

Optimal. Leaf size=18 \[ -x^3 \left (x-4 e^{-4 x} (-2+x) x\right ) \]

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Rubi [A]  time = 0.46, antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 21, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6741, 12, 6742, 2196, 2176, 2194} \begin {gather*} 4 e^{-4 x} x^5-8 e^{-4 x} x^4-x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*x^3 - 4*E^(4*x)*x^3 + 52*x^4 - 16*x^5)/E^(4*x),x]

[Out]

-x^4 - (8*x^4)/E^(4*x) + (4*x^5)/E^(4*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6741

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

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 4 e^{-4 x} x^3 \left (-8-e^{4 x}+13 x-4 x^2\right ) \, dx\\ &=4 \int e^{-4 x} x^3 \left (-8-e^{4 x}+13 x-4 x^2\right ) \, dx\\ &=4 \int \left (-x^3-e^{-4 x} x^3 \left (8-13 x+4 x^2\right )\right ) \, dx\\ &=-x^4-4 \int e^{-4 x} x^3 \left (8-13 x+4 x^2\right ) \, dx\\ &=-x^4-4 \int \left (8 e^{-4 x} x^3-13 e^{-4 x} x^4+4 e^{-4 x} x^5\right ) \, dx\\ &=-x^4-16 \int e^{-4 x} x^5 \, dx-32 \int e^{-4 x} x^3 \, dx+52 \int e^{-4 x} x^4 \, dx\\ &=8 e^{-4 x} x^3-x^4-13 e^{-4 x} x^4+4 e^{-4 x} x^5-20 \int e^{-4 x} x^4 \, dx-24 \int e^{-4 x} x^2 \, dx+52 \int e^{-4 x} x^3 \, dx\\ &=6 e^{-4 x} x^2-5 e^{-4 x} x^3-x^4-8 e^{-4 x} x^4+4 e^{-4 x} x^5-12 \int e^{-4 x} x \, dx-20 \int e^{-4 x} x^3 \, dx+39 \int e^{-4 x} x^2 \, dx\\ &=3 e^{-4 x} x-\frac {15}{4} e^{-4 x} x^2-x^4-8 e^{-4 x} x^4+4 e^{-4 x} x^5-3 \int e^{-4 x} \, dx-15 \int e^{-4 x} x^2 \, dx+\frac {39}{2} \int e^{-4 x} x \, dx\\ &=\frac {3 e^{-4 x}}{4}-\frac {15}{8} e^{-4 x} x-x^4-8 e^{-4 x} x^4+4 e^{-4 x} x^5+\frac {39}{8} \int e^{-4 x} \, dx-\frac {15}{2} \int e^{-4 x} x \, dx\\ &=-\frac {15}{32} e^{-4 x}-x^4-8 e^{-4 x} x^4+4 e^{-4 x} x^5-\frac {15}{8} \int e^{-4 x} \, dx\\ &=-x^4-8 e^{-4 x} x^4+4 e^{-4 x} x^5\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 20, normalized size = 1.11 \begin {gather*} -e^{-4 x} \left (8+e^{4 x}-4 x\right ) x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x^3 - 4*E^(4*x)*x^3 + 52*x^4 - 16*x^5)/E^(4*x),x]

[Out]

-(((8 + E^(4*x) - 4*x)*x^4)/E^(4*x))

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fricas [A]  time = 0.60, size = 25, normalized size = 1.39 \begin {gather*} {\left (4 \, x^{5} - x^{4} e^{\left (4 \, x\right )} - 8 \, x^{4}\right )} e^{\left (-4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3*exp(4*x)-16*x^5+52*x^4-32*x^3)/exp(4*x),x, algorithm="fricas")

[Out]

(4*x^5 - x^4*e^(4*x) - 8*x^4)*e^(-4*x)

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giac [A]  time = 0.23, size = 24, normalized size = 1.33 \begin {gather*} 4 \, x^{5} e^{\left (-4 \, x\right )} - 8 \, x^{4} e^{\left (-4 \, x\right )} - x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3*exp(4*x)-16*x^5+52*x^4-32*x^3)/exp(4*x),x, algorithm="giac")

[Out]

4*x^5*e^(-4*x) - 8*x^4*e^(-4*x) - x^4

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maple [A]  time = 0.03, size = 23, normalized size = 1.28

method	result	size
risch	\(-x^{4}+\left (4 x^{5}-8 x^{4}\right ) {\mathrm e}^{-4 x}\)	\(23\)
norman	\(\left (-8 x^{4}+4 x^{5}-x^{4} {\mathrm e}^{4 x}\right ) {\mathrm e}^{-4 x}\)	\(28\)
derivativedivides	\(-x^{4}-8 \,{\mathrm e}^{-4 x} x^{4}+4 \,{\mathrm e}^{-4 x} x^{5}\)	\(29\)
default	\(-x^{4}-8 \,{\mathrm e}^{-4 x} x^{4}+4 \,{\mathrm e}^{-4 x} x^{5}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^3*exp(4*x)-16*x^5+52*x^4-32*x^3)/exp(4*x),x,method=_RETURNVERBOSE)

[Out]

-x^4+(4*x^5-8*x^4)*exp(-4*x)

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maxima [B]  time = 0.42, size = 84, normalized size = 4.67 \begin {gather*} -x^{4} + \frac {1}{32} \, {\left (128 \, x^{5} + 160 \, x^{4} + 160 \, x^{3} + 120 \, x^{2} + 60 \, x + 15\right )} e^{\left (-4 \, x\right )} - \frac {13}{32} \, {\left (32 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 12 \, x + 3\right )} e^{\left (-4 \, x\right )} + \frac {1}{4} \, {\left (32 \, x^{3} + 24 \, x^{2} + 12 \, x + 3\right )} e^{\left (-4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3*exp(4*x)-16*x^5+52*x^4-32*x^3)/exp(4*x),x, algorithm="maxima")

[Out]

-x^4 + 1/32*(128*x^5 + 160*x^4 + 160*x^3 + 120*x^2 + 60*x + 15)*e^(-4*x) - 13/32*(32*x^4 + 32*x^3 + 24*x^2 + 1
2*x + 3)*e^(-4*x) + 1/4*(32*x^3 + 24*x^2 + 12*x + 3)*e^(-4*x)

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mupad [B]  time = 0.06, size = 20, normalized size = 1.11 \begin {gather*} -x^4\,\left (8\,{\mathrm {e}}^{-4\,x}-4\,x\,{\mathrm {e}}^{-4\,x}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-4*x)*(4*x^3*exp(4*x) + 32*x^3 - 52*x^4 + 16*x^5),x)

[Out]

-x^4*(8*exp(-4*x) - 4*x*exp(-4*x) + 1)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.94 \begin {gather*} - x^{4} + \left (4 x^{5} - 8 x^{4}\right ) e^{- 4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**3*exp(4*x)-16*x**5+52*x**4-32*x**3)/exp(4*x),x)

[Out]

-x**4 + (4*x**5 - 8*x**4)*exp(-4*x)

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