3.708 \(\int e^{8-2 x} x^3 \, dx\)

Optimal. Leaf size=52 \[ -\frac {1}{2} e^{8-2 x} x^3-\frac {3}{4} e^{8-2 x} x^2-\frac {3}{4} e^{8-2 x} x-\frac {3}{8} e^{8-2 x} \]

[Out]

-3/8*exp(8-2*x)-3/4*exp(8-2*x)*x-3/4*exp(8-2*x)*x^2-1/2*exp(8-2*x)*x^3

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2176, 2194} \[ -\frac {1}{2} e^{8-2 x} x^3-\frac {3}{4} e^{8-2 x} x^2-\frac {3}{4} e^{8-2 x} x-\frac {3}{8} e^{8-2 x} \]

Antiderivative was successfully verified.

[In]

Int[E^(8 - 2*x)*x^3,x]

[Out]

(-3*E^(8 - 2*x))/8 - (3*E^(8 - 2*x)*x)/4 - (3*E^(8 - 2*x)*x^2)/4 - (E^(8 - 2*x)*x^3)/2

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]

Rubi steps

\begin {align*} \int e^{8-2 x} x^3 \, dx &=-\frac {1}{2} e^{8-2 x} x^3+\frac {3}{2} \int e^{8-2 x} x^2 \, dx\\ &=-\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3+\frac {3}{2} \int e^{8-2 x} x \, dx\\ &=-\frac {3}{4} e^{8-2 x} x-\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3+\frac {3}{4} \int e^{8-2 x} \, dx\\ &=-\frac {3}{8} e^{8-2 x}-\frac {3}{4} e^{8-2 x} x-\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.50 \[ -\frac {1}{8} e^{8-2 x} \left (4 x^3+6 x^2+6 x+3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(8 - 2*x)*x^3,x]

[Out]

-1/8*(E^(8 - 2*x)*(3 + 6*x + 6*x^2 + 4*x^3))

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fricas [A]  time = 0.39, size = 23, normalized size = 0.44 \[ -\frac {1}{8} \, {\left (4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x + 8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(8-2*x)*x^3,x, algorithm="fricas")

[Out]

-1/8*(4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x + 8)

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giac [A]  time = 0.21, size = 23, normalized size = 0.44 \[ -\frac {1}{8} \, {\left (4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x + 8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(8-2*x)*x^3,x, algorithm="giac")

[Out]

-1/8*(4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x + 8)

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maple [A]  time = 0.02, size = 24, normalized size = 0.46 \[ -\frac {\left (4 x^{3}+6 x^{2}+6 x +3\right ) {\mathrm e}^{-2 x +8}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(8-2*x)*x^3,x)

[Out]

-1/8*(4*x^3+6*x^2+6*x+3)*exp(8-2*x)

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maxima [A]  time = 0.97, size = 30, normalized size = 0.58 \[ -\frac {1}{8} \, {\left (4 \, x^{3} e^{8} + 6 \, x^{2} e^{8} + 6 \, x e^{8} + 3 \, e^{8}\right )} e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(8-2*x)*x^3,x, algorithm="maxima")

[Out]

-1/8*(4*x^3*e^8 + 6*x^2*e^8 + 6*x*e^8 + 3*e^8)*e^(-2*x)

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mupad [B]  time = 0.05, size = 23, normalized size = 0.44 \[ -{\mathrm {e}}^{8-2\,x}\,\left (\frac {x^3}{2}+\frac {3\,x^2}{4}+\frac {3\,x}{4}+\frac {3}{8}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*exp(8 - 2*x),x)

[Out]

-exp(8 - 2*x)*((3*x)/4 + (3*x^2)/4 + x^3/2 + 3/8)

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sympy [A]  time = 0.09, size = 24, normalized size = 0.46 \[ \frac {\left (- 4 x^{3} - 6 x^{2} - 6 x - 3\right ) e^{8 - 2 x}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(8-2*x)*x**3,x)

[Out]

(-4*x**3 - 6*x**2 - 6*x - 3)*exp(8 - 2*x)/8

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