3.535 \(\int e^{-x/2} x^3 \, dx\)

Optimal. Leaf size=44 \[ -2 e^{-x/2} x^3-12 e^{-x/2} x^2-48 e^{-x/2} x-96 e^{-x/2} \]

[Out]

-96/E^(x/2) - (48*x)/E^(x/2) - (12*x^2)/E^(x/2) - (2*x^3)/E^(x/2)

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Rubi [A]  time = 0.0317714, antiderivative size = 44, normalized size of antiderivative = 1., 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} \[ -2 e^{-x/2} x^3-12 e^{-x/2} x^2-48 e^{-x/2} x-96 e^{-x/2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^(x/2),x]

[Out]

-96/E^(x/2) - (48*x)/E^(x/2) - (12*x^2)/E^(x/2) - (2*x^3)/E^(x/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^{-x/2} x^3 \, dx &=-2 e^{-x/2} x^3+6 \int e^{-x/2} x^2 \, dx\\ &=-12 e^{-x/2} x^2-2 e^{-x/2} x^3+24 \int e^{-x/2} x \, dx\\ &=-48 e^{-x/2} x-12 e^{-x/2} x^2-2 e^{-x/2} x^3+48 \int e^{-x/2} \, dx\\ &=-96 e^{-x/2}-48 e^{-x/2} x-12 e^{-x/2} x^2-2 e^{-x/2} x^3\\ \end{align*}

Mathematica [A]  time = 0.0071218, size = 23, normalized size = 0.52 \[ e^{-x/2} \left (-2 x^3-12 x^2-48 x-96\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/E^(x/2),x]

[Out]

(-96 - 48*x - 12*x^2 - 2*x^3)/E^(x/2)

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Maple [A]  time = 0.002, size = 22, normalized size = 0.5 \begin{align*} -2\,{\frac{{x}^{3}+6\,{x}^{2}+24\,x+48}{{{\rm e}^{x/2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/exp(1/2*x),x)

[Out]

-2*(x^3+6*x^2+24*x+48)/exp(1/2*x)

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Maxima [A]  time = 0.937732, size = 26, normalized size = 0.59 \begin{align*} -2 \,{\left (x^{3} + 6 \, x^{2} + 24 \, x + 48\right )} e^{\left (-\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/exp(1/2*x),x, algorithm="maxima")

[Out]

-2*(x^3 + 6*x^2 + 24*x + 48)*e^(-1/2*x)

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Fricas [A]  time = 1.75255, size = 55, normalized size = 1.25 \begin{align*} -2 \,{\left (x^{3} + 6 \, x^{2} + 24 \, x + 48\right )} e^{\left (-\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/exp(1/2*x),x, algorithm="fricas")

[Out]

-2*(x^3 + 6*x^2 + 24*x + 48)*e^(-1/2*x)

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Sympy [A]  time = 0.084769, size = 20, normalized size = 0.45 \begin{align*} \left (- 2 x^{3} - 12 x^{2} - 48 x - 96\right ) e^{- \frac{x}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/exp(1/2*x),x)

[Out]

(-2*x**3 - 12*x**2 - 48*x - 96)*exp(-x/2)

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Giac [A]  time = 1.08907, size = 26, normalized size = 0.59 \begin{align*} -2 \,{\left (x^{3} + 6 \, x^{2} + 24 \, x + 48\right )} e^{\left (-\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/exp(1/2*x),x, algorithm="giac")

[Out]

-2*(x^3 + 6*x^2 + 24*x + 48)*e^(-1/2*x)