3.77 \(\int e^{-2 x} x^2 \, dx\)

Optimal. Leaf size=32 \[ -\frac{1}{2} e^{-2 x} x^2-\frac{1}{2} e^{-2 x} x-\frac{e^{-2 x}}{4} \]

[Out]

-1/(4*E^(2*x)) - x/(2*E^(2*x)) - x^2/(2*E^(2*x))

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Rubi [A]  time = 0.0176593, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2176, 2194} \[ -\frac{1}{2} e^{-2 x} x^2-\frac{1}{2} e^{-2 x} x-\frac{e^{-2 x}}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*E^(2*x)) - x/(2*E^(2*x)) - x^2/(2*E^(2*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]

Rubi steps

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

Mathematica [A]  time = 0.0063322, size = 19, normalized size = 0.59 \[ -\frac{1}{4} e^{-2 x} \left (2 x^2+2 x+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 + 2*x + 2*x^2)/(4*E^(2*x))

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Maple [A]  time = 0.002, size = 19, normalized size = 0.6 \begin{align*} -{\frac{2\,{x}^{2}+2\,x+1}{4\,{{\rm e}^{2\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.96247, size = 22, normalized size = 0.69 \begin{align*} -\frac{1}{4} \,{\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x^2 + 2*x + 1)*e^(-2*x)

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Fricas [A]  time = 0.486861, size = 45, normalized size = 1.41 \begin{align*} -\frac{1}{4} \,{\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x^2 + 2*x + 1)*e^(-2*x)

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Sympy [A]  time = 0.081091, size = 17, normalized size = 0.53 \begin{align*} \frac{\left (- 2 x^{2} - 2 x - 1\right ) e^{- 2 x}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*x**2 - 2*x - 1)*exp(-2*x)/4

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Giac [A]  time = 1.08076, size = 22, normalized size = 0.69 \begin{align*} -\frac{1}{4} \,{\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x^2 + 2*x + 1)*e^(-2*x)