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

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, 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]

-1/4*(1 + 2*x + 2*x^2)/E^(2*x)

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fricas [A]  time = 0.38, size = 16, normalized size = 0.50 \[ -\frac {1}{4} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \]

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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giac [A]  time = 0.01, size = 16, normalized size = 0.50 \[ -\frac {1}{4} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \]

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)

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maple [A]  time = 0.00, size = 19, normalized size = 0.59 \[ -\frac {\left (2 x^{2}+2 x +1\right ) {\mathrm e}^{-2 x}}{4} \]

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 = 0.65, size = 16, normalized size = 0.50 \[ -\frac {1}{4} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \]

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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mupad [B]  time = 0.08, size = 16, normalized size = 0.50 \[ -\frac {{\mathrm {e}}^{-2\,x}\,\left (4\,x^2+4\,x+2\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(-2*x)*(4*x + 4*x^2 + 2))/8

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sympy [A]  time = 0.09, size = 17, normalized size = 0.53 \[ \frac {\left (- 2 x^{2} - 2 x - 1\right ) e^{- 2 x}}{4} \]

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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