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

Optimal. Leaf size=14 \[ \frac {80}{27}-e^{-2 x} x^2 \]

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Rubi [A]  time = 0.06, antiderivative size = 10, normalized size of antiderivative = 0.71, number of steps used = 8, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} -e^{-2 x} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x + 2*x^2)/E^(2*x),x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{-2 x} x (-2+2 x) \, dx\\ &=\int \left (-2 e^{-2 x} x+2 e^{-2 x} x^2\right ) \, dx\\ &=-\left (2 \int e^{-2 x} x \, dx\right )+2 \int e^{-2 x} x^2 \, dx\\ &=e^{-2 x} x-e^{-2 x} x^2+2 \int e^{-2 x} x \, dx-\int e^{-2 x} \, dx\\ &=\frac {e^{-2 x}}{2}-e^{-2 x} x^2+\int e^{-2 x} \, dx\\ &=-e^{-2 x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.71 \begin {gather*} -e^{-2 x} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 2*x^2)/E^(2*x),x]

[Out]

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

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fricas [A]  time = 0.56, size = 9, normalized size = 0.64 \begin {gather*} -x^{2} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*e^(-2*x)

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giac [A]  time = 0.25, size = 9, normalized size = 0.64 \begin {gather*} -x^{2} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*e^(-2*x)

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maple [A]  time = 0.02, size = 10, normalized size = 0.71

method	result	size
gosper	\(-x^{2} {\mathrm e}^{-2 x}\)	\(10\)
default	\(-x^{2} {\mathrm e}^{-2 x}\)	\(10\)
norman	\(-x^{2} {\mathrm e}^{-2 x}\)	\(10\)
risch	\(-x^{2} {\mathrm e}^{-2 x}\)	\(10\)
meijerg	\(-\frac {\left (12 x^{2}+12 x +6\right ) {\mathrm e}^{-2 x}}{12}+\frac {\left (4 x +2\right ) {\mathrm e}^{-2 x}}{4}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2-2*x)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x^2/exp(x)^2

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maxima [B]  time = 0.41, size = 28, normalized size = 2.00 \begin {gather*} -\frac {1}{2} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {1}{2} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x^2 + 2*x + 1)*e^(-2*x) + 1/2*(2*x + 1)*e^(-2*x)

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mupad [B]  time = 1.39, size = 9, normalized size = 0.64 \begin {gather*} -x^2\,{\mathrm {e}}^{-2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-2*x)*(2*x - 2*x^2),x)

[Out]

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

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sympy [A]  time = 0.08, size = 8, normalized size = 0.57 \begin {gather*} - x^{2} e^{- 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*exp(-2*x)

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