3.741 \(\int e^{-4 x} (2-3 x+x^2) \, dx\)

Optimal. Leaf size=32 \[ -\frac{1}{4} e^{-4 x} x^2+\frac{5}{8} e^{-4 x} x-\frac{11 e^{-4 x}}{32} \]

[Out]

-11/(32*E^(4*x)) + (5*x)/(8*E^(4*x)) - x^2/(4*E^(4*x))

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Rubi [A]  time = 0.0439118, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2196, 2194, 2176} \[ -\frac{1}{4} e^{-4 x} x^2+\frac{5}{8} e^{-4 x} x-\frac{11 e^{-4 x}}{32} \]

Antiderivative was successfully verified.

[In]

Int[(2 - 3*x + x^2)/E^(4*x),x]

[Out]

-11/(32*E^(4*x)) + (5*x)/(8*E^(4*x)) - x^2/(4*E^(4*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

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 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

Rubi steps

\begin{align*} \int e^{-4 x} \left (2-3 x+x^2\right ) \, dx &=\int \left (2 e^{-4 x}-3 e^{-4 x} x+e^{-4 x} x^2\right ) \, dx\\ &=2 \int e^{-4 x} \, dx-3 \int e^{-4 x} x \, dx+\int e^{-4 x} x^2 \, dx\\ &=-\frac{1}{2} e^{-4 x}+\frac{3}{4} e^{-4 x} x-\frac{1}{4} e^{-4 x} x^2+\frac{1}{2} \int e^{-4 x} x \, dx-\frac{3}{4} \int e^{-4 x} \, dx\\ &=-\frac{5}{16} e^{-4 x}+\frac{5}{8} e^{-4 x} x-\frac{1}{4} e^{-4 x} x^2+\frac{1}{8} \int e^{-4 x} \, dx\\ &=-\frac{11}{32} e^{-4 x}+\frac{5}{8} e^{-4 x} x-\frac{1}{4} e^{-4 x} x^2\\ \end{align*}

Mathematica [A]  time = 0.0194691, size = 19, normalized size = 0.59 \[ -\frac{1}{32} e^{-4 x} \left (8 x^2-20 x+11\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - 3*x + x^2)/E^(4*x),x]

[Out]

-(11 - 20*x + 8*x^2)/(32*E^(4*x))

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Maple [A]  time = 0.019, size = 19, normalized size = 0.6 \begin{align*} -{\frac{8\,{x}^{2}-20\,x+11}{32\,{{\rm e}^{4\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(8*x^2-20*x+11)/exp(4*x)

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Maxima [A]  time = 1.00812, size = 46, normalized size = 1.44 \begin{align*} -\frac{1}{32} \,{\left (8 \, x^{2} + 4 \, x + 1\right )} e^{\left (-4 \, x\right )} + \frac{3}{16} \,{\left (4 \, x + 1\right )} e^{\left (-4 \, x\right )} - \frac{1}{2} \, e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x+2)/exp(4*x),x, algorithm="maxima")

[Out]

-1/32*(8*x^2 + 4*x + 1)*e^(-4*x) + 3/16*(4*x + 1)*e^(-4*x) - 1/2*e^(-4*x)

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Fricas [A]  time = 0.701215, size = 49, normalized size = 1.53 \begin{align*} -\frac{1}{32} \,{\left (8 \, x^{2} - 20 \, x + 11\right )} e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x+2)/exp(4*x),x, algorithm="fricas")

[Out]

-1/32*(8*x^2 - 20*x + 11)*e^(-4*x)

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Sympy [A]  time = 0.09161, size = 15, normalized size = 0.47 \begin{align*} \frac{\left (- 8 x^{2} + 20 x - 11\right ) e^{- 4 x}}{32} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-3*x+2)/exp(4*x),x)

[Out]

(-8*x**2 + 20*x - 11)*exp(-4*x)/32

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Giac [A]  time = 1.2642, size = 22, normalized size = 0.69 \begin{align*} -\frac{1}{32} \,{\left (8 \, x^{2} - 20 \, x + 11\right )} e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x+2)/exp(4*x),x, algorithm="giac")

[Out]

-1/32*(8*x^2 - 20*x + 11)*e^(-4*x)