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/exp(4*x)+5/8*x/exp(4*x)-1/4*x^2/exp(4*x)

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Rubi [A]  time = 0.04, antiderivative size = 32, normalized size of antiderivative = 1.00, 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 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 {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*}

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Mathematica [A]  time = 0.02, 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]

-1/32*(11 - 20*x + 8*x^2)/E^(4*x)

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fricas [A]  time = 0.39, size = 16, normalized size = 0.50 \[ -\frac {1}{32} \, {\left (8 \, x^{2} - 20 \, x + 11\right )} e^{\left (-4 \, x\right )} \]

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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giac [A]  time = 0.21, size = 16, normalized size = 0.50 \[ -\frac {1}{32} \, {\left (8 \, x^{2} - 20 \, x + 11\right )} e^{\left (-4 \, x\right )} \]

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)

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maple [A]  time = 0.02, size = 19, normalized size = 0.59 \[ -\frac {\left (8 x^{2}-20 x +11\right ) {\mathrm e}^{-4 x}}{32} \]

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 = 0.53, size = 34, normalized size = 1.06 \[ -\frac {1}{32} \, {\left (8 \, x^{2} + 4 \, x + 1\right )} e^{\left (-4 \, x\right )} + \frac {3}{16} \, {\left (4 \, x + 1\right )} e^{\left (-4 \, x\right )} - \frac {1}{2} \, e^{\left (-4 \, x\right )} \]

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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mupad [B]  time = 0.05, size = 16, normalized size = 0.50 \[ -\frac {{\mathrm {e}}^{-4\,x}\,\left (8\,x^2-20\,x+11\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-4*x)*(x^2 - 3*x + 2),x)

[Out]

-(exp(-4*x)*(8*x^2 - 20*x + 11))/32

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sympy [A]  time = 0.09, size = 15, normalized size = 0.47 \[ \frac {\left (- 8 x^{2} + 20 x - 11\right ) e^{- 4 x}}{32} \]

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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