3.8.41 \(\int e^{-4 x} (2-3 x+x^2) \, dx\) [741]

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

[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 = 0.03, 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 = {2227, 2225, 2207} \begin {gather*} -\frac {1}{4} e^{-4 x} x^2+\frac {5}{8} e^{-4 x} x-\frac {11 e^{-4 x}}{32} \end {gather*}

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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2227

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] &&  !TrueQ[$UseGamma]

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 = 0.05, size = 19, normalized size = 0.59 \begin {gather*} -\frac {1}{32} e^{-4 x} \left (11-20 x+8 x^2\right ) \end {gather*}

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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Maple [A]
time = 0.02, size = 30, normalized size = 0.94

method	result	size
risch	\(\left (-\frac {1}{4} x^{2}+\frac {5}{8} x -\frac {11}{32}\right ) {\mathrm e}^{-4 x}\)	\(16\)
norman	\(\left (-\frac {1}{4} x^{2}+\frac {5}{8} x -\frac {11}{32}\right ) {\mathrm e}^{-4 x}\)	\(18\)
gosper	\(-\frac {\left (8 x^{2}-20 x +11\right ) {\mathrm e}^{-4 x}}{32}\)	\(19\)
derivativedivides	\(-\frac {11 \,{\mathrm e}^{-4 x}}{32}+\frac {5 x \,{\mathrm e}^{-4 x}}{8}-\frac {x^{2} {\mathrm e}^{-4 x}}{4}\)	\(30\)
default	\(-\frac {11 \,{\mathrm e}^{-4 x}}{32}+\frac {5 x \,{\mathrm e}^{-4 x}}{8}-\frac {x^{2} {\mathrm e}^{-4 x}}{4}\)	\(30\)
meijerg	\(\frac {11}{32}-\frac {\left (48 x^{2}+24 x +6\right ) {\mathrm e}^{-4 x}}{192}+\frac {3 \left (2+8 x \right ) {\mathrm e}^{-4 x}}{32}-\frac {{\mathrm e}^{-4 x}}{2}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/32/exp(4*x)+5/8*x/exp(4*x)-1/4*x^2/exp(4*x)

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Maxima [A]
time = 0.28, size = 34, normalized size = 1.06 \begin {gather*} -\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 {gather*}

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 = 0.36, size = 16, normalized size = 0.50 \begin {gather*} -\frac {1}{32} \, {\left (8 \, x^{2} - 20 \, x + 11\right )} e^{\left (-4 \, x\right )} \end {gather*}

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 = 0.02, size = 15, normalized size = 0.47 \begin {gather*} \frac {\left (- 8 x^{2} + 20 x - 11\right ) e^{- 4 x}}{32} \end {gather*}

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 = 4.95, size = 16, normalized size = 0.50 \begin {gather*} -\frac {1}{32} \, {\left (8 \, x^{2} - 20 \, x + 11\right )} e^{\left (-4 \, x\right )} \end {gather*}

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

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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