3.55.3 \(\int \frac {1}{16} e^{-2 x} (-145+16 e^{2 x}-34 x-2 x^2) \, dx\)

Optimal. Leaf size=20 \[ x+\frac {1}{64} e^{-2 x} \left (2+4 (9+x)^2\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 33, normalized size of antiderivative = 1.65, number of steps used = 9, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 6742, 2194, 2176} \begin {gather*} \frac {1}{16} e^{-2 x} x^2+\frac {9}{8} e^{-2 x} x+x+\frac {163 e^{-2 x}}{32} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-145 + 16*E^(2*x) - 34*x - 2*x^2)/(16*E^(2*x)),x]

[Out]

163/(32*E^(2*x)) + x + (9*x)/(8*E^(2*x)) + x^2/(16*E^(2*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int e^{-2 x} \left (-145+16 e^{2 x}-34 x-2 x^2\right ) \, dx\\ &=\frac {1}{16} \int \left (16-145 e^{-2 x}-34 e^{-2 x} x-2 e^{-2 x} x^2\right ) \, dx\\ &=x-\frac {1}{8} \int e^{-2 x} x^2 \, dx-\frac {17}{8} \int e^{-2 x} x \, dx-\frac {145}{16} \int e^{-2 x} \, dx\\ &=\frac {145 e^{-2 x}}{32}+x+\frac {17}{16} e^{-2 x} x+\frac {1}{16} e^{-2 x} x^2-\frac {1}{8} \int e^{-2 x} x \, dx-\frac {17}{16} \int e^{-2 x} \, dx\\ &=\frac {81 e^{-2 x}}{16}+x+\frac {9}{8} e^{-2 x} x+\frac {1}{16} e^{-2 x} x^2-\frac {1}{16} \int e^{-2 x} \, dx\\ &=\frac {163 e^{-2 x}}{32}+x+\frac {9}{8} e^{-2 x} x+\frac {1}{16} e^{-2 x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 1.20 \begin {gather*} \frac {1}{16} \left (16 x+e^{-2 x} \left (\frac {163}{2}+18 x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-145 + 16*E^(2*x) - 34*x - 2*x^2)/(16*E^(2*x)),x]

[Out]

(16*x + (163/2 + 18*x + x^2)/E^(2*x))/16

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fricas [A]  time = 0.48, size = 23, normalized size = 1.15 \begin {gather*} \frac {1}{32} \, {\left (2 \, x^{2} + 32 \, x e^{\left (2 \, x\right )} + 36 \, x + 163\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*exp(x)^2-2*x^2-34*x-145)/exp(x)^2,x, algorithm="fricas")

[Out]

1/32*(2*x^2 + 32*x*e^(2*x) + 36*x + 163)*e^(-2*x)

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giac [A]  time = 0.13, size = 24, normalized size = 1.20 \begin {gather*} \frac {1}{16} \, x^{2} e^{\left (-2 \, x\right )} + \frac {9}{8} \, x e^{\left (-2 \, x\right )} + x + \frac {163}{32} \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*exp(x)^2-2*x^2-34*x-145)/exp(x)^2,x, algorithm="giac")

[Out]

1/16*x^2*e^(-2*x) + 9/8*x*e^(-2*x) + x + 163/32*e^(-2*x)

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maple [A]  time = 0.03, size = 17, normalized size = 0.85

method	result	size
risch	\(x +\frac {\left (\frac {163}{2}+18 x +x^{2}\right ) {\mathrm e}^{-2 x}}{16}\)	\(17\)
norman	\(\left (\frac {163}{32}+x \,{\mathrm e}^{2 x}+\frac {9 x}{8}+\frac {x^{2}}{16}\right ) {\mathrm e}^{-2 x}\)	\(22\)
default	\(x +\frac {163 \,{\mathrm e}^{-2 x}}{32}+\frac {9 x \,{\mathrm e}^{-2 x}}{8}+\frac {x^{2} {\mathrm e}^{-2 x}}{16}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(16*exp(x)^2-2*x^2-34*x-145)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

x+1/16*(163/2+18*x+x^2)*exp(-2*x)

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maxima [B]  time = 0.59, size = 35, normalized size = 1.75 \begin {gather*} \frac {1}{32} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {17}{32} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + x + \frac {145}{32} \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*exp(x)^2-2*x^2-34*x-145)/exp(x)^2,x, algorithm="maxima")

[Out]

1/32*(2*x^2 + 2*x + 1)*e^(-2*x) + 17/32*(2*x + 1)*e^(-2*x) + x + 145/32*e^(-2*x)

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mupad [B]  time = 0.07, size = 24, normalized size = 1.20 \begin {gather*} x+\frac {163\,{\mathrm {e}}^{-2\,x}}{32}+\frac {9\,x\,{\mathrm {e}}^{-2\,x}}{8}+\frac {x^2\,{\mathrm {e}}^{-2\,x}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-2*x)*((17*x)/8 - exp(2*x) + x^2/8 + 145/16),x)

[Out]

x + (163*exp(-2*x))/32 + (9*x*exp(-2*x))/8 + (x^2*exp(-2*x))/16

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sympy [A]  time = 0.10, size = 17, normalized size = 0.85 \begin {gather*} x + \frac {\left (2 x^{2} + 36 x + 163\right ) e^{- 2 x}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*exp(x)**2-2*x**2-34*x-145)/exp(x)**2,x)

[Out]

x + (2*x**2 + 36*x + 163)*exp(-2*x)/32

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