3.68.1 \(\int -\frac {51 e^{-\frac {3}{-4-17 x+\log (4)}}}{16+136 x+289 x^2+(-8-34 x) \log (4)+\log ^2(4)} \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {1}{1+6 x+\frac {1}{3} (1-x-\log (4))}} \]

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Rubi [A]  time = 0.10, antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 6688, 2209} \begin {gather*} e^{\frac {3}{17 x+4-\log (4)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-51/(E^(3/(-4 - 17*x + Log[4]))*(16 + 136*x + 289*x^2 + (-8 - 34*x)*Log[4] + Log[4]^2)),x]

[Out]

E^(3/(4 + 17*x - Log[4]))

Rule 12

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (51 \int \frac {e^{-\frac {3}{-4-17 x+\log (4)}}}{16+136 x+289 x^2+(-8-34 x) \log (4)+\log ^2(4)} \, dx\right )\\ &=-\left (51 \int \frac {e^{\frac {3}{4+17 x-\log (4)}}}{(4+17 x-\log (4))^2} \, dx\right )\\ &=e^{\frac {3}{4+17 x-\log (4)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 0.68 \begin {gather*} e^{\frac {3}{4+17 x-\log (4)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-51/(E^(3/(-4 - 17*x + Log[4]))*(16 + 136*x + 289*x^2 + (-8 - 34*x)*Log[4] + Log[4]^2)),x]

[Out]

E^(3/(4 + 17*x - Log[4]))

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fricas [A]  time = 0.50, size = 14, normalized size = 0.64 \begin {gather*} e^{\left (\frac {3}{17 \, x - 2 \, \log \relax (2) + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-51*exp(-3/(2*log(2)-17*x-4))/(4*log(2)^2+2*(-34*x-8)*log(2)+289*x^2+136*x+16),x, algorithm="fricas"
)

[Out]

e^(3/(17*x - 2*log(2) + 4))

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giac [A]  time = 0.22, size = 14, normalized size = 0.64 \begin {gather*} e^{\left (\frac {3}{17 \, x - 2 \, \log \relax (2) + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-51*exp(-3/(2*log(2)-17*x-4))/(4*log(2)^2+2*(-34*x-8)*log(2)+289*x^2+136*x+16),x, algorithm="giac")

[Out]

e^(3/(17*x - 2*log(2) + 4))

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maple [A]  time = 0.11, size = 15, normalized size = 0.68




method result size



gosper \({\mathrm e}^{-\frac {3}{2 \ln \relax (2)-17 x -4}}\) \(15\)
derivativedivides \({\mathrm e}^{\frac {3}{-2 \ln \relax (2)+17 x +4}}\) \(15\)
default \({\mathrm e}^{\frac {3}{-2 \ln \relax (2)+17 x +4}}\) \(15\)
risch \({\mathrm e}^{-\frac {3}{2 \ln \relax (2)-17 x -4}}\) \(15\)
norman \(\frac {\left (2 \ln \relax (2)-4\right ) {\mathrm e}^{-\frac {3}{2 \ln \relax (2)-17 x -4}}-17 x \,{\mathrm e}^{-\frac {3}{2 \ln \relax (2)-17 x -4}}}{2 \ln \relax (2)-17 x -4}\) \(52\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-51*exp(-3/(2*ln(2)-17*x-4))/(4*ln(2)^2+2*(-34*x-8)*ln(2)+289*x^2+136*x+16),x,method=_RETURNVERBOSE)

[Out]

exp(-3/(2*ln(2)-17*x-4))

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maxima [A]  time = 0.46, size = 14, normalized size = 0.64 \begin {gather*} e^{\left (\frac {3}{17 \, x - 2 \, \log \relax (2) + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-51*exp(-3/(2*log(2)-17*x-4))/(4*log(2)^2+2*(-34*x-8)*log(2)+289*x^2+136*x+16),x, algorithm="maxima"
)

[Out]

e^(3/(17*x - 2*log(2) + 4))

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mupad [B]  time = 4.39, size = 14, normalized size = 0.64 \begin {gather*} {\mathrm {e}}^{\frac {3}{17\,x-\ln \relax (4)+4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(51*exp(3/(17*x - 2*log(2) + 4)))/(136*x - 2*log(2)*(34*x + 8) + 4*log(2)^2 + 289*x^2 + 16),x)

[Out]

exp(3/(17*x - log(4) + 4))

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sympy [A]  time = 0.27, size = 14, normalized size = 0.64 \begin {gather*} e^{- \frac {3}{- 17 x - 4 + 2 \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-51*exp(-3/(2*ln(2)-17*x-4))/(4*ln(2)**2+2*(-34*x-8)*ln(2)+289*x**2+136*x+16),x)

[Out]

exp(-3/(-17*x - 4 + 2*log(2)))

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