\(\int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx\) [2692]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 36, antiderivative size = 17 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\frac {16}{9 x^2 \left (-\frac {5}{2}+\log (3)\right )}} \]

[Out]

exp(16/9/x^2/(ln(3)-5/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 12, 6838} \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{-\frac {32}{9 \left (5 x^2-2 x^2 \log (3)\right )}} \]

[In]

Int[(-64*E^(32/(-45*x^2 + 18*x^2*Log[3])))/(-45*x^3 + 18*x^3*Log[3]),x]

[Out]

E^(-32/(9*(5*x^2 - 2*x^2*Log[3])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{x^3 (-45+18 \log (3))} \, dx \\ & = \frac {64 \int \frac {e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{x^3} \, dx}{9 (5-\log (9))} \\ & = e^{-\frac {32}{9 \left (5 x^2-2 x^2 \log (3)\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\frac {32}{9 x^2 (-5+\log (9))}} \]

[In]

Integrate[(-64*E^(32/(-45*x^2 + 18*x^2*Log[3])))/(-45*x^3 + 18*x^3*Log[3]),x]

[Out]

E^(32/(9*x^2*(-5 + Log[9])))

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method result size
gosper \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) \(15\)
derivativedivides \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) \(15\)
default \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) \(15\)
risch \({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}\) \(15\)
norman \({\mathrm e}^{\frac {32}{18 x^{2} \ln \left (3\right )-45 x^{2}}}\) \(19\)
parallelrisch \(\frac {18 \ln \left (3\right ) {\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}-45 \,{\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \left (3\right )-5\right )}}}{18 \ln \left (3\right )-45}\) \(46\)

[In]

int(-64*exp(32/(18*x^2*ln(3)-45*x^2))/(18*x^3*ln(3)-45*x^3),x,method=_RETURNVERBOSE)

[Out]

exp(32/9/x^2/(2*ln(3)-5))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\left (\frac {32}{9 \, {\left (2 \, x^{2} \log \left (3\right ) - 5 \, x^{2}\right )}}\right )} \]

[In]

integrate(-64*exp(32/(18*x^2*log(3)-45*x^2))/(18*x^3*log(3)-45*x^3),x, algorithm="fricas")

[Out]

e^(32/9/(2*x^2*log(3) - 5*x^2))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\frac {32}{- 45 x^{2} + 18 x^{2} \log {\left (3 \right )}}} \]

[In]

integrate(-64*exp(32/(18*x**2*ln(3)-45*x**2))/(18*x**3*ln(3)-45*x**3),x)

[Out]

exp(32/(-45*x**2 + 18*x**2*log(3)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\left (\frac {32}{9 \, x^{2} {\left (2 \, \log \left (3\right ) - 5\right )}}\right )} \]

[In]

integrate(-64*exp(32/(18*x^2*log(3)-45*x^2))/(18*x^3*log(3)-45*x^3),x, algorithm="maxima")

[Out]

e^(32/9/(x^2*(2*log(3) - 5)))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx=e^{\left (\frac {32}{9 \, {\left (2 \, x^{2} \log \left (3\right ) - 5 \, x^{2}\right )}}\right )} \]

[In]

integrate(-64*exp(32/(18*x^2*log(3)-45*x^2))/(18*x^3*log(3)-45*x^3),x, algorithm="giac")

[Out]

e^(32/9/(2*x^2*log(3) - 5*x^2))

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx={\mathrm {e}}^{\frac {32}{9\,x^2\,\left (2\,\ln \left (3\right )-5\right )}} \]

[In]

int(-(64*exp(32/(18*x^2*log(3) - 45*x^2)))/(18*x^3*log(3) - 45*x^3),x)

[Out]

exp(32/(9*x^2*(2*log(3) - 5)))