\(\int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx\) [5945]

Optimal result

Integrand size = 20, antiderivative size = 11 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{-\frac {8}{x^3 \log (3)}} \]

[Out]

exp(-8/x^3/ln(3))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2240} \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{-\frac {8}{x^3 \log (3)}} \]

[In]

Int[24/(E^(8/(x^3*Log[3]))*x^4*Log[3]),x]

[Out]

E^(-8/(x^3*Log[3]))

Rule 12

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

Rule 2240

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]

Rubi steps \begin{align*} \text {integral}& = \frac {24 \int \frac {e^{-\frac {8}{x^3 \log (3)}}}{x^4} \, dx}{\log (3)} \\ & = e^{-\frac {8}{x^3 \log (3)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{-\frac {8}{x^3 \log (3)}} \]

[In]

Integrate[24/(E^(8/(x^3*Log[3]))*x^4*Log[3]),x]

[Out]

E^(-8/(x^3*Log[3]))

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00

method	result	size
gosper	\({\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(11\)
derivativedivides	\({\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(11\)
default	\({\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(11\)
norman	\({\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(11\)
risch	\({\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(11\)
parallelrisch	\({\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(11\)
meijerg	\(-1+{\mathrm e}^{-\frac {8}{x^{3} \ln \left (3\right )}}\)	\(13\)

[In]

int(24*exp(-8/x^3/ln(3))/x^4/ln(3),x,method=_RETURNVERBOSE)

[Out]

exp(-8/x^3/ln(3))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{\left (-\frac {8}{x^{3} \log \left (3\right )}\right )} \]

[In]

integrate(24*exp(-8/x^3/log(3))/x^4/log(3),x, algorithm="fricas")

[Out]

e^(-8/(x^3*log(3)))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{- \frac {8}{x^{3} \log {\left (3 \right )}}} \]

[In]

integrate(24*exp(-8/x**3/ln(3))/x**4/ln(3),x)

[Out]

exp(-8/(x**3*log(3)))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{\left (-\frac {8}{x^{3} \log \left (3\right )}\right )} \]

[In]

integrate(24*exp(-8/x^3/log(3))/x^4/log(3),x, algorithm="maxima")

[Out]

e^(-8/(x^3*log(3)))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx=e^{\left (-\frac {8}{x^{3} \log \left (3\right )}\right )} \]

[In]

integrate(24*exp(-8/x^3/log(3))/x^4/log(3),x, algorithm="giac")

[Out]

e^(-8/(x^3*log(3)))

Mupad [B] (verification not implemented)

Time = 11.21 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {24 e^{-\frac {8}{x^3 \log (3)}}}{x^4 \log (3)} \, dx={\mathrm {e}}^{-\frac {8}{x^3\,\ln \left (3\right )}} \]

[In]

int((24*exp(-8/(x^3*log(3))))/(x^4*log(3)),x)

[Out]

exp(-8/(x^3*log(3)))