\(\int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx\) [9177]

Optimal result

Integrand size = 27, antiderivative size = 13 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=e^{-2+\frac {16}{x^2 \log ^2(x)}} \]

[Out]

exp(8/x^2/ln(x)^2)^2/exp(1)^2

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6838} \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=e^{\frac {16}{x^2 \log ^2(x)}-2} \]

[In]

Int[(E^(-2 + 16/(x^2*Log[x]^2))*(-32 - 32*Log[x]))/(x^3*Log[x]^3),x]

[Out]

E^(-2 + 16/(x^2*Log[x]^2))

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}& = e^{-2+\frac {16}{x^2 \log ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=e^{-2+\frac {16}{x^2 \log ^2(x)}} \]

[In]

Integrate[(E^(-2 + 16/(x^2*Log[x]^2))*(-32 - 32*Log[x]))/(x^3*Log[x]^3),x]

[Out]

E^(-2 + 16/(x^2*Log[x]^2))

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.62

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (x^{2} \ln \left (x \right )^{2}-8\right )}{\ln \left (x \right )^{2} x^{2}}}\)	\(21\)

[In]

int((-32*ln(x)-32)*exp(8/x^2/ln(x)^2)^2/x^3/exp(1)^2/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

exp(-2*(x^2*ln(x)^2-8)/ln(x)^2/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=e^{\left (-\frac {2 \, {\left (x^{2} \log \left (x\right )^{2} - 8\right )}}{x^{2} \log \left (x\right )^{2}}\right )} \]

[In]

integrate((-32*log(x)-32)*exp(8/x^2/log(x)^2)^2/x^3/exp(1)^2/log(x)^3,x, algorithm="fricas")

[Out]

e^(-2*(x^2*log(x)^2 - 8)/(x^2*log(x)^2))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=\frac {e^{\frac {16}{x^{2} \log {\left (x \right )}^{2}}}}{e^{2}} \]

[In]

integrate((-32*ln(x)-32)*exp(8/x**2/ln(x)**2)**2/x**3/exp(1)**2/ln(x)**3,x)

[Out]

exp(-2)*exp(16/(x**2*log(x)**2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=e^{\left (\frac {16}{x^{2} \log \left (x\right )^{2}} - 2\right )} \]

[In]

integrate((-32*log(x)-32)*exp(8/x^2/log(x)^2)^2/x^3/exp(1)^2/log(x)^3,x, algorithm="maxima")

[Out]

e^(16/(x^2*log(x)^2) - 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx=e^{\left (\frac {16}{x^{2} \log \left (x\right )^{2}} - 2\right )} \]

[In]

integrate((-32*log(x)-32)*exp(8/x^2/log(x)^2)^2/x^3/exp(1)^2/log(x)^3,x, algorithm="giac")

[Out]

e^(16/(x^2*log(x)^2) - 2)

Mupad [B] (verification not implemented)

Time = 12.65 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx={\mathrm {e}}^{-2}\,{\mathrm {e}}^{\frac {16}{x^2\,{\ln \left (x\right )}^2}} \]

[In]

int(-(exp(-2)*exp(16/(x^2*log(x)^2))*(32*log(x) + 32))/(x^3*log(x)^3),x)

[Out]

exp(-2)*exp(16/(x^2*log(x)^2))