\(\int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx\) [3020]

Optimal result

Integrand size = 14, antiderivative size = 15 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=3 \left (-2-16\ 3^{-1-\frac {2}{x}}\right ) \]

[Out]

-6-16/exp(2*ln(3)/x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.60, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2240} \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=-16 3^{-2/x} \]

[In]

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

[Out]

-16/3^(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 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}& = -\left ((32 \log (3)) \int \frac {3^{-2/x}}{x^2} \, dx\right ) \\ & = -16 3^{-2/x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=-\frac {32\ 9^{-1/x} \log (3)}{\log (9)} \]

[In]

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

[Out]

(-32*Log[3])/(9^x^(-1)*Log[9])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67

method	result	size
risch	\(-16 \,9^{-\frac {1}{x}}\)	\(10\)
gosper	\(-16 \,{\mathrm e}^{-\frac {2 \ln \left (3\right )}{x}}\)	\(13\)
derivativedivides	\(-16 \,{\mathrm e}^{-\frac {2 \ln \left (3\right )}{x}}\)	\(13\)
default	\(-16 \,{\mathrm e}^{-\frac {2 \ln \left (3\right )}{x}}\)	\(13\)
norman	\(-16 \,{\mathrm e}^{-\frac {2 \ln \left (3\right )}{x}}\)	\(13\)
meijerg	\(16-16 \,{\mathrm e}^{-\frac {2 \ln \left (3\right )}{x}}\)	\(13\)
parallelrisch	\(-16 \,{\mathrm e}^{-\frac {2 \ln \left (3\right )}{x}}\)	\(13\)

[In]

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

[Out]

-16/(9^(1/x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=-\frac {16}{3^{\frac {2}{x}}} \]

[In]

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

[Out]

-16/3^(2/x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=- 16 e^{- \frac {2 \log {\left (3 \right )}}{x}} \]

[In]

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

[Out]

-16*exp(-2*log(3)/x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=-\frac {16}{3^{\frac {2}{x}}} \]

[In]

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

[Out]

-16/3^(2/x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=-\frac {16}{3^{\frac {2}{x}}} \]

[In]

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

[Out]

-16/3^(2/x)

Mupad [B] (verification not implemented)

Time = 9.32 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int -\frac {32\ 3^{-2/x} \log (3)}{x^2} \, dx=-\frac {16}{3^{2/x}} \]

[In]

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

[Out]

-16/3^(2/x)