\(\int \frac {e^{-18+2 \log ^{2 x}(2)} (-1+4 x \log ^{2 x}(2) \log (\log (2)))}{x^2} \, dx\) [4821]

Optimal result

Integrand size = 30, antiderivative size = 16 \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {e^{-18+2 \log ^{2 x}(2)}}{x} \]

[Out]

exp(-18)/x*exp(exp(2*x*ln(ln(2))))^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2326} \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {e^{2 \log ^{2 x}(2)-18}}{x} \]

[In]

Int[(E^(-18 + 2*Log[2]^(2*x))*(-1 + 4*x*Log[2]^(2*x)*Log[Log[2]]))/x^2,x]

[Out]

E^(-18 + 2*Log[2]^(2*x))/x

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{-18+2 \log ^{2 x}(2)}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {e^{2 \left (-9+\log ^{2 x}(2)\right )}}{x} \]

[In]

Integrate[(E^(-18 + 2*Log[2]^(2*x))*(-1 + 4*x*Log[2]^(2*x)*Log[Log[2]]))/x^2,x]

[Out]

E^(2*(-9 + Log[2]^(2*x)))/x

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{-18+2 \ln \left (2\right )^{2 x}}}{x}\)	\(16\)
norman	\(\frac {{\mathrm e}^{-18} {\mathrm e}^{2 \,{\mathrm e}^{2 x \ln \left (\ln \left (2\right )\right )}}}{x}\)	\(19\)
parallelrisch	\(\frac {{\mathrm e}^{-18} {\mathrm e}^{2 \,{\mathrm e}^{2 x \ln \left (\ln \left (2\right )\right )}}}{x}\)	\(19\)

[In]

int((4*x*ln(ln(2))*exp(2*x*ln(ln(2)))-1)*exp(exp(2*x*ln(ln(2))))^2/x^2/exp(18),x,method=_RETURNVERBOSE)

[Out]

1/x*exp(-18+2*ln(2)^(2*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {e^{\left (2 \, \log \left (2\right )^{2 \, x} - 18\right )}}{x} \]

[In]

integrate((4*x*log(log(2))*exp(2*x*log(log(2)))-1)*exp(exp(2*x*log(log(2))))^2/x^2/exp(18),x, algorithm="frica
s")

[Out]

e^(2*log(2)^(2*x) - 18)/x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {e^{2 e^{2 x \log {\left (\log {\left (2 \right )} \right )}}}}{x e^{18}} \]

[In]

integrate((4*x*ln(ln(2))*exp(2*x*ln(ln(2)))-1)*exp(exp(2*x*ln(ln(2))))**2/x**2/exp(18),x)

[Out]

exp(-18)*exp(2*exp(2*x*log(log(2))))/x

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {e^{\left (2 \, \log \left (2\right )^{2 \, x} - 18\right )}}{x} \]

[In]

integrate((4*x*log(log(2))*exp(2*x*log(log(2)))-1)*exp(exp(2*x*log(log(2))))^2/x^2/exp(18),x, algorithm="maxim
a")

[Out]

e^(2*log(2)^(2*x) - 18)/x

Giac [F]

\[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\int { \frac {{\left (4 \, x \log \left (2\right )^{2 \, x} \log \left (\log \left (2\right )\right ) - 1\right )} e^{\left (2 \, \log \left (2\right )^{2 \, x} - 18\right )}}{x^{2}} \,d x } \]

[In]

integrate((4*x*log(log(2))*exp(2*x*log(log(2)))-1)*exp(exp(2*x*log(log(2))))^2/x^2/exp(18),x, algorithm="giac"
)

[Out]

integrate((4*x*log(2)^(2*x)*log(log(2)) - 1)*e^(2*log(2)^(2*x) - 18)/x^2, x)

Mupad [B] (verification not implemented)

Time = 10.79 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-18+2 \log ^{2 x}(2)} \left (-1+4 x \log ^{2 x}(2) \log (\log (2))\right )}{x^2} \, dx=\frac {{\mathrm {e}}^{-18}\,{\mathrm {e}}^{2\,{\ln \left (2\right )}^{2\,x}}}{x} \]

[In]

int((exp(-18)*exp(2*exp(2*x*log(log(2))))*(4*x*exp(2*x*log(log(2)))*log(log(2)) - 1))/x^2,x)

[Out]

(exp(-18)*exp(2*log(2)^(2*x)))/x