\(\int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx\) [9780]

Optimal result

Integrand size = 81, antiderivative size = 24 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {e^{40+e^4-x^2 \log ^2(3 x)}}{\log (\log (x))} \]

[Out]

exp(-x^2*ln(3*x)^2+exp(4)+40)/ln(ln(x))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).

Time = 0.68 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6820, 2326} \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {x e^{-x^2 \log ^2(3 x)+e^4+40} \log (3 x) (\log (3 x)+1)}{\left (x \log ^2(3 x)+x \log (3 x)\right ) \log (\log (x))} \]

[In]

Int[(-E^(40 + E^4 - x^2*Log[3*x]^2) + E^(40 + E^4 - x^2*Log[3*x]^2)*(-2*x^2*Log[x]*Log[3*x] - 2*x^2*Log[x]*Log
[3*x]^2)*Log[Log[x]])/(x*Log[x]*Log[Log[x]]^2),x]

[Out]

(E^(40 + E^4 - x^2*Log[3*x]^2)*x*Log[3*x]*(1 + Log[3*x]))/((x*Log[3*x] + x*Log[3*x]^2)*Log[Log[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]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{40 \left (1+\frac {e^4}{40}\right )-x^2 \log ^2(3 x)} \left (-1-2 x^2 \log (x) \log (3 x) (1+\log (3 x)) \log (\log (x))\right )}{x \log (x) \log ^2(\log (x))} \, dx \\ & = \frac {e^{40+e^4-x^2 \log ^2(3 x)} x \log (3 x) (1+\log (3 x))}{\left (x \log (3 x)+x \log ^2(3 x)\right ) \log (\log (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {e^{40+e^4-x^2 \log ^2(3 x)}}{\log (\log (x))} \]

[In]

Integrate[(-E^(40 + E^4 - x^2*Log[3*x]^2) + E^(40 + E^4 - x^2*Log[3*x]^2)*(-2*x^2*Log[x]*Log[3*x] - 2*x^2*Log[
x]*Log[3*x]^2)*Log[Log[x]])/(x*Log[x]*Log[Log[x]]^2),x]

[Out]

E^(40 + E^4 - x^2*Log[3*x]^2)/Log[Log[x]]

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{-x^{2} \ln \left (3 x \right )^{2}+{\mathrm e}^{4}+40}}{\ln \left (\ln \left (x \right )\right )}\)	\(23\)
risch	\(\frac {x^{-2 x^{2} \ln \left (3\right )} {\mathrm e}^{-x^{2} \ln \left (x \right )^{2}+40-x^{2} \ln \left (3\right )^{2}+{\mathrm e}^{4}}}{\ln \left (\ln \left (x \right )\right )}\)	\(39\)

[In]

int(((-2*x^2*ln(x)*ln(3*x)^2-2*x^2*ln(x)*ln(3*x))*exp(-x^2*ln(3*x)^2+exp(4)+40)*ln(ln(x))-exp(-x^2*ln(3*x)^2+e
xp(4)+40))/x/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

exp(-x^2*ln(3*x)^2+exp(4)+40)/ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {e^{\left (-x^{2} \log \left (3\right )^{2} - 2 \, x^{2} \log \left (3\right ) \log \left (x\right ) - x^{2} \log \left (x\right )^{2} + e^{4} + 40\right )}}{\log \left (\log \left (x\right )\right )} \]

[In]

integrate(((-2*x^2*log(x)*log(3*x)^2-2*x^2*log(x)*log(3*x))*exp(-x^2*log(3*x)^2+exp(4)+40)*log(log(x))-exp(-x^
2*log(3*x)^2+exp(4)+40))/x/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

e^(-x^2*log(3)^2 - 2*x^2*log(3)*log(x) - x^2*log(x)^2 + e^4 + 40)/log(log(x))

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {e^{- x^{2} \left (\log {\left (x \right )} + \log {\left (3 \right )}\right )^{2} + 40 + e^{4}}}{\log {\left (\log {\left (x \right )} \right )}} \]

[In]

integrate(((-2*x**2*ln(x)*ln(3*x)**2-2*x**2*ln(x)*ln(3*x))*exp(-x**2*ln(3*x)**2+exp(4)+40)*ln(ln(x))-exp(-x**2
*ln(3*x)**2+exp(4)+40))/x/ln(x)/ln(ln(x))**2,x)

[Out]

exp(-x**2*(log(x) + log(3))**2 + 40 + exp(4))/log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {e^{\left (-x^{2} \log \left (3\right )^{2} - 2 \, x^{2} \log \left (3\right ) \log \left (x\right ) - x^{2} \log \left (x\right )^{2} + e^{4} + 40\right )}}{\log \left (\log \left (x\right )\right )} \]

[In]

integrate(((-2*x^2*log(x)*log(3*x)^2-2*x^2*log(x)*log(3*x))*exp(-x^2*log(3*x)^2+exp(4)+40)*log(log(x))-exp(-x^
2*log(3*x)^2+exp(4)+40))/x/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

e^(-x^2*log(3)^2 - 2*x^2*log(3)*log(x) - x^2*log(x)^2 + e^4 + 40)/log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {e^{\left (-x^{2} \log \left (3\right )^{2} - 2 \, x^{2} \log \left (3\right ) \log \left (x\right ) - x^{2} \log \left (x\right )^{2} + e^{4} + 40\right )}}{\log \left (\log \left (x\right )\right )} \]

[In]

integrate(((-2*x^2*log(x)*log(3*x)^2-2*x^2*log(x)*log(3*x))*exp(-x^2*log(3*x)^2+exp(4)+40)*log(log(x))-exp(-x^
2*log(3*x)^2+exp(4)+40))/x/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

e^(-x^2*log(3)^2 - 2*x^2*log(3)*log(x) - x^2*log(x)^2 + e^4 + 40)/log(log(x))

Mupad [B] (verification not implemented)

Time = 15.36 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-e^{40+e^4-x^2 \log ^2(3 x)}+e^{40+e^4-x^2 \log ^2(3 x)} \left (-2 x^2 \log (x) \log (3 x)-2 x^2 \log (x) \log ^2(3 x)\right ) \log (\log (x))}{x \log (x) \log ^2(\log (x))} \, dx=\frac {{\mathrm {e}}^{40}\,{\mathrm {e}}^{-x^2\,{\ln \left (3\right )}^2}\,{\mathrm {e}}^{{\mathrm {e}}^4}\,{\mathrm {e}}^{-x^2\,{\ln \left (x\right )}^2}}{x^{2\,x^2\,\ln \left (3\right )}\,\ln \left (\ln \left (x\right )\right )} \]

[In]

int(-(exp(exp(4) - x^2*log(3*x)^2 + 40) + log(log(x))*exp(exp(4) - x^2*log(3*x)^2 + 40)*(2*x^2*log(3*x)*log(x)
+ 2*x^2*log(3*x)^2*log(x)))/(x*log(log(x))^2*log(x)),x)

[Out]

(exp(40)*exp(-x^2*log(3)^2)*exp(exp(4))*exp(-x^2*log(x)^2))/(x^(2*x^2*log(3))*log(log(x)))