\(\int (e^x x)^{4 x^2} (\frac {2 \log (3)}{9})^{4 x^2} (4 x+4 x^2+8 x \log (\frac {2}{9} e^x x \log (3))) \, dx\) [8664]

Optimal result

Integrand size = 47, antiderivative size = 24 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=\left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \]

[Out]

exp(4*x^2*ln(2/9*x*ln(3)*exp(x)))

Rubi [F(-1)]

Timed out. \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=\text {\$Aborted} \]

[In]

Int[(E^x*x)^(4*x^2)*((2*Log[3])/9)^(4*x^2)*(4*x + 4*x^2 + 8*x*Log[(2*E^x*x*Log[3])/9]),x]

[Out]

$Aborted

Rubi steps \begin{align*} \text {integral}& = \int 2^{2+4 x^2} x \left (e^x x\right )^{4 x^2} \left (\frac {\log (3)}{9}\right )^{4 x^2} \left (1+x+2 \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(24)=48\).

Time = 0.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=-\frac {4 e^{x^2 \left (-\log \left (\frac {6561}{16}\right )+4 \log (\log (3))\right )} \left (e^x x\right )^{4 x^2} \left (-\log \left (\frac {9}{2}\right )+\log (\log (3))\right )}{\log \left (\frac {6561}{16}\right )-4 \log (\log (3))} \]

[In]

Integrate[(E^x*x)^(4*x^2)*((2*Log[3])/9)^(4*x^2)*(4*x + 4*x^2 + 8*x*Log[(2*E^x*x*Log[3])/9]),x]

[Out]

(-4*E^(x^2*(-Log[6561/16] + 4*Log[Log[3]]))*(E^x*x)^(4*x^2)*(-Log[9/2] + Log[Log[3]]))/(Log[6561/16] - 4*Log[L
og[3]])

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\({\mathrm e}^{4 x^{2} \ln \left (\frac {2 x \ln \left (3\right ) {\mathrm e}^{x}}{9}\right )}\)	\(15\)
risch	\(x^{4 x^{2}} \ln \left (3\right )^{4 x^{2}} 16^{x^{2}} \left (\frac {1}{6561}\right )^{x^{2}} \left ({\mathrm e}^{x}\right )^{4 x^{2}} {\mathrm e}^{-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} x \right ) x^{2} \left (-\operatorname {csgn}\left (i {\mathrm e}^{x} x \right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{x} x \right )+\operatorname {csgn}\left (i x \right )\right )}\)	\(81\)

[In]

int((8*x*ln(2/9*x*ln(3)*exp(x))+4*x^2+4*x)*exp(4*x^2*ln(2/9*x*ln(3)*exp(x))),x,method=_RETURNVERBOSE)

[Out]

exp(4*x^2*ln(2/9*x*ln(3)*exp(x)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=\left (\frac {2}{9} \, x e^{x} \log \left (3\right )\right )^{4 \, x^{2}} \]

[In]

integrate((8*x*log(2/9*x*log(3)*exp(x))+4*x^2+4*x)*exp(4*x^2*log(2/9*x*log(3)*exp(x))),x, algorithm="fricas")

[Out]

(2/9*x*e^x*log(3))^(4*x^2)

Sympy [A] (verification not implemented)

Time = 81.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=e^{4 x^{2} \log {\left (\frac {2 x e^{x} \log {\left (3 \right )}}{9} \right )}} \]

[In]

integrate((8*x*ln(2/9*x*ln(3)*exp(x))+4*x**2+4*x)*exp(4*x**2*ln(2/9*x*ln(3)*exp(x))),x)

[Out]

exp(4*x**2*log(2*x*exp(x)*log(3)/9))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (13) = 26\).

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} \log \left (3\right ) + 4 \, x^{2} \log \left (2\right ) + 4 \, x^{2} \log \left (x\right ) + 4 \, x^{2} \log \left (\log \left (3\right )\right )\right )} \]

[In]

integrate((8*x*log(2/9*x*log(3)*exp(x))+4*x^2+4*x)*exp(4*x^2*log(2/9*x*log(3)*exp(x))),x, algorithm="maxima")

[Out]

e^(4*x^3 - 8*x^2*log(3) + 4*x^2*log(2) + 4*x^2*log(x) + 4*x^2*log(log(3)))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} \log \left (3\right ) + 4 \, x^{2} \log \left (2 \, x \log \left (3\right )\right )\right )} \]

[In]

integrate((8*x*log(2/9*x*log(3)*exp(x))+4*x^2+4*x)*exp(4*x^2*log(2/9*x*log(3)*exp(x))),x, algorithm="giac")

[Out]

e^(4*x^3 - 8*x^2*log(3) + 4*x^2*log(2*x*log(3)))

Mupad [B] (verification not implemented)

Time = 14.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=x^{4\,x^2}\,{\mathrm {e}}^{4\,x^3}\,{\left (\frac {16\,{\ln \left (3\right )}^4}{6561}\right )}^{x^2} \]

[In]

int(exp(4*x^2*log((2*x*exp(x)*log(3))/9))*(4*x + 8*x*log((2*x*exp(x)*log(3))/9) + 4*x^2),x)

[Out]

x^(4*x^2)*exp(4*x^3)*((16*log(3)^4)/6561)^(x^2)