Integrand size = 79, antiderivative size = 21 \[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx=e^{x^{-8-2 x+2 \log \left (3 e^x \log ^2(x)\right )}} \]
Time = 0.99 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx=e^{x^{-8-2 x+2 \log \left (3 e^x \log ^2(x)\right )}} \]
Integrate[E^(E^(2*(-5 - x)*Log[x] + 2*Log[x]*Log[3*E^x*Log[x]^2])*x^2 + 2* (-5 - x)*Log[x] + 2*Log[x]*Log[3*E^x*Log[x]^2])*(-4*x - 2*x^2 + 2*x*Log[3* E^x*Log[x]^2]),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (-2 x^2-4 x+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \exp \left (x^2 \exp \left (2 \log \left (3 e^x \log ^2(x)\right ) \log (x)+2 (-x-5) \log (x)\right )+2 \log (x) \log \left (3 e^x \log ^2(x)\right )+2 (-x-5) \log (x)\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int 2 x \left (-x+\log \left (3 e^x \log ^2(x)\right )-2\right ) \exp \left (x^2 \exp \left (2 \log \left (3 e^x \log ^2(x)\right ) \log (x)+2 (-x-5) \log (x)\right )+2 \log (x) \log \left (3 e^x \log ^2(x)\right )+2 (-x-5) \log (x)\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{2-2 (x+5)}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1} \left (x-\log \left (3 e^x \log ^2(x)\right )+2\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{2-2 (x+5)}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1} \left (x-\log \left (3 e^x \log ^2(x)\right )+2\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -2 \int \exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1} \left (x-\log \left (3 e^x \log ^2(x)\right )+2\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (2 \exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1}-\exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) \log \left (3 e^x \log ^2(x)\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1}+\exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\int \exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) x^{-2 \left (x-\log \left (3 e^x \log ^2(x)\right )+4\right )}dx+2 \int \exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1}dx-\int \exp \left (e^{2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^{-2 x-8}\right ) x^{-2 (x+5)+2 \log \left (3 e^x \log ^2(x)\right )+1} \log \left (3 e^x \log ^2(x)\right )dx\right )\) |
Int[E^(E^(2*(-5 - x)*Log[x] + 2*Log[x]*Log[3*E^x*Log[x]^2])*x^2 + 2*(-5 - x)*Log[x] + 2*Log[x]*Log[3*E^x*Log[x]^2])*(-4*x - 2*x^2 + 2*x*Log[3*E^x*Lo g[x]^2]),x]
3.17.87.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 2.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \({\mathrm e}^{x^{2} {\mathrm e}^{\left (2 \ln \left (3 \,{\mathrm e}^{x} \ln \left (x \right )^{2}\right )-2 x -10\right ) \ln \left (x \right )}}\) | \(26\) |
risch | \({\mathrm e}^{x^{2} x^{4 \ln \left (\ln \left (x \right )\right )-i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right )-i \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (x \right )^{2}\right ) \pi -i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+2 \ln \left (3 \,{\mathrm e}^{x}\right )+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right )} x^{-2 x -10}}\) | \(104\) |
int((2*x*ln(3*exp(x)*ln(x)^2)-2*x^2-4*x)*exp(ln(x)*ln(3*exp(x)*ln(x)^2)+(- x-5)*ln(x))^2*exp(x^2*exp(ln(x)*ln(3*exp(x)*ln(x)^2)+(-x-5)*ln(x))^2),x,me thod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx=e^{\left (x^{2} e^{\left (-2 \, {\left (x + 5\right )} \log \left (x\right ) + 2 \, \log \left (3 \, e^{x} \log \left (x\right )^{2}\right ) \log \left (x\right )\right )}\right )} \]
integrate((2*x*log(3*exp(x)*log(x)^2)-2*x^2-4*x)*exp(log(x)*log(3*exp(x)*l og(x)^2)+(-x-5)*log(x))^2*exp(x^2*exp(log(x)*log(3*exp(x)*log(x)^2)+(-x-5) *log(x))^2),x, algorithm=\
Timed out. \[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx=\text {Timed out} \]
integrate((2*x*ln(3*exp(x)*ln(x)**2)-2*x**2-4*x)*exp(ln(x)*ln(3*exp(x)*ln( x)**2)+(-x-5)*ln(x))**2*exp(x**2*exp(ln(x)*ln(3*exp(x)*ln(x)**2)+(-x-5)*ln (x))**2),x)
Time = 0.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx=e^{\left (\frac {e^{\left (2 \, \log \left (3\right ) \log \left (x\right ) + 4 \, \log \left (x\right ) \log \left (\log \left (x\right )\right )\right )}}{x^{8}}\right )} \]
integrate((2*x*log(3*exp(x)*log(x)^2)-2*x^2-4*x)*exp(log(x)*log(3*exp(x)*l og(x)^2)+(-x-5)*log(x))^2*exp(x^2*exp(log(x)*log(3*exp(x)*log(x)^2)+(-x-5) *log(x))^2),x, algorithm=\
\[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx=\int { -2 \, {\left (x^{2} - x \log \left (3 \, e^{x} \log \left (x\right )^{2}\right ) + 2 \, x\right )} e^{\left (x^{2} e^{\left (-2 \, {\left (x + 5\right )} \log \left (x\right ) + 2 \, \log \left (3 \, e^{x} \log \left (x\right )^{2}\right ) \log \left (x\right )\right )} - 2 \, {\left (x + 5\right )} \log \left (x\right ) + 2 \, \log \left (3 \, e^{x} \log \left (x\right )^{2}\right ) \log \left (x\right )\right )} \,d x } \]
integrate((2*x*log(3*exp(x)*log(x)^2)-2*x^2-4*x)*exp(log(x)*log(3*exp(x)*l og(x)^2)+(-x-5)*log(x))^2*exp(x^2*exp(log(x)*log(3*exp(x)*log(x)^2)+(-x-5) *log(x))^2),x, algorithm=\
Time = 8.61 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int e^{e^{2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} x^2+2 (-5-x) \log (x)+2 \log (x) \log \left (3 e^x \log ^2(x)\right )} \left (-4 x-2 x^2+2 x \log \left (3 e^x \log ^2(x)\right )\right ) \, dx={\mathrm {e}}^{\frac {x^{2\,\ln \left (3\right )}\,x^{2\,\ln \left ({\ln \left (x\right )}^2\right )}}{x^8}} \]