Integrand size = 191, antiderivative size = 30 \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\frac {e^{x^8} x}{-1+e^2+\frac {e^{\frac {x}{5 \log (x)}}}{x}+x} \]
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\frac {e^{x^8} x^2}{e^{\frac {x}{5 \log (x)}}-x+e^2 x+x^2} \]
Integrate[(E^x^8*(-5*x^2 - 40*x^10 + 40*x^11 + E^2*(5*x^2 + 40*x^10))*Log[ x]^2 + E^(x/(5*Log[x]))*(E^x^8*x^2 - E^x^8*x^2*Log[x] + E^x^8*(10*x + 40*x ^9)*Log[x]^2))/(5*E^((2*x)/(5*Log[x]))*Log[x]^2 + E^(x/(5*Log[x]))*(-10*x + 10*E^2*x + 10*x^2)*Log[x]^2 + (5*x^2 + 5*E^4*x^2 - 10*x^3 + 5*x^4 + E^2* (-10*x^2 + 10*x^3))*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.
| \(\displaystyle \int \frac {e^{\frac {x}{5 \log (x)}} \left (e^{x^8} \left (40 x^9+10 x\right ) \log ^2(x)+e^{x^8} x^2-e^{x^8} x^2 \log (x)\right )+e^{x^8} \left (40 x^{11}-40 x^{10}-5 x^2+e^2 \left (40 x^{10}+5 x^2\right )\right ) \log ^2(x)}{\left (10 x^2+10 e^2 x-10 x\right ) e^{\frac {x}{5 \log (x)}} \log ^2(x)+\left (5 x^4-10 x^3+5 e^4 x^2+5 x^2+e^2 \left (10 x^3-10 x^2\right )\right ) \log ^2(x)+5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {x}{5 \log (x)}} \left (e^{x^8} \left (40 x^9+10 x\right ) \log ^2(x)+e^{x^8} x^2-e^{x^8} x^2 \log (x)\right )+e^{x^8} \left (40 x^{11}-40 x^{10}-5 x^2+e^2 \left (40 x^{10}+5 x^2\right )\right ) \log ^2(x)}{5 \left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int -\frac {5 e^{x^8} \left (-8 x^{11}+8 x^{10}+x^2-e^2 \left (8 x^{10}+x^2\right )\right ) \log ^2(x)-e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} \log (x) x^2+10 e^{x^8} \left (4 x^9+x\right ) \log ^2(x)\right )}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{5} \int \frac {5 e^{x^8} \left (-8 x^{11}+8 x^{10}+x^2-e^2 \left (8 x^{10}+x^2\right )\right ) \log ^2(x)-e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} \log (x) x^2+10 e^{x^8} \left (4 x^9+x\right ) \log ^2(x)\right )}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{5} \int \left (\frac {e^{x^8} \left (-\log (x) x^2+x^2+10 \log ^2(x) x+\left (1-e^2\right ) \log (x) x-\left (1-e^2\right ) x-5 \left (1-e^2\right ) \log ^2(x)\right ) x^2}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log ^2(x)}+\frac {e^{x^8} \left (-40 \log ^2(x) x^8+\log (x) x-x-10 \log ^2(x)\right ) x}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right ) \log ^2(x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (-\int \frac {e^{x^8} x^2}{\left (-x^2+\left (1-e^2\right ) x-e^{\frac {x}{5 \log (x)}}\right ) \log ^2(x)}dx-10 \int \frac {e^{x^8} x}{-x^2+\left (1-e^2\right ) x-e^{\frac {x}{5 \log (x)}}}dx+5 \left (1-e^2\right ) \int \frac {e^{x^8} x^2}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2}dx-\int \frac {e^{x^8} x^2}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right ) \log (x)}dx-40 \int \frac {e^{x^8} x^9}{-x^2+\left (1-e^2\right ) x-e^{\frac {x}{5 \log (x)}}}dx-\int \frac {e^{x^8} x^4}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log ^2(x)}dx+\int \frac {e^{x^8} x^4}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log (x)}dx+\left (1-e^2\right ) \int \frac {e^{x^8} x^3}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log ^2(x)}dx-10 \int \frac {e^{x^8} x^3}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2}dx-\left (1-e^2\right ) \int \frac {e^{x^8} x^3}{\left (x^2-\left (1-e^2\right ) x+e^{\frac {x}{5 \log (x)}}\right )^2 \log (x)}dx\right )\) |
Int[(E^x^8*(-5*x^2 - 40*x^10 + 40*x^11 + E^2*(5*x^2 + 40*x^10))*Log[x]^2 + E^(x/(5*Log[x]))*(E^x^8*x^2 - E^x^8*x^2*Log[x] + E^x^8*(10*x + 40*x^9)*Lo g[x]^2))/(5*E^((2*x)/(5*Log[x]))*Log[x]^2 + E^(x/(5*Log[x]))*(-10*x + 10*E ^2*x + 10*x^2)*Log[x]^2 + (5*x^2 + 5*E^4*x^2 - 10*x^3 + 5*x^4 + E^2*(-10*x ^2 + 10*x^3))*Log[x]^2),x]
3.11.97.3.1 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 = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00
\[\frac {x^{2} {\mathrm e}^{x^{8}}}{{\mathrm e}^{2} x +x^{2}+{\mathrm e}^{\frac {x}{5 \ln \left (x \right )}}-x}\]
int((((40*x^9+10*x)*exp(x^8)*ln(x)^2-x^2*exp(x^8)*ln(x)+x^2*exp(x^8))*exp( 1/5*x/ln(x))+((40*x^10+5*x^2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x^8)*ln(x) ^2)/(5*ln(x)^2*exp(1/5*x/ln(x))^2+(10*exp(2)*x+10*x^2-10*x)*ln(x)^2*exp(1/ 5*x/ln(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x^3+5*x^2)*ln(x )^2),x)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x e^{2} - x + e^{\left (\frac {x}{5 \, \log \left (x\right )}\right )}} \]
integrate((((40*x^9+10*x)*exp(x^8)*log(x)^2-x^2*exp(x^8)*log(x)+x^2*exp(x^ 8))*exp(1/5*x/log(x))+((40*x^10+5*x^2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x ^8)*log(x)^2)/(5*log(x)^2*exp(1/5*x/log(x))^2+(10*exp(2)*x+10*x^2-10*x)*lo g(x)^2*exp(1/5*x/log(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x ^3+5*x^2)*log(x)^2),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{x^{8}}}{x^{2} - x + x e^{2} + e^{\frac {x}{5 \log {\left (x \right )}}}} \]
integrate((((40*x**9+10*x)*exp(x**8)*ln(x)**2-x**2*exp(x**8)*ln(x)+x**2*ex p(x**8))*exp(1/5*x/ln(x))+((40*x**10+5*x**2)*exp(2)+40*x**11-40*x**10-5*x* *2)*exp(x**8)*ln(x)**2)/(5*ln(x)**2*exp(1/5*x/ln(x))**2+(10*exp(2)*x+10*x* *2-10*x)*ln(x)**2*exp(1/5*x/ln(x))+(5*x**2*exp(2)**2+(10*x**3-10*x**2)*exp (2)+5*x**4-10*x**3+5*x**2)*ln(x)**2),x)
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x {\left (e^{2} - 1\right )} + e^{\left (\frac {x}{5 \, \log \left (x\right )}\right )}} \]
integrate((((40*x^9+10*x)*exp(x^8)*log(x)^2-x^2*exp(x^8)*log(x)+x^2*exp(x^ 8))*exp(1/5*x/log(x))+((40*x^10+5*x^2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x ^8)*log(x)^2)/(5*log(x)^2*exp(1/5*x/log(x))^2+(10*exp(2)*x+10*x^2-10*x)*lo g(x)^2*exp(1/5*x/log(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x ^3+5*x^2)*log(x)^2),x, algorithm=\
Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x e^{2} - x + e^{\left (\frac {x}{5 \, \log \left (x\right )}\right )}} \]
integrate((((40*x^9+10*x)*exp(x^8)*log(x)^2-x^2*exp(x^8)*log(x)+x^2*exp(x^ 8))*exp(1/5*x/log(x))+((40*x^10+5*x^2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x ^8)*log(x)^2)/(5*log(x)^2*exp(1/5*x/log(x))^2+(10*exp(2)*x+10*x^2-10*x)*lo g(x)^2*exp(1/5*x/log(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x ^3+5*x^2)*log(x)^2),x, algorithm=\
Timed out. \[ \int \frac {e^{x^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {x}{5\,\ln \left (x\right )}}\,\left (x^2\,{\mathrm {e}}^{x^8}+{\mathrm {e}}^{x^8}\,{\ln \left (x\right )}^2\,\left (40\,x^9+10\,x\right )-x^2\,{\mathrm {e}}^{x^8}\,\ln \left (x\right )\right )+{\mathrm {e}}^{x^8}\,{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^2\,\left (40\,x^{10}+5\,x^2\right )-5\,x^2-40\,x^{10}+40\,x^{11}\right )}{5\,{\mathrm {e}}^{\frac {2\,x}{5\,\ln \left (x\right )}}\,{\ln \left (x\right )}^2+{\ln \left (x\right )}^2\,\left (5\,x^2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (10\,x^2-10\,x^3\right )+5\,x^2-10\,x^3+5\,x^4\right )+{\mathrm {e}}^{\frac {x}{5\,\ln \left (x\right )}}\,{\ln \left (x\right )}^2\,\left (10\,x\,{\mathrm {e}}^2-10\,x+10\,x^2\right )} \,d x \]
int((exp(x/(5*log(x)))*(x^2*exp(x^8) + exp(x^8)*log(x)^2*(10*x + 40*x^9) - x^2*exp(x^8)*log(x)) + exp(x^8)*log(x)^2*(exp(2)*(5*x^2 + 40*x^10) - 5*x^ 2 - 40*x^10 + 40*x^11))/(5*exp((2*x)/(5*log(x)))*log(x)^2 + log(x)^2*(5*x^ 2*exp(4) - exp(2)*(10*x^2 - 10*x^3) + 5*x^2 - 10*x^3 + 5*x^4) + exp(x/(5*l og(x)))*log(x)^2*(10*x*exp(2) - 10*x + 10*x^2)),x)
int((exp(x/(5*log(x)))*(x^2*exp(x^8) + exp(x^8)*log(x)^2*(10*x + 40*x^9) - x^2*exp(x^8)*log(x)) + exp(x^8)*log(x)^2*(exp(2)*(5*x^2 + 40*x^10) - 5*x^ 2 - 40*x^10 + 40*x^11))/(5*exp((2*x)/(5*log(x)))*log(x)^2 + log(x)^2*(5*x^ 2*exp(4) - exp(2)*(10*x^2 - 10*x^3) + 5*x^2 - 10*x^3 + 5*x^4) + exp(x/(5*l og(x)))*log(x)^2*(10*x*exp(2) - 10*x + 10*x^2)), x)