Integrand size = 102, antiderivative size = 25 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {e^{e^{\frac {1}{x}}} x^2}{2 \log (1-\log (25+x))} \]
Time = 4.58 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {e^{e^{\frac {1}{x}}} x^2}{2 \log (1-\log (25+x))} \]
Integrate[(-(E^E^x^(-1)*x^2) + E^E^x^(-1)*(-50*x - 2*x^2 + E^x^(-1)*(25 + x) + (E^x^(-1)*(-25 - x) + 50*x + 2*x^2)*Log[25 + x])*Log[1 - Log[25 + x]] )/((-50 - 2*x + (50 + 2*x)*Log[25 + x])*Log[1 - Log[25 + 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.
\(\displaystyle \int \frac {e^{e^{\frac {1}{x}}} \left (-2 x^2+\left (2 x^2+50 x+e^{\frac {1}{x}} (-x-25)\right ) \log (x+25)-50 x+e^{\frac {1}{x}} (x+25)\right ) \log (1-\log (x+25))-e^{e^{\frac {1}{x}}} x^2}{(-2 x+(2 x+50) \log (x+25)-50) \log ^2(1-\log (x+25))} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{e^{\frac {1}{x}}} \left (-\frac {x^2}{(x+25) (\log (x+25)-1)}-\left (e^{\frac {1}{x}}-2 x\right ) \log (1-\log (x+25))\right )}{2 \log ^2(1-\log (x+25))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} \left (\frac {x^2}{(x+25) (1-\log (x+25))}-\left (e^{\frac {1}{x}}-2 x\right ) \log (1-\log (x+25))\right )}{\log ^2(1-\log (x+25))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {e^{e^{\frac {1}{x}}} x (2 \log (x+25) \log (1-\log (x+25)) x-2 \log (1-\log (x+25)) x-x+50 \log (x+25) \log (1-\log (x+25))-50 \log (1-\log (x+25)))}{(x+25) (\log (x+25)-1) \log ^2(1-\log (x+25))}-\frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (x+25))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (25 \int \frac {e^{e^{\frac {1}{x}}}}{(\log (x+25)-1) \log ^2(1-\log (x+25))}dx-\int \frac {e^{e^{\frac {1}{x}}} x}{(\log (x+25)-1) \log ^2(1-\log (x+25))}dx-625 \int \frac {e^{e^{\frac {1}{x}}}}{(x+25) (\log (x+25)-1) \log ^2(1-\log (x+25))}dx-\int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (x+25))}dx+2 \int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (x+25))}dx\right )\) |
Int[(-(E^E^x^(-1)*x^2) + E^E^x^(-1)*(-50*x - 2*x^2 + E^x^(-1)*(25 + x) + ( E^x^(-1)*(-25 - x) + 50*x + 2*x^2)*Log[25 + x])*Log[1 - Log[25 + x]])/((-5 0 - 2*x + (50 + 2*x)*Log[25 + x])*Log[1 - Log[25 + x]]^2),x]
3.13.73.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 43.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{\frac {1}{x}}}}{2 \ln \left (-\ln \left (x +25\right )+1\right )}\) | \(22\) |
parallelrisch | \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{\frac {1}{x}}}}{2 \ln \left (-\ln \left (x +25\right )+1\right )}\) | \(22\) |
int(((((-x-25)*exp(1/x)+2*x^2+50*x)*ln(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*e xp(exp(1/x))*ln(-ln(x+25)+1)-x^2*exp(exp(1/x)))/((2*x+50)*ln(x+25)-2*x-50) /ln(-ln(x+25)+1)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \]
integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2- 50*x)*exp(exp(1/x))*log(-log(x+25)+1)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+2 5)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm=\
Time = 0.68 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {x^{2} e^{e^{\frac {1}{x}}}}{2 \log {\left (1 - \log {\left (x + 25 \right )} \right )}} \]
integrate(((((-x-25)*exp(1/x)+2*x**2+50*x)*ln(x+25)+(x+25)*exp(1/x)-2*x**2 -50*x)*exp(exp(1/x))*ln(-ln(x+25)+1)-x**2*exp(exp(1/x)))/((2*x+50)*ln(x+25 )-2*x-50)/ln(-ln(x+25)+1)**2,x)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \]
integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2- 50*x)*exp(exp(1/x))*log(-log(x+25)+1)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+2 5)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \]
integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2- 50*x)*exp(exp(1/x))*log(-log(x+25)+1)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+2 5)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm=\
Time = 11.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx=\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/x}}}{2\,\ln \left (1-\ln \left (x+25\right )\right )} \]