Integrand size = 89, antiderivative size = 26 \[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=2-\frac {1}{5} \log \left (-2+x+\frac {x}{3 e^{-e^x}+\log (5)}\right ) \]
\[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=\int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx \]
Integrate[(-9 + E^E^x*(-3 - 3*E^x*x - 6*Log[5]) + E^(2*E^x)*(-Log[5] - Log [5]^2))/(-90 + 45*x + E^E^x*(15*x + (-60 + 30*x)*Log[5]) + E^(2*E^x)*(5*x* Log[5] + (-10 + 5*x)*Log[5]^2)),x]
Integrate[(-9 + E^E^x*(-3 - 3*E^x*x - 6*Log[5]) + E^(2*E^x)*(-Log[5] - Log [5]^2))/(-90 + 45*x + E^E^x*(15*x + (-60 + 30*x)*Log[5]) + E^(2*E^x)*(5*x* Log[5] + (-10 + 5*x)*Log[5]^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.
| \(\displaystyle \int \frac {e^{2 e^x} \left (-\log ^2(5)-\log (5)\right )+e^{e^x} \left (-3 e^x x-3-6 \log (5)\right )-9}{45 x+e^{2 e^x} \left ((5 x-10) \log ^2(5)+5 x \log (5)\right )+e^{e^x} (15 x+(30 x-60) \log (5))-90} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-e^{2 e^x} \left (-\log ^2(5)-\log (5)\right )-e^{e^x} \left (-3 e^x x-3-6 \log (5)\right )+9}{5 \left (e^{e^x} \log (5)+3\right ) \left (-3 x-e^{e^x} x (1+\log (5))+2 e^{e^x} \log (5)+6\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {3 e^{e^x} \left (e^x x+2 \log (5)+1\right )+e^{2 e^x} \log (5) (1+\log (5))+9}{\left (3+e^{e^x} \log (5)\right ) \left (-e^{e^x} (1+\log (5)) x-3 x+2 e^{e^x} \log (5)+6\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{5} \int \left (\frac {3 e^{x+e^x} x}{\left (3+e^{e^x} \log (5)\right ) \left (-e^{e^x} (1+\log (5)) x-3 x+2 e^{e^x} \log (5)+6\right )}+\frac {e^{2 e^x} \log (5) (1+\log (5))}{\left (3+e^{e^x} \log (5)\right ) \left (-e^{e^x} (1+\log (5)) x-3 x+2 e^{e^x} \log (5)+6\right )}+\frac {9}{\left (3+e^{e^x} \log (5)\right ) \left (-e^{e^x} (1+\log (5)) x-3 x+2 e^{e^x} \log (5)+6\right )}+\frac {3 e^{e^x} (1+\log (25))}{\left (3+e^{e^x} \log (5)\right ) \left (-e^{e^x} (1+\log (5)) x-3 x+2 e^{e^x} \log (5)+6\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\log ^2(5) (1+\log (5)) \int \frac {e^{2 e^x}}{x \left (9+e^{e^x} \log (125)\right )}dx+\log (25) \int \frac {e^{x+e^x}}{e^{e^x} (1+\log (5)) x+3 x-2 e^{e^x} \log (5)-6}dx+3 \log (25) \int \frac {1}{x \left (e^{e^x} (1+\log (5)) x+3 x-2 e^{e^x} \log (5)-6\right )}dx+\log (25) (1+\log (25)) \int \frac {e^{e^x}}{x \left (e^{e^x} (1+\log (5)) x+3 x-2 e^{e^x} \log (5)-6\right )}dx+\frac {1}{3} \log (5) (1+\log (5)) \log (25) \int \frac {e^{2 e^x}}{x \left (e^{e^x} (1+\log (5)) x+3 x-2 e^{e^x} \log (5)-6\right )}dx+3 (1+\log (5)) \int \frac {1}{-e^{e^x} (1+\log (5)) x-3 x+e^{e^x} \log (25)+6}dx+(1+\log (5)) (1+\log (25)) \int \frac {e^{e^x}}{-e^{e^x} (1+\log (5)) x-3 x+e^{e^x} \log (25)+6}dx+\frac {1}{3} \log (5) (1+\log (5))^2 \int \frac {e^{2 e^x}}{-e^{e^x} (1+\log (5)) x-3 x+e^{e^x} \log (25)+6}dx+(1+\log (5)) \int \frac {e^{x+e^x} x}{-e^{e^x} (1+\log (5)) x-3 x+e^{e^x} \log (25)+6}dx+9 \log (5) \int \frac {1}{x \left (9+e^{e^x} \log (125)\right )}dx+3 \log (5) (1+\log (25)) \int \frac {e^{e^x}}{x \left (9+e^{e^x} \log (125)\right )}dx+\log \left (e^{e^x} \log (5)+3\right )\right )\) |
Int[(-9 + E^E^x*(-3 - 3*E^x*x - 6*Log[5]) + E^(2*E^x)*(-Log[5] - Log[5]^2) )/(-90 + 45*x + E^E^x*(15*x + (-60 + 30*x)*Log[5]) + E^(2*E^x)*(5*x*Log[5] + (-10 + 5*x)*Log[5]^2)),x]
3.17.79.3.1 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 = 0.64 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54
| method | result | size |
| norman | \(\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}} \ln \left (5\right )+3\right )}{5}-\frac {\ln \left (\ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{x}} x -2 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (5\right )+x \,{\mathrm e}^{{\mathrm e}^{x}}+3 x -6\right )}{5}\) | \(40\) |
| parallelrisch | \(\frac {\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{x}} \ln \left (5\right )+3}{\ln \left (5\right )}\right )}{5}-\frac {\ln \left (\frac {\ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{x}} x -2 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (5\right )+x \,{\mathrm e}^{{\mathrm e}^{x}}+3 x -6}{\ln \left (5\right )+1}\right )}{5}\) | \(52\) |
| risch | \(-\frac {\ln \left (\left (\ln \left (5\right )+1\right ) x -2 \ln \left (5\right )\right )}{5}-\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {-6+3 x}{x \ln \left (5\right )-2 \ln \left (5\right )+x}\right )}{5}+\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {3}{\ln \left (5\right )}\right )}{5}\) | \(53\) |
int(((-ln(5)^2-ln(5))*exp(exp(x))^2+(-3*exp(x)*x-6*ln(5)-3)*exp(exp(x))-9) /(((5*x-10)*ln(5)^2+5*x*ln(5))*exp(exp(x))^2+((30*x-60)*ln(5)+15*x)*exp(ex p(x))+45*x-90),x,method=_RETURNVERBOSE)
1/5*ln(exp(exp(x))*ln(5)+3)-1/5*ln(ln(5)*exp(exp(x))*x-2*exp(exp(x))*ln(5) +x*exp(exp(x))+3*x-6)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.44 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=-\frac {1}{5} \, \log \left ({\left (x - 2\right )} \log \left (5\right ) + x\right ) + \frac {1}{5} \, \log \left (e^{\left (e^{x}\right )} \log \left (5\right ) + 3\right ) - \frac {1}{5} \, \log \left (\frac {{\left ({\left (x - 2\right )} \log \left (5\right ) + x\right )} e^{\left (e^{x}\right )} + 3 \, x - 6}{{\left (x - 2\right )} \log \left (5\right ) + x}\right ) \]
integrate(((-log(5)^2-log(5))*exp(exp(x))^2+(-3*exp(x)*x-6*log(5)-3)*exp(e xp(x))-9)/(((5*x-10)*log(5)^2+5*x*log(5))*exp(exp(x))^2+((30*x-60)*log(5)+ 15*x)*exp(exp(x))+45*x-90),x, algorithm=\
-1/5*log((x - 2)*log(5) + x) + 1/5*log(e^(e^x)*log(5) + 3) - 1/5*log((((x - 2)*log(5) + x)*e^(e^x) + 3*x - 6)/((x - 2)*log(5) + x))
Exception generated. \[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=\text {Exception raised: PolynomialError} \]
integrate(((-ln(5)**2-ln(5))*exp(exp(x))**2+(-3*exp(x)*x-6*ln(5)-3)*exp(ex p(x))-9)/(((5*x-10)*ln(5)**2+5*x*ln(5))*exp(exp(x))**2+((30*x-60)*ln(5)+15 *x)*exp(exp(x))+45*x-90),x)
Exception raised: PolynomialError >> 1/(5*x**2*log(5) + 5*x**2*log(5)**3 + 10*x**2*log(5)**2 - 20*x*log(5)**3 - 20*x*log(5)**2 + 20*log(5)**3) conta ins an element of the set of generators.
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=-\frac {1}{5} \, \log \left (x {\left (\log \left (5\right ) + 1\right )} - 2 \, \log \left (5\right )\right ) - \frac {1}{5} \, \log \left (\frac {{\left (x {\left (\log \left (5\right ) + 1\right )} - 2 \, \log \left (5\right )\right )} e^{\left (e^{x}\right )} + 3 \, x - 6}{x {\left (\log \left (5\right ) + 1\right )} - 2 \, \log \left (5\right )}\right ) + \frac {1}{5} \, \log \left (\frac {e^{\left (e^{x}\right )} \log \left (5\right ) + 3}{\log \left (5\right )}\right ) \]
integrate(((-log(5)^2-log(5))*exp(exp(x))^2+(-3*exp(x)*x-6*log(5)-3)*exp(e xp(x))-9)/(((5*x-10)*log(5)^2+5*x*log(5))*exp(exp(x))^2+((30*x-60)*log(5)+ 15*x)*exp(exp(x))+45*x-90),x, algorithm=\
-1/5*log(x*(log(5) + 1) - 2*log(5)) - 1/5*log(((x*(log(5) + 1) - 2*log(5)) *e^(e^x) + 3*x - 6)/(x*(log(5) + 1) - 2*log(5))) + 1/5*log((e^(e^x)*log(5) + 3)/log(5))
Time = 0.41 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=-\frac {1}{5} \, e^{x} - \frac {1}{5} \, \log \left (x e^{\left (e^{x}\right )} \log \left (5\right ) + x e^{\left (e^{x}\right )} - 2 \, e^{\left (e^{x}\right )} \log \left (5\right ) + 3 \, x - 6\right ) + \frac {1}{5} \, \log \left (e^{\left (e^{x}\right )} \log \left (5\right ) + 3\right ) \]
integrate(((-log(5)^2-log(5))*exp(exp(x))^2+(-3*exp(x)*x-6*log(5)-3)*exp(e xp(x))-9)/(((5*x-10)*log(5)^2+5*x*log(5))*exp(exp(x))^2+((30*x-60)*log(5)+ 15*x)*exp(exp(x))+45*x-90),x, algorithm=\
-1/5*e^x - 1/5*log(x*e^(e^x)*log(5) + x*e^(e^x) - 2*e^(e^x)*log(5) + 3*x - 6) + 1/5*log(e^(e^x)*log(5) + 3)
Timed out. \[ \int \frac {-9+e^{e^x} \left (-3-3 e^x x-6 \log (5)\right )+e^{2 e^x} \left (-\log (5)-\log ^2(5)\right )}{-90+45 x+e^{e^x} (15 x+(-60+30 x) \log (5))+e^{2 e^x} \left (5 x \log (5)+(-10+5 x) \log ^2(5)\right )} \, dx=\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (6\,\ln \left (5\right )+3\,x\,{\mathrm {e}}^x+3\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (\ln \left (5\right )+{\ln \left (5\right )}^2\right )+9}{45\,x+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (5\,x\,\ln \left (5\right )+{\ln \left (5\right )}^2\,\left (5\,x-10\right )\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (15\,x+\ln \left (5\right )\,\left (30\,x-60\right )\right )-90} \,d x \]
int(-(exp(exp(x))*(6*log(5) + 3*x*exp(x) + 3) + exp(2*exp(x))*(log(5) + lo g(5)^2) + 9)/(45*x + exp(2*exp(x))*(5*x*log(5) + log(5)^2*(5*x - 10)) + ex p(exp(x))*(15*x + log(5)*(30*x - 60)) - 90),x)