Integrand size = 121, antiderivative size = 19 \[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=e^{9-\frac {\log \left (-\frac {100}{9}+\log (\log (-2+x))\right )}{x}} \]
\[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=\int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx \]
Integrate[(E^((5*x - Log[(-100 + 9*Log[Log[-2 + x]])/9])/x)*(-9*E^4*x + (E ^4*(200 - 100*x)*Log[-2 + x] + E^4*(-18 + 9*x)*Log[-2 + x]*Log[Log[-2 + x] ])*Log[(-100 + 9*Log[Log[-2 + x]])/9]))/((200*x^2 - 100*x^3)*Log[-2 + x] + (-18*x^2 + 9*x^3)*Log[-2 + x]*Log[Log[-2 + x]]),x]
Integrate[(E^((5*x - Log[(-100 + 9*Log[Log[-2 + x]])/9])/x)*(-9*E^4*x + (E ^4*(200 - 100*x)*Log[-2 + x] + E^4*(-18 + 9*x)*Log[-2 + x]*Log[Log[-2 + x] ])*Log[(-100 + 9*Log[Log[-2 + x]])/9]))/((200*x^2 - 100*x^3)*Log[-2 + x] + (-18*x^2 + 9*x^3)*Log[-2 + x]*Log[Log[-2 + x]]), x]
Time = 0.50 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (9 \log (\log (x-2))-100)\right )}{x}} \left (\left (e^4 (200-100 x) \log (x-2)+e^4 (9 x-18) \log (\log (x-2)) \log (x-2)\right ) \log \left (\frac {1}{9} (9 \log (\log (x-2))-100)\right )-9 e^4 x\right )}{\left (200 x^2-100 x^3\right ) \log (x-2)+\left (9 x^3-18 x^2\right ) \log (\log (x-2)) \log (x-2)} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle \exp \left (\frac {5 x-\log \left (\frac {1}{9} (9 \log (\log (x-2))-100)\right )}{x}+4\right )\) |
Int[(E^((5*x - Log[(-100 + 9*Log[Log[-2 + x]])/9])/x)*(-9*E^4*x + (E^4*(20 0 - 100*x)*Log[-2 + x] + E^4*(-18 + 9*x)*Log[-2 + x]*Log[Log[-2 + x]])*Log [(-100 + 9*Log[Log[-2 + x]])/9]))/((200*x^2 - 100*x^3)*Log[-2 + x] + (-18* x^2 + 9*x^3)*Log[-2 + x]*Log[Log[-2 + x]]),x]
3.12.37.3.1 Defintions of rubi rules used
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89
\[\left (\ln \left (\ln \left (-2+x \right )\right )-\frac {100}{9}\right )^{-\frac {1}{x}} {\mathrm e}^{9}\]
int((((9*x-18)*exp(4)*ln(-2+x)*ln(ln(-2+x))+(-100*x+200)*exp(4)*ln(-2+x))* ln(ln(ln(-2+x))-100/9)-9*x*exp(4))*exp((-ln(ln(ln(-2+x))-100/9)+5*x)/x)/(( 9*x^3-18*x^2)*ln(-2+x)*ln(ln(-2+x))+(-100*x^3+200*x^2)*ln(-2+x)),x)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=e^{\left (\frac {5 \, x - \log \left (\log \left (\log \left (x - 2\right )\right ) - \frac {100}{9}\right )}{x} + 4\right )} \]
integrate((((9*x-18)*exp(4)*log(-2+x)*log(log(-2+x))+(-100*x+200)*exp(4)*l og(-2+x))*log(log(log(-2+x))-100/9)-9*x*exp(4))*exp((-log(log(log(-2+x))-1 00/9)+5*x)/x)/((9*x^3-18*x^2)*log(-2+x)*log(log(-2+x))+(-100*x^3+200*x^2)* log(-2+x)),x, algorithm=\
Time = 1.66 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=e^{4} e^{\frac {5 x - \log {\left (\log {\left (\log {\left (x - 2 \right )} \right )} - \frac {100}{9} \right )}}{x}} \]
integrate((((9*x-18)*exp(4)*ln(-2+x)*ln(ln(-2+x))+(-100*x+200)*exp(4)*ln(- 2+x))*ln(ln(ln(-2+x))-100/9)-9*x*exp(4))*exp((-ln(ln(ln(-2+x))-100/9)+5*x) /x)/((9*x**3-18*x**2)*ln(-2+x)*ln(ln(-2+x))+(-100*x**3+200*x**2)*ln(-2+x)) ,x)
Time = 0.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=e^{\left (\frac {2 \, \log \left (3\right )}{x} - \frac {\log \left (9 \, \log \left (\log \left (x - 2\right )\right ) - 100\right )}{x} + 9\right )} \]
integrate((((9*x-18)*exp(4)*log(-2+x)*log(log(-2+x))+(-100*x+200)*exp(4)*l og(-2+x))*log(log(log(-2+x))-100/9)-9*x*exp(4))*exp((-log(log(log(-2+x))-1 00/9)+5*x)/x)/((9*x^3-18*x^2)*log(-2+x)*log(log(-2+x))+(-100*x^3+200*x^2)* log(-2+x)),x, algorithm=\
Time = 2.59 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=e^{\left (-\frac {\log \left (\log \left (\log \left (x - 2\right )\right ) - \frac {100}{9}\right )}{x} + 9\right )} \]
integrate((((9*x-18)*exp(4)*log(-2+x)*log(log(-2+x))+(-100*x+200)*exp(4)*l og(-2+x))*log(log(log(-2+x))-100/9)-9*x*exp(4))*exp((-log(log(log(-2+x))-1 00/9)+5*x)/x)/((9*x^3-18*x^2)*log(-2+x)*log(log(-2+x))+(-100*x^3+200*x^2)* log(-2+x)),x, algorithm=\
Time = 8.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {5 x-\log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )}{x}} \left (-9 e^4 x+\left (e^4 (200-100 x) \log (-2+x)+e^4 (-18+9 x) \log (-2+x) \log (\log (-2+x))\right ) \log \left (\frac {1}{9} (-100+9 \log (\log (-2+x)))\right )\right )}{\left (200 x^2-100 x^3\right ) \log (-2+x)+\left (-18 x^2+9 x^3\right ) \log (-2+x) \log (\log (-2+x))} \, dx=\frac {{\mathrm {e}}^9}{{\left (\ln \left (\ln \left (x-2\right )\right )-\frac {100}{9}\right )}^{1/x}} \]