Integrand size = 138, antiderivative size = 30 \begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\left (-5+e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}}\right ) \log (2) \end {dmath*}
\begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx \end {dmath*}
Integrate[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2 *E^x*x*Log[2] + (-10*Log[2] - 2*E^x*Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E ^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2] ]))/((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2),x]
Integrate[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2 *E^x*x*Log[2] + (-10*Log[2] - 2*E^x*Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E ^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2] ]))/((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^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.
\(\displaystyle \int \frac {e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (\left (e^x \left (x^2-x+1\right ) \log (2)+(5-5 x) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 e^x x \log (2)+10 x \log (2)+\left (-2 e^x \log (2)-10 \log (2)\right ) \log (x)\right )}{\left (10 e^x x+e^{2 x} x+25 x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (\left (e^x \left (x^2-x+1\right ) \log (2)+(5-5 x) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 e^x x \log (2)+10 x \log (2)+\left (-2 e^x \log (2)-10 \log (2)\right ) \log (x)\right )}{\left (e^x+5\right )^2 x \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log (2) e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (x^2 \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log (x) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+\log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 x-2 \log (x)\right )}{\left (e^x+5\right ) x \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}-\frac {5 \log (2) e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} (x-\log (x))}{\left (e^x+5\right )^2 \log \left (\log \left (25 x^2\right )\right )}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}-1} \left (x^2 \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log (x) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+\log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 x-2 \log (x)\right )}{\left (e^x+5\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}-\frac {5 \log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} (x-\log (x))}{\left (e^x+5\right )^2 \log \left (\log \left (25 x^2\right )\right )}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {\log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}-1} \left (x^2 \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log (x) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+\log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 x-2 \log (x)\right )}{\left (e^x+5\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}-\frac {5 \log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} (x-\log (x))}{\left (e^x+5\right )^2 \log \left (\log \left (25 x^2\right )\right )}\right )dx\) |
Int[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2*E^x*x *Log[2] + (-10*Log[2] - 2*E^x*Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2]]))/(( 25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2),x]
3.6.66.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.84 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93
\[\ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (x \right )-x}{\left ({\mathrm e}^{x}+5\right ) \ln \left (2 \ln \left (x \right )+2 \ln \left (5\right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )}}\]
int(((-x*ln(2)*exp(x)*ln(x)+(x^2-x+1)*ln(2)*exp(x)+(-5*x+5)*ln(2))*ln(25*x ^2)*ln(ln(25*x^2))+(-2*exp(x)*ln(2)-10*ln(2))*ln(x)+2*x*ln(2)*exp(x)+10*x* ln(2))*exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x^2)))/(x*exp(x)^2+10*exp(x)*x+25 *x)/ln(25*x^2)/ln(ln(25*x^2))^2,x)
ln(2)*exp((ln(x)-x)/(exp(x)+5)/ln(2*ln(x)+2*ln(5)-1/2*I*Pi*csgn(I*x^2)*(-c sgn(I*x^2)+csgn(I*x))^2))
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=e^{\left (-\frac {x - \log \left (x\right )}{{\left (e^{x} + 5\right )} \log \left (2 \, \log \left (5\right ) + 2 \, \log \left (x\right )\right )}\right )} \log \left (2\right ) \end {dmath*}
integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2 ))*log(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*lo g(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*ex p(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm=\
Time = 33.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=e^{\frac {- x + \log {\left (x \right )}}{\left (e^{x} + 5\right ) \log {\left (2 \log {\left (x \right )} + \log {\left (25 \right )} \right )}}} \log {\left (2 \right )} \end {dmath*}
integrate(((-x*ln(2)*exp(x)*ln(x)+(x**2-x+1)*ln(2)*exp(x)+(-5*x+5)*ln(2))* ln(25*x**2)*ln(ln(25*x**2))+(-2*exp(x)*ln(2)-10*ln(2))*ln(x)+2*x*ln(2)*exp (x)+10*x*ln(2))*exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x**2)))/(x*exp(x)**2+10* exp(x)*x+25*x)/ln(25*x**2)/ln(ln(25*x**2))**2,x)
Exception generated. \begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\text {Exception raised: RuntimeError} \end {dmath*}
integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2 ))*log(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*lo g(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*ex p(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm=\
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
\begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\int { \frac {{\left (2 \, x e^{x} \log \left (2\right ) - {\left (x e^{x} \log \left (2\right ) \log \left (x\right ) - {\left (x^{2} - x + 1\right )} e^{x} \log \left (2\right ) + 5 \, {\left (x - 1\right )} \log \left (2\right )\right )} \log \left (25 \, x^{2}\right ) \log \left (\log \left (25 \, x^{2}\right )\right ) + 10 \, x \log \left (2\right ) - 2 \, {\left (e^{x} \log \left (2\right ) + 5 \, \log \left (2\right )\right )} \log \left (x\right )\right )} e^{\left (-\frac {x - \log \left (x\right )}{{\left (e^{x} + 5\right )} \log \left (\log \left (25 \, x^{2}\right )\right )}\right )}}{{\left (x e^{\left (2 \, x\right )} + 10 \, x e^{x} + 25 \, x\right )} \log \left (25 \, x^{2}\right ) \log \left (\log \left (25 \, x^{2}\right )\right )^{2}} \,d x } \end {dmath*}
integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2 ))*log(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*lo g(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*ex p(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm=\
Time = 14.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \begin {dmath*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=x^{\frac {1}{5\,\ln \left (\ln \left (25\,x^2\right )\right )+\ln \left (\ln \left (25\,x^2\right )\right )\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {x}{5\,\ln \left (\ln \left (25\,x^2\right )\right )+\ln \left (\ln \left (25\,x^2\right )\right )\,{\mathrm {e}}^x}}\,\ln \left (2\right ) \end {dmath*}
int(-(exp(-(x - log(x))/(log(log(25*x^2))*(exp(x) + 5)))*(log(x)*(10*log(2 ) + 2*exp(x)*log(2)) - 10*x*log(2) - 2*x*exp(x)*log(2) + log(log(25*x^2))* log(25*x^2)*(log(2)*(5*x - 5) - exp(x)*log(2)*(x^2 - x + 1) + x*exp(x)*log (2)*log(x))))/(log(log(25*x^2))^2*log(25*x^2)*(25*x + x*exp(2*x) + 10*x*ex p(x))),x)