Integrand size = 95, antiderivative size = 24 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\left (-e^{16}+(2+2 x)^5\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right ) \]
Time = 1.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\left (-e^{16}+32 (1+x)^5\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right ) \]
Integrate[((E^17*(1 + x))/x + (E*(-1 - x)*(2 + 2*x)^5)/x + (5*E - 25*x)*(2 + 2*x)^5*Log[5 - E/x]*Log[Log[5 - E/x]])/((-5*x - 5*x^2 + (E*(x + x^2))/x )*Log[5 - E/x]),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.
\(\displaystyle \int \frac {\frac {e (-x-1) (2 x+2)^5}{x}+\frac {e^{17} (x+1)}{x}+(5 e-25 x) (2 x+2)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x^2+\frac {e \left (x^2+x\right )}{x}-5 x\right ) \log \left (5-\frac {e}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\frac {e (-x-1) (2 x+2)^5}{x}+\frac {e^{17} (x+1)}{x}+(5 e-25 x) (2 x+2)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{(e-5 x) (x+1) \log \left (5-\frac {e}{x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e \left (-32 x^5-160 x^4-320 x^3-320 x^2-160 x-32 \left (1-\frac {e^{16}}{32}\right )\right )}{(e-5 x) x \log \left (5-\frac {e}{x}\right )}+160 (x+1)^4 \log \left (\log \left (5-\frac {e}{x}\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 160 \int x^4 \log \left (\log \left (5-\frac {e}{x}\right )\right )dx+640 \int x^3 \log \left (\log \left (5-\frac {e}{x}\right )\right )dx+960 \int x^2 \log \left (\log \left (5-\frac {e}{x}\right )\right )dx+e \int \frac {-32 x^5-160 x^4-320 x^3-320 x^2-160 x-32 \left (1-\frac {e^{16}}{32}\right )}{(e-5 x) x \log \left (\frac {5 x-e}{x}\right )}dx+160 \int \log \left (\log \left (5-\frac {e}{x}\right )\right )dx+640 \int x \log \left (\log \left (5-\frac {e}{x}\right )\right )dx\) |
Int[((E^17*(1 + x))/x + (E*(-1 - x)*(2 + 2*x)^5)/x + (5*E - 25*x)*(2 + 2*x )^5*Log[5 - E/x]*Log[Log[5 - E/x]])/((-5*x - 5*x^2 + (E*(x + x^2))/x)*Log[ 5 - E/x]),x]
3.18.45.3.1 Defintions of rubi rules used
\[\int \frac {\left (5 x \,{\mathrm e}^{1-\ln \left (x \right )}-25 x \right ) \left (2+2 x \right )^{5} \ln \left (-{\mathrm e}^{1-\ln \left (x \right )}+5\right ) \ln \left (\ln \left (-{\mathrm e}^{1-\ln \left (x \right )}+5\right )\right )+\left (-1-x \right ) {\mathrm e}^{1-\ln \left (x \right )} \left (2+2 x \right )^{5}+\left (1+x \right ) {\mathrm e}^{16} {\mathrm e}^{1-\ln \left (x \right )}}{\left (\left (x^{2}+x \right ) {\mathrm e}^{1-\ln \left (x \right )}-5 x^{2}-5 x \right ) \ln \left (-{\mathrm e}^{1-\ln \left (x \right )}+5\right )}d x\]
int(((5*x*exp(1-ln(x))-25*x)*(2+2*x)^5*ln(-exp(1-ln(x))+5)*ln(ln(-exp(1-ln (x))+5))+(-1-x)*exp(1-ln(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-ln(x)))/((x^2+x )*exp(1-ln(x))-5*x^2-5*x)/ln(-exp(1-ln(x))+5),x)
int(((5*x*exp(1-ln(x))-25*x)*(2+2*x)^5*ln(-exp(1-ln(x))+5)*ln(ln(-exp(1-ln (x))+5))+(-1-x)*exp(1-ln(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-ln(x)))/((x^2+x )*exp(1-ln(x))-5*x^2-5*x)/ln(-exp(1-ln(x))+5),x)
Time = 0.42 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx={\left (32 \, x^{5} + 160 \, x^{4} + 320 \, x^{3} + 320 \, x^{2} + 160 \, x - e^{16} + 32\right )} \log \left (\log \left (\frac {5 \, x - e}{x}\right )\right ) \]
integrate(((5*x*exp(1-log(x))-25*x)*(2+2*x)^5*log(-exp(1-log(x))+5)*log(lo g(-exp(1-log(x))+5))+(-1-x)*exp(1-log(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-lo g(x)))/((x^2+x)*exp(1-log(x))-5*x^2-5*x)/log(-exp(1-log(x))+5),x, algorith m=\
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (19) = 38\).
Time = 0.70 (sec) , antiderivative size = 110, normalized size of antiderivative = 4.58 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\left (32 x^{5} + 160 x^{4} + 320 x^{3} + 320 x^{2} + 160 x - \frac {32 e^{2}}{15} - \frac {16 e}{3} - \frac {48 e^{3}}{125} - \frac {64 e^{4}}{1875} - \frac {16 e^{5}}{13125}\right ) \log {\left (\log {\left (5 - \frac {e}{x} \right )} \right )} - \frac {\left (-420000 - 28000 e^{2} - 70000 e - 5040 e^{3} - 448 e^{4} - 16 e^{5} + 13125 e^{16}\right ) \log {\left (\log {\left (5 - \frac {e}{x} \right )} \right )}}{13125} \]
integrate(((5*x*exp(1-ln(x))-25*x)*(2+2*x)**5*ln(-exp(1-ln(x))+5)*ln(ln(-e xp(1-ln(x))+5))+(-1-x)*exp(1-ln(x))*(2+2*x)**5+(1+x)*exp(16)*exp(1-ln(x))) /((x**2+x)*exp(1-ln(x))-5*x**2-5*x)/ln(-exp(1-ln(x))+5),x)
(32*x**5 + 160*x**4 + 320*x**3 + 320*x**2 + 160*x - 32*exp(2)/15 - 16*E/3 - 48*exp(3)/125 - 64*exp(4)/1875 - 16*exp(5)/13125)*log(log(5 - E/x)) - (- 420000 - 28000*exp(2) - 70000*E - 5040*exp(3) - 448*exp(4) - 16*exp(5) + 1 3125*exp(16))*log(log(5 - E/x))/13125
Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx={\left (32 \, x^{5} + 160 \, x^{4} + 320 \, x^{3} + 320 \, x^{2} + 160 \, x - e^{16} + 32\right )} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) \]
integrate(((5*x*exp(1-log(x))-25*x)*(2+2*x)^5*log(-exp(1-log(x))+5)*log(lo g(-exp(1-log(x))+5))+(-1-x)*exp(1-log(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-lo g(x)))/((x^2+x)*exp(1-log(x))-5*x^2-5*x)/log(-exp(1-log(x))+5),x, algorith m=\
Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (26) = 52\).
Time = 0.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 5.62 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=32 \, x^{5} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 160 \, x^{4} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 320 \, x^{3} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 320 \, x^{2} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 160 \, x \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) - e^{16} \log \left (-\log \left (5 \, x - e\right ) + \log \left (x\right )\right ) + 32 \, \log \left (-\log \left (5 \, x - e\right ) + \log \left (x\right )\right ) \]
integrate(((5*x*exp(1-log(x))-25*x)*(2+2*x)^5*log(-exp(1-log(x))+5)*log(lo g(-exp(1-log(x))+5))+(-1-x)*exp(1-log(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-lo g(x)))/((x^2+x)*exp(1-log(x))-5*x^2-5*x)/log(-exp(1-log(x))+5),x, algorith m=\
32*x^5*log(log(5*x - e) - log(x)) + 160*x^4*log(log(5*x - e) - log(x)) + 3 20*x^3*log(log(5*x - e) - log(x)) + 320*x^2*log(log(5*x - e) - log(x)) + 1 60*x*log(log(5*x - e) - log(x)) - e^16*log(-log(5*x - e) + log(x)) + 32*lo g(-log(5*x - e) + log(x))
Time = 12.87 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\ln \left (\ln \left (5-\frac {\mathrm {e}}{x}\right )\right )\,\left (32\,x^5+160\,x^4+320\,x^3+320\,x^2+160\,x-{\mathrm {e}}^{16}+32\right ) \]