Integrand size = 79, antiderivative size = 25 \[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=\log \left (-x+\frac {25 \log ^2(3)}{e^8 \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )}\right ) \]
Time = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=-2 \log \left (\log \left (5 \log \left (-\frac {x}{2}\right )\right )\right )+\log \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right ) \]
Integrate[(50*Log[3]^2 + E^8*x*Log[-1/2*x]*Log[5*Log[-1/2*x]]^3)/(-25*x*Lo g[3]^2*Log[-1/2*x]*Log[5*Log[-1/2*x]] + E^8*x^2*Log[-1/2*x]*Log[5*Log[-1/2 *x]]^3),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.
| \(\displaystyle \int \frac {e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )+50 \log ^2(3)}{e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )-50 \log ^2(3)}{x \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right ) \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {1}{x}+\frac {25 \log ^2(3) \log \left (-\frac {x}{2}\right )+2 e^8 x \log \left (5 \log \left (-\frac {x}{2}\right )\right )}{x \log \left (-\frac {x}{2}\right ) \left (e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )-25 \log ^2(3)\right )}-\frac {2}{x \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 e^8 \int \frac {\log \left (5 \log \left (-\frac {x}{2}\right )\right )}{\log \left (-\frac {x}{2}\right ) \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )}dx+25 \log ^2(3) \int \frac {1}{x \left (e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )-25 \log ^2(3)\right )}dx+\log (x)-2 \log \left (\log \left (5 \log \left (-\frac {x}{2}\right )\right )\right )\) |
Int[(50*Log[3]^2 + E^8*x*Log[-1/2*x]*Log[5*Log[-1/2*x]]^3)/(-25*x*Log[3]^2 *Log[-1/2*x]*Log[5*Log[-1/2*x]] + E^8*x^2*Log[-1/2*x]*Log[5*Log[-1/2*x]]^3 ),x]
3.14.17.3.1 Defintions of rubi rules used
Time = 3.92 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
| method | result | size |
| risch | \(\ln \left (x \right )-2 \ln \left (\ln \left (5 \ln \left (-\frac {x}{2}\right )\right )\right )+\ln \left (\ln \left (5 \ln \left (-\frac {x}{2}\right )\right )^{2}-\frac {25 \ln \left (3\right )^{2} {\mathrm e}^{-8}}{x}\right )\) | \(36\) |
| parallelrisch | \(-2 \ln \left (\ln \left (5 \ln \left (-\frac {x}{2}\right )\right )\right )+\ln \left (\left (x \,{\mathrm e}^{8} \ln \left (5 \ln \left (-\frac {x}{2}\right )\right )^{2}-25 \ln \left (3\right )^{2}\right ) {\mathrm e}^{-8}\right )\) | \(40\) |
int((x*exp(4)^2*ln(-1/2*x)*ln(5*ln(-1/2*x))^3+50*ln(3)^2)/(x^2*exp(4)^2*ln (-1/2*x)*ln(5*ln(-1/2*x))^3-25*x*ln(3)^2*ln(-1/2*x)*ln(5*ln(-1/2*x))),x,me thod=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=\log \left (-\frac {1}{2} \, x\right ) + \log \left (\frac {x e^{8} \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )^{2} - 25 \, \log \left (3\right )^{2}}{x}\right ) - 2 \, \log \left (\log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )\right ) \]
integrate((x*exp(4)^2*log(-1/2*x)*log(5*log(-1/2*x))^3+50*log(3)^2)/(x^2*e xp(4)^2*log(-1/2*x)*log(5*log(-1/2*x))^3-25*x*log(3)^2*log(-1/2*x)*log(5*l og(-1/2*x))),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (5 \log {\left (- \frac {x}{2} \right )} \right )}^{2} - \frac {25 \log {\left (3 \right )}^{2}}{x e^{8}} \right )} - 2 \log {\left (\log {\left (5 \log {\left (- \frac {x}{2} \right )} \right )} \right )} \]
integrate((x*exp(4)**2*ln(-1/2*x)*ln(5*ln(-1/2*x))**3+50*ln(3)**2)/(x**2*e xp(4)**2*ln(-1/2*x)*ln(5*ln(-1/2*x))**3-25*x*ln(3)**2*ln(-1/2*x)*ln(5*ln(- 1/2*x))),x)
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (22) = 44\).
Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.96 \[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=\log \left (x\right ) + \log \left (\frac {{\left (x e^{8} \log \left (5\right )^{2} + 2 \, x e^{8} \log \left (5\right ) \log \left (-\log \left (2\right ) + \log \left (-x\right )\right ) + x e^{8} \log \left (-\log \left (2\right ) + \log \left (-x\right )\right )^{2} - 25 \, \log \left (3\right )^{2}\right )} e^{\left (-8\right )}}{x}\right ) - 2 \, \log \left (\log \left (5\right ) + \log \left (-\log \left (2\right ) + \log \left (-x\right )\right )\right ) \]
integrate((x*exp(4)^2*log(-1/2*x)*log(5*log(-1/2*x))^3+50*log(3)^2)/(x^2*e xp(4)^2*log(-1/2*x)*log(5*log(-1/2*x))^3-25*x*log(3)^2*log(-1/2*x)*log(5*l og(-1/2*x))),x, algorithm=\
log(x) + log((x*e^8*log(5)^2 + 2*x*e^8*log(5)*log(-log(2) + log(-x)) + x*e ^8*log(-log(2) + log(-x))^2 - 25*log(3)^2)*e^(-8)/x) - 2*log(log(5) + log( -log(2) + log(-x)))
\[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=\int { \frac {x e^{8} \log \left (-\frac {1}{2} \, x\right ) \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )^{3} + 50 \, \log \left (3\right )^{2}}{x^{2} e^{8} \log \left (-\frac {1}{2} \, x\right ) \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )^{3} - 25 \, x \log \left (3\right )^{2} \log \left (-\frac {1}{2} \, x\right ) \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )} \,d x } \]
integrate((x*exp(4)^2*log(-1/2*x)*log(5*log(-1/2*x))^3+50*log(3)^2)/(x^2*e xp(4)^2*log(-1/2*x)*log(5*log(-1/2*x))^3-25*x*log(3)^2*log(-1/2*x)*log(5*l og(-1/2*x))),x, algorithm=\
integrate((x*e^8*log(-1/2*x)*log(5*log(-1/2*x))^3 + 50*log(3)^2)/(x^2*e^8* log(-1/2*x)*log(5*log(-1/2*x))^3 - 25*x*log(3)^2*log(-1/2*x)*log(5*log(-1/ 2*x))), x)
Time = 8.90 (sec) , antiderivative size = 138, normalized size of antiderivative = 5.52 \[ \int \frac {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx=2\,\ln \left (\frac {4\,x\,{\mathrm {e}}^8-25\,{\ln \left (-\frac {x}{2}\right )}^2\,{\ln \left (3\right )}^2}{x}\right )+\ln \left (16\,x\,\ln \left (-\frac {x}{2}\right )\,{\mathrm {e}}^{16}-400\,\ln \left (-\frac {x}{2}\right )\,{\mathrm {e}}^8\,{\ln \left (3\right )}^2\right )-2\,\ln \left (5625\,\ln \left (5\,\ln \left (-\frac {x}{2}\right )\right )\,{\ln \left (-\frac {x}{2}\right )}^2\,{\ln \left (3\right )}^4-900\,x\,\ln \left (5\,\ln \left (-\frac {x}{2}\right )\right )\,{\mathrm {e}}^8\,{\ln \left (3\right )}^2\right )-\ln \left (x-25\,{\mathrm {e}}^{-8}\,{\ln \left (3\right )}^2\right )+4\,\ln \left (x\right )+\ln \left (\frac {25\,{\ln \left (3\right )}^2-x\,{\mathrm {e}}^8\,{\left (\ln \left (5\right )+\ln \left (\ln \left (-\frac {x}{2}\right )\right )\right )}^2}{x^2\,\ln \left (-\frac {x}{2}\right )}\right ) \]
int((50*log(3)^2 + x*log(5*log(-x/2))^3*log(-x/2)*exp(8))/(x^2*log(5*log(- x/2))^3*log(-x/2)*exp(8) - 25*x*log(5*log(-x/2))*log(-x/2)*log(3)^2),x)
2*log((4*x*exp(8) - 25*log(-x/2)^2*log(3)^2)/x) + log(16*x*log(-x/2)*exp(1 6) - 400*log(-x/2)*exp(8)*log(3)^2) - 2*log(5625*log(5*log(-x/2))*log(-x/2 )^2*log(3)^4 - 900*x*log(5*log(-x/2))*exp(8)*log(3)^2) - log(x - 25*exp(-8 )*log(3)^2) + 4*log(x) + log((25*log(3)^2 - x*exp(8)*(log(5) + log(log(-x/ 2)))^2)/(x^2*log(-x/2)))