Integrand size = 58, antiderivative size = 35 \[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \] Output:
exp(5*x*ln(1/4/x)/ln(-4*ln(5))^2)^2
\[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=\int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx \] Input:
Integrate[((x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2)*(-10 + 10*Log[1/(4*x )]))/(4^((10*x)/(I*Pi + Log[Log[625]])^2)*(I*Pi + Log[Log[625]])^2),x]
Output:
Integrate[((x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2)*(-10 + 10*Log[1/(4*x )]))/4^((10*x)/(I*Pi + Log[Log[625]])^2), x]/(I*Pi + Log[Log[625]])^2
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.
| \(\displaystyle \int \frac {4^{-\frac {10 x}{(\log (\log (625))+i \pi )^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(\log (\log (625))+i \pi )^2}} \left (10 \log \left (\frac {1}{4 x}\right )-10\right )}{(\log (\log (625))+i \pi )^2} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -5 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (1-\log \left (\frac {1}{4 x}\right )\right )dx}{(\log (\log (625))+i \pi )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {5 \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (1-\log \left (\frac {1}{4 x}\right )\right )dx}{(\log (\log (625))+i \pi )^2}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {5 \int \left (2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} (1+\log (4))-2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \log \left (\frac {1}{x}\right )\right )dx}{(\log (\log (625))+i \pi )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {5 \left (-\log \left (\frac {1}{x}\right ) \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}}dx+(1+\log (4)) \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}}dx-\int \frac {\int 2^{\frac {20 x}{(\pi -i \log (\log (625)))^2}+1} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}}dx}{x}dx\right )}{(\log (\log (625))+i \pi )^2}\) |
Input:
Int[((x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2)*(-10 + 10*Log[1/(4*x)]))/( 4^((10*x)/(I*Pi + Log[Log[625]])^2)*(I*Pi + Log[Log[625]])^2),x]
Output:
$Aborted
Time = 0.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.57
| method | result | size |
| default | \({\mathrm e}^{\frac {10 x \ln \left (\frac {1}{4 x}\right )}{\ln \left (-4 \ln \left (5\right )\right )^{2}}}\) | \(20\) |
| parallelrisch | \({\mathrm e}^{\frac {10 x \ln \left (\frac {1}{4 x}\right )}{\ln \left (-4 \ln \left (5\right )\right )^{2}}}\) | \(20\) |
| norman | \(-\frac {\left (-2 i \ln \left (\ln \left (5\right )\right ) \pi -4 i \ln \left (2\right ) \pi +\pi ^{2}-\ln \left (\ln \left (5\right )\right )^{2}-4 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right )-4 \ln \left (2\right )^{2}\right ) {\mathrm e}^{\frac {10 x \ln \left (\frac {1}{4 x}\right )}{\ln \left (-4 \ln \left (5\right )\right )^{2}}}}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right ) \ln \left (-4 \ln \left (5\right )\right )}\) | \(80\) |
| risch | \(\frac {\left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \left (5\right )}\right )\right )^{2}}} \ln \left (\ln \left (5\right )\right )^{2}}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right )^{2}}+\frac {4 \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \left (5\right )}\right )\right )^{2}}} \ln \left (\ln \left (5\right )\right ) \ln \left (2\right )}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right )^{2}}+\frac {4 \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \left (5\right )}\right )\right )^{2}}} \ln \left (2\right )^{2}}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right )^{2}}+\frac {2 i \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \left (5\right )}\right )\right )^{2}}} \ln \left (\ln \left (5\right )\right ) \pi }{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right )^{2}}+\frac {4 i \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \left (5\right )}\right )\right )^{2}}} \ln \left (2\right ) \pi }{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right )^{2}}-\frac {\left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \left (5\right )}\right )\right )^{2}}} \pi ^{2}}{\left (2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+i \pi \right )^{2}}\) | \(273\) |
Input:
int((10*ln(1/4/x)-10)*exp(5*x*ln(1/4/x)/ln(-4*ln(5))^2)^2/ln(-4*ln(5))^2,x ,method=_RETURNVERBOSE)
Output:
exp(5*x*ln(1/4/x)/ln(-4*ln(5))^2)^2
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.46 \[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=\left (\frac {1}{4 \, x}\right )^{\frac {10 \, x}{\log \left (-4 \, \log \left (5\right )\right )^{2}}} \] Input:
integrate((10*log(1/4/x)-10)*exp(5*x*log(1/4/x)/log(-4*log(5))^2)^2/log(-4 *log(5))^2,x, algorithm="fricas")
Output:
(1/4/x)^(10*x/log(-4*log(5))^2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (22) = 44\).
Time = 10.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.11 \[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=e^{- \frac {20 x \log {\left (2 \right )}}{- \pi ^{2} + \log {\left (\log {\left (5 \right )} \right )}^{2} + 4 \log {\left (2 \right )} \log {\left (\log {\left (5 \right )} \right )} + 4 \log {\left (2 \right )}^{2} + 2 i \pi \log {\left (\log {\left (5 \right )} \right )} + 4 i \pi \log {\left (2 \right )}}} e^{\frac {10 x \log {\left (\frac {1}{x} \right )}}{- \pi ^{2} + \log {\left (\log {\left (5 \right )} \right )}^{2} + 4 \log {\left (2 \right )} \log {\left (\log {\left (5 \right )} \right )} + 4 \log {\left (2 \right )}^{2} + 2 i \pi \log {\left (\log {\left (5 \right )} \right )} + 4 i \pi \log {\left (2 \right )}}} \] Input:
integrate((10*ln(1/4/x)-10)*exp(5*x*ln(1/4/x)/ln(-4*ln(5))**2)**2/ln(-4*ln (5))**2,x)
Output:
exp(-20*x*log(2)/(-pi**2 + log(log(5))**2 + 4*log(2)*log(log(5)) + 4*log(2 )**2 + 2*I*pi*log(log(5)) + 4*I*pi*log(2)))*exp(10*x*log(1/x)/(-pi**2 + lo g(log(5))**2 + 4*log(2)*log(log(5)) + 4*log(2)**2 + 2*I*pi*log(log(5)) + 4 *I*pi*log(2)))
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (16) = 32\).
Time = 0.16 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.66 \[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=\frac {{\left (4 \, \log \left (2\right )^{2} + 4 \, \log \left (2\right ) \log \left (-\log \left (5\right )\right ) + \log \left (-\log \left (5\right )\right )^{2}\right )} e^{\left (-\frac {20 \, x \log \left (2\right )}{4 \, \log \left (2\right )^{2} + 4 \, \log \left (2\right ) \log \left (-\log \left (5\right )\right ) + \log \left (-\log \left (5\right )\right )^{2}} - \frac {10 \, x \log \left (x\right )}{4 \, \log \left (2\right )^{2} + 4 \, \log \left (2\right ) \log \left (-\log \left (5\right )\right ) + \log \left (-\log \left (5\right )\right )^{2}}\right )}}{\log \left (-4 \, \log \left (5\right )\right )^{2}} \] Input:
integrate((10*log(1/4/x)-10)*exp(5*x*log(1/4/x)/log(-4*log(5))^2)^2/log(-4 *log(5))^2,x, algorithm="maxima")
Output:
(4*log(2)^2 + 4*log(2)*log(-log(5)) + log(-log(5))^2)*e^(-20*x*log(2)/(4*l og(2)^2 + 4*log(2)*log(-log(5)) + log(-log(5))^2) - 10*x*log(x)/(4*log(2)^ 2 + 4*log(2)*log(-log(5)) + log(-log(5))^2))/log(-4*log(5))^2
\[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=\int { \frac {10 \, \left (\frac {1}{4 \, x}\right )^{\frac {10 \, x}{\log \left (-4 \, \log \left (5\right )\right )^{2}}} {\left (\log \left (\frac {1}{4 \, x}\right ) - 1\right )}}{\log \left (-4 \, \log \left (5\right )\right )^{2}} \,d x } \] Input:
integrate((10*log(1/4/x)-10)*exp(5*x*log(1/4/x)/log(-4*log(5))^2)^2/log(-4 *log(5))^2,x, algorithm="giac")
Output:
integrate(10*(1/4/x)^(10*x/log(-4*log(5))^2)*(log(1/4/x) - 1)/log(-4*log(5 ))^2, x)
Time = 3.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.66 \[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx={\mathrm {e}}^{-\frac {20\,x\,\ln \left (2\right )}{{\ln \left (\ln \left (5\right )\right )}^2+4\,\ln \left (2\right )\,\ln \left (\ln \left (5\right )\right )-\pi ^2+4\,{\ln \left (2\right )}^2+\pi \,\ln \left (2\right )\,4{}\mathrm {i}+\pi \,\ln \left (\ln \left (5\right )\right )\,2{}\mathrm {i}}}\,{\mathrm {e}}^{\frac {10\,x\,\ln \left (\frac {1}{x}\right )}{{\ln \left (\ln \left (5\right )\right )}^2+4\,\ln \left (2\right )\,\ln \left (\ln \left (5\right )\right )-\pi ^2+4\,{\ln \left (2\right )}^2+\pi \,\ln \left (2\right )\,4{}\mathrm {i}+\pi \,\ln \left (\ln \left (5\right )\right )\,2{}\mathrm {i}}} \] Input:
int((exp((10*x*log(1/(4*x)))/log(-4*log(5))^2)*(10*log(1/(4*x)) - 10))/log (-4*log(5))^2,x)
Output:
exp(-(20*x*log(2))/(pi*log(2)*4i + log(log(5))^2 + pi*log(log(5))*2i + 4*l og(2)*log(log(5)) - pi^2 + 4*log(2)^2))*exp((10*x*log(1/x))/(pi*log(2)*4i + log(log(5))^2 + pi*log(log(5))*2i + 4*log(2)*log(log(5)) - pi^2 + 4*log( 2)^2))
Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.51 \[ \int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx=\frac {1}{e^{\frac {10 \,\mathrm {log}\left (4 x \right ) x}{\mathrm {log}\left (-4 \,\mathrm {log}\left (5\right )\right )^{2}}}} \] Input:
int((10*log(1/4/x)-10)*exp(5*x*log(1/4/x)/log(-4*log(5))^2)^2/log(-4*log(5 ))^2,x)
Output:
1/e**((10*log(4*x)*x)/log( - 4*log(5))**2)