Integrand size = 113, antiderivative size = 32 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {6+\frac {1}{3} e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{2 x} \] Output:
1/2*(6+1/3*ln(ln(x*ln(1/2*x)/(ln(2)-4)))*exp(x))/x
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {18+e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x} \] Input:
Integrate[(E^x + E^x*Log[x/2] - 18*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2]) ] + E^x*(-1 + x)*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]*Log[Log[(x*Log[x /2])/(-4 + Log[2])]])/(6*x^2*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]),x]
Output:
(18 + E^x*Log[Log[(x*Log[x/2])/(-4 + Log[2])]])/(6*x)
Time = 1.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {27, 7293, 2009}
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.
\(\displaystyle \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right ) \log \left (\frac {x}{2}\right )+e^x (x-1) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right )\right ) \log \left (\frac {x}{2}\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {e^x \log \left (\frac {x}{2}\right )-18 \log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right ) \log \left (\frac {x}{2}\right )-e^x (1-x) \log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right ) \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right ) \log \left (\frac {x}{2}\right )+e^x}{x^2 \log \left (\frac {x}{2}\right ) \log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{6} \int \left (\frac {e^x \left (x \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right ) \log \left (\frac {x}{2}\right )-\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right ) \log \left (\frac {x}{2}\right )+\log \left (\frac {x}{2}\right )+1\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )}-\frac {18}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (\frac {18}{x}+\frac {e^x \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right )}{x}\right )\) |
Input:
Int[(E^x + E^x*Log[x/2] - 18*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])] + E^ x*(-1 + x)*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]*Log[Log[(x*Log[x/2])/( -4 + Log[2])]])/(6*x^2*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]),x]
Output:
(18/x + (E^x*Log[Log[-((x*Log[x/2])/(4 - Log[2]))]])/x)/6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
parallelrisch | \(\frac {18+\ln \left (\ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\right ) {\mathrm e}^{x}}{6 x}\) | \(25\) |
Input:
int(1/6*((-1+x)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*ln(1/2* x)/(ln(2)-4)))-18*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))+exp(x)*ln(1/2*x)+exp (x))/x^2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x,method=_RETURNVERBOSE)
Output:
1/6/x*(18+ln(ln(x*ln(1/2*x)/(ln(2)-4)))*exp(x))
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (\frac {x \log \left (\frac {1}{2} \, x\right )}{\log \left (2\right ) - 4}\right )\right ) + 18}{6 \, x} \] Input:
integrate(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(l og(x*log(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+ex p(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, alg orithm="fricas")
Output:
1/6*(e^x*log(log(x*log(1/2*x)/(log(2) - 4))) + 18)/x
Time = 0.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log {\left (\log {\left (\frac {x \log {\left (\frac {x}{2} \right )}}{-4 + \log {\left (2 \right )}} \right )} \right )}}{6 x} + \frac {3}{x} \] Input:
integrate(1/6*((-1+x)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*l n(1/2*x)/(ln(2)-4)))-18*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))+exp(x)*ln(1/2* x)+exp(x))/x**2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x)
Output:
exp(x)*log(log(x*log(x/2)/(-4 + log(2))))/(6*x) + 3/x
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (x\right ) - \log \left (\log \left (2\right ) - 4\right ) + \log \left (-\log \left (2\right ) + \log \left (x\right )\right )\right )}{6 \, x} + \frac {3}{x} \] Input:
integrate(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(l og(x*log(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+ex p(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, alg orithm="maxima")
Output:
1/6*e^x*log(log(x) - log(log(2) - 4) + log(-log(2) + log(x)))/x + 3/x
Time = 0.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (x\right ) - \log \left (\log \left (2\right ) - 4\right ) + \log \left (-\log \left (2\right ) + \log \left (x\right )\right )\right ) + 18}{6 \, x} \] Input:
integrate(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(l og(x*log(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+ex p(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, alg orithm="giac")
Output:
1/6*(e^x*log(log(x) - log(log(2) - 4) + log(-log(2) + log(x))) + 18)/x
Timed out. \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\int \frac {\frac {{\mathrm {e}}^x}{6}-3\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )+\frac {\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x}{6}+\frac {\ln \left (\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\right )\,\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\,\left (x-1\right )}{6}}{x^2\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )} \,d x \] Input:
int((exp(x)/6 - 3*log(x/2)*log((x*log(x/2))/(log(2) - 4)) + (log(x/2)*exp( x))/6 + (log(log((x*log(x/2))/(log(2) - 4)))*log(x/2)*exp(x)*log((x*log(x/ 2))/(log(2) - 4))*(x - 1))/6)/(x^2*log(x/2)*log((x*log(x/2))/(log(2) - 4)) ),x)
Output:
int((exp(x)/6 - 3*log(x/2)*log((x*log(x/2))/(log(2) - 4)) + (log(x/2)*exp( x))/6 + (log(log((x*log(x/2))/(log(2) - 4)))*log(x/2)*exp(x)*log((x*log(x/ 2))/(log(2) - 4))*(x - 1))/6)/(x^2*log(x/2)*log((x*log(x/2))/(log(2) - 4)) ), x)
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{2}\right ) x}{\mathrm {log}\left (2\right )-4}\right )\right )+18}{6 x} \] Input:
int(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(log(x*l og(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+exp(x)*l og(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x)
Output:
(e**x*log(log((log(x/2)*x)/(log(2) - 4))) + 18)/(6*x)