Integrand size = 57, antiderivative size = 21 \[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\log \left (x \log \left (x-x \left (1-\log \left (-\frac {x}{\log (16)}\right )\right )\right )\right ) \] Output:
ln(ln(x-x*(1-ln(-1/4*x/ln(2))))*x)
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\log \left (-\frac {x}{\log (16)}\right )+\log \left (\log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )\right ) \] Input:
Integrate[(1 + Log[-(x/Log[16])] + Log[-(x/Log[16])]*Log[x*Log[-(x/Log[16] )]])/(x*Log[-(x/Log[16])]*Log[x*Log[-(x/Log[16])]]),x]
Output:
Log[-(x/Log[16])] + Log[Log[x*Log[-(x/Log[16])]]]
Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {7236}
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.
\(\displaystyle \int \frac {\log \left (x \log \left (-\frac {x}{\log (16)}\right )\right ) \log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right )+1}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx\) |
\(\Big \downarrow \) 7236 |
\(\displaystyle \log \left (x \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )\right )\) |
Input:
Int[(1 + Log[-(x/Log[16])] + Log[-(x/Log[16])]*Log[x*Log[-(x/Log[16])]])/( x*Log[-(x/Log[16])]*Log[x*Log[-(x/Log[16])]]),x]
Output:
Log[x*Log[x*Log[-(x/Log[16])]]]
Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]], x] /; !FalseQ[q]]
Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76
method | result | size |
parallelrisch | \(\ln \left (x \right )+\ln \left (\ln \left (x \ln \left (-\frac {x}{4 \ln \left (2\right )}\right )\right )\right )\) | \(16\) |
default | \(\ln \left (x \right )+\ln \left (\ln \left (x \left (-2 \ln \left (2\right )-\ln \left (\ln \left (2\right )\right )+\ln \left (-x \right )\right )\right )\right )\) | \(22\) |
norman | \(\ln \left (-\frac {x}{4 \ln \left (2\right )}\right )+\ln \left (\ln \left (x \ln \left (-\frac {x}{4 \ln \left (2\right )}\right )\right )\right )\) | \(22\) |
parts | \(\ln \left (x \right )+\ln \left (\ln \left (x \left (-2 \ln \left (2\right )-\ln \left (\ln \left (2\right )\right )+\ln \left (-x \right )\right )\right )\right )\) | \(22\) |
Input:
int((ln(-1/4*x/ln(2))*ln(x*ln(-1/4*x/ln(2)))+ln(-1/4*x/ln(2))+1)/x/ln(-1/4 *x/ln(2))/ln(x*ln(-1/4*x/ln(2))),x,method=_RETURNVERBOSE)
Output:
ln(x)+ln(ln(x*ln(-1/4*x/ln(2))))
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\log \left (-\frac {x}{4 \, \log \left (2\right )}\right ) + \log \left (\log \left (x \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )\right )\right ) \] Input:
integrate((log(-1/4*x/log(2))*log(x*log(-1/4*x/log(2)))+log(-1/4*x/log(2)) +1)/x/log(-1/4*x/log(2))/log(x*log(-1/4*x/log(2))),x, algorithm="fricas")
Output:
log(-1/4*x/log(2)) + log(log(x*log(-1/4*x/log(2))))
Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x \log {\left (- \frac {x}{4 \log {\left (2 \right )}} \right )} \right )} \right )} \] Input:
integrate((ln(-1/4*x/ln(2))*ln(x*ln(-1/4*x/ln(2)))+ln(-1/4*x/ln(2))+1)/x/l n(-1/4*x/ln(2))/ln(x*ln(-1/4*x/ln(2))),x)
Output:
log(x) + log(log(x*log(-x/(4*log(2)))))
\[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\int { \frac {\log \left (x \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )\right ) \log \left (-\frac {x}{4 \, \log \left (2\right )}\right ) + \log \left (-\frac {x}{4 \, \log \left (2\right )}\right ) + 1}{x \log \left (x \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )\right ) \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )} \,d x } \] Input:
integrate((log(-1/4*x/log(2))*log(x*log(-1/4*x/log(2)))+log(-1/4*x/log(2)) +1)/x/log(-1/4*x/log(2))/log(x*log(-1/4*x/log(2))),x, algorithm="maxima")
Output:
integrate((2*log(2) - log(x) + log(-log(2)) - 1)/(x*(2*log(2) + log(-log(2 )))*log(x) - x*log(x)^2 + (x*(2*log(2) + log(-log(2))) - x*log(x))*log(-2* log(2) + log(-x) - log(log(2)))), x) + log(x)
\[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\int { \frac {\log \left (x \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )\right ) \log \left (-\frac {x}{4 \, \log \left (2\right )}\right ) + \log \left (-\frac {x}{4 \, \log \left (2\right )}\right ) + 1}{x \log \left (x \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )\right ) \log \left (-\frac {x}{4 \, \log \left (2\right )}\right )} \,d x } \] Input:
integrate((log(-1/4*x/log(2))*log(x*log(-1/4*x/log(2)))+log(-1/4*x/log(2)) +1)/x/log(-1/4*x/log(2))/log(x*log(-1/4*x/log(2))),x, algorithm="giac")
Output:
integrate((log(x*log(-1/4*x/log(2)))*log(-1/4*x/log(2)) + log(-1/4*x/log(2 )) + 1)/(x*log(x*log(-1/4*x/log(2)))*log(-1/4*x/log(2))), x)
Time = 3.44 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\ln \left (\ln \left (x\,\ln \left (-\frac {x}{4}\right )-x\,\ln \left (\ln \left (2\right )\right )\right )\right )+\ln \left (x\right ) \] Input:
int((log(-x/(4*log(2))) + log(-x/(4*log(2)))*log(x*log(-x/(4*log(2)))) + 1 )/(x*log(-x/(4*log(2)))*log(x*log(-x/(4*log(2))))),x)
Output:
log(log(x*log(-x/4) - x*log(log(2)))) + log(x)
Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1+\log \left (-\frac {x}{\log (16)}\right )+\log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )}{x \log \left (-\frac {x}{\log (16)}\right ) \log \left (x \log \left (-\frac {x}{\log (16)}\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (-\frac {x}{4 \,\mathrm {log}\left (2\right )}\right ) x \right )\right )+\mathrm {log}\left (x \right ) \] Input:
int((log(-1/4*x/log(2))*log(x*log(-1/4*x/log(2)))+log(-1/4*x/log(2))+1)/x/ log(-1/4*x/log(2))/log(x*log(-1/4*x/log(2))),x)
Output:
log(log(log(( - x)/(4*log(2)))*x)) + log(x)