Integrand size = 39, antiderivative size = 26 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (\frac {1}{2} \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right )\right ) \] Output:
ln(1/2*ln(1/10*x^3/exp(4)/exp(x)/ln(x))+1/2*x)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (x+\log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )\right ) \] Input:
Integrate[(-1 + 3*Log[x])/(x^2*Log[x] + x*Log[x]*Log[(E^(-4 - x)*x^3)/(10* Log[x])]),x]
Output:
Log[x + Log[(E^(-4 - x)*x^3)/(10*Log[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.
\(\displaystyle \int \frac {3 \log (x)-1}{x \log (x) \log \left (\frac {e^{-x-4} x^3}{10 \log (x)}\right )+x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 \log (x)-1}{x \log (x) \left (\log \left (\frac {e^{-x-4} x^3}{10 \log (x)}\right )+x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3}{x \left (\log \left (\frac {e^{-x-4} x^3}{10 \log (x)}\right )+x\right )}-\frac {1}{x \log (x) \left (\log \left (\frac {e^{-x-4} x^3}{10 \log (x)}\right )+x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {1}{x \left (x+\log \left (\frac {e^{-x-4} x^3}{10 \log (x)}\right )\right )}dx-\int \frac {1}{x \log (x) \left (x+\log \left (\frac {e^{-x-4} x^3}{10 \log (x)}\right )\right )}dx\) |
Input:
Int[(-1 + 3*Log[x])/(x^2*Log[x] + x*Log[x]*Log[(E^(-4 - x)*x^3)/(10*Log[x] )]),x]
Output:
$Aborted
Time = 0.95 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
default | \(\ln \left (\ln \left (\frac {x^{3} {\mathrm e}^{-4} {\mathrm e}^{-x}}{10 \ln \left (x \right )}\right )+x \right )\) | \(22\) |
parallelrisch | \(\ln \left (\ln \left (\frac {x^{3} {\mathrm e}^{-4} {\mathrm e}^{-x}}{10 \ln \left (x \right )}\right )+x \right )\) | \(22\) |
risch | \(\ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {i \left (2 i x -2 i \ln \left (2\right )+6 i \ln \left (x \right )-2 i \ln \left (\ln \left (x \right )\right )-8 i+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )^{3}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right )+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )+\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{3} {\mathrm e}^{-x}}{\ln \left (x \right )}\right )-2 i \ln \left (5\right )\right )}{2}\right )\) | \(359\) |
Input:
int((3*ln(x)-1)/(x*ln(x)*ln(1/10*x^3/exp(4)/exp(x)/ln(x))+x^2*ln(x)),x,met hod=_RETURNVERBOSE)
Output:
ln(ln(1/10*x^3/exp(4)/exp(x)/ln(x))+x)
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (x + \log \left (\frac {x^{3} e^{\left (-x - 4\right )}}{10 \, \log \left (x\right )}\right )\right ) \] Input:
integrate((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*lo g(x)),x, algorithm="fricas")
Output:
log(x + log(1/10*x^3*e^(-x - 4)/log(x)))
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log {\left (x + \log {\left (\frac {x^{3} e^{- x}}{10 e^{4} \log {\left (x \right )}} \right )} \right )} \] Input:
integrate((3*ln(x)-1)/(x*ln(x)*ln(1/10*x**3/exp(4)/exp(x)/ln(x))+x**2*ln(x )),x)
Output:
log(x + log(x**3*exp(-4)*exp(-x)/(10*log(x))))
Time = 0.16 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.54 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (\log \left (5\right ) + \log \left (2\right ) - 3 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) + 4\right ) \] Input:
integrate((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*lo g(x)),x, algorithm="maxima")
Output:
log(log(5) + log(2) - 3*log(x) + log(log(x)) + 4)
Time = 0.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.46 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\log \left (\log \left (10\right ) - 3 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) + 4\right ) \] Input:
integrate((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*lo g(x)),x, algorithm="giac")
Output:
log(log(10) - 3*log(x) + log(log(x)) + 4)
Time = 4.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.50 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\ln \left (\ln \left (\frac {x^3}{10\,\ln \left (x\right )}\right )-4\right ) \] Input:
int((3*log(x) - 1)/(x^2*log(x) + x*log((x^3*exp(-x)*exp(-4))/(10*log(x)))* log(x)),x)
Output:
log(log(x^3/(10*log(x))) - 4)
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-1+3 \log (x)}{x^2 \log (x)+x \log (x) \log \left (\frac {e^{-4-x} x^3}{10 \log (x)}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {x^{3}}{10 e^{x} \mathrm {log}\left (x \right ) e^{4}}\right )+x \right ) \] Input:
int((3*log(x)-1)/(x*log(x)*log(1/10*x^3/exp(4)/exp(x)/log(x))+x^2*log(x)), x)
Output:
log(log(x**3/(10*e**x*log(x)*e**4)) + x)