Integrand size = 114, antiderivative size = 34 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\log \left (\frac {x}{\left (4-e^{2-\frac {3}{\log (x)}}-x\right ) \left (3-\frac {1}{5 \log ^2(x)}\right )}\right ) \]
Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\frac {3}{\log (x)}+\log (x)-\log \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right )+2 \log (\log (x))-\log \left (1-15 \log ^2(x)\right ) \]
Integrate[((8 - 2*x)*Log[x] + 4*Log[x]^2 - 60*Log[x]^4 + E^((-3 + 2*Log[x] )/Log[x])*(3 - 2*Log[x] - 46*Log[x]^2 + 15*Log[x]^4))/((4*x - x^2)*Log[x]^ 2 + (-60*x + 15*x^2)*Log[x]^4 + E^((-3 + 2*Log[x])/Log[x])*(-(x*Log[x]^2) + 15*x*Log[x]^4)),x]
3/Log[x] + Log[x] - Log[E^2 - 4*E^(3/Log[x]) + E^(3/Log[x])*x] + 2*Log[Log [x]] - Log[1 - 15*Log[x]^2]
Time = 4.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {7292, 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.
\(\displaystyle \int \frac {-60 \log ^4(x)+4 \log ^2(x)+e^{\frac {2 \log (x)-3}{\log (x)}} \left (15 \log ^4(x)-46 \log ^2(x)-2 \log (x)+3\right )+(8-2 x) \log (x)}{\left (15 x^2-60 x\right ) \log ^4(x)+\left (4 x-x^2\right ) \log ^2(x)+e^{\frac {2 \log (x)-3}{\log (x)}} \left (15 x \log ^4(x)-x \log ^2(x)\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {3}{\log (x)}} \left (60 \log ^4(x)-4 \log ^2(x)-e^{\frac {2 \log (x)-3}{\log (x)}} \left (15 \log ^4(x)-46 \log ^2(x)-2 \log (x)+3\right )-(8-2 x) \log (x)\right )}{x \left (x e^{\frac {3}{\log (x)}}-4 e^{\frac {3}{\log (x)}}+e^2\right ) \log ^2(x) \left (1-15 \log ^2(x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {15 \log ^4(x)-46 \log ^2(x)-2 \log (x)+3}{x \log ^2(x) \left (15 \log ^2(x)-1\right )}-\frac {e^{\frac {3}{\log (x)}} \left (-3 x+x \log ^2(x)+12\right )}{x \left (x e^{\frac {3}{\log (x)}}-4 e^{\frac {3}{\log (x)}}+e^2\right ) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log \left (1-15 \log ^2(x)\right )+\log (x)-\log \left (x e^{\frac {3}{\log (x)}}-4 e^{\frac {3}{\log (x)}}+e^2\right )+2 \log (\log (x))+\frac {3}{\log (x)}\) |
Int[((8 - 2*x)*Log[x] + 4*Log[x]^2 - 60*Log[x]^4 + E^((-3 + 2*Log[x])/Log[ x])*(3 - 2*Log[x] - 46*Log[x]^2 + 15*Log[x]^4))/((4*x - x^2)*Log[x]^2 + (- 60*x + 15*x^2)*Log[x]^4 + E^((-3 + 2*Log[x])/Log[x])*(-(x*Log[x]^2) + 15*x *Log[x]^4)),x]
3/Log[x] + Log[x] - Log[E^2 - 4*E^(3/Log[x]) + E^(3/Log[x])*x] + 2*Log[Log [x]] - Log[1 - 15*Log[x]^2]
3.1.40.3.1 Defintions of rubi rules used
Time = 0.86 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(2 \ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (x \right )^{2}-\frac {1}{15}\right )-\ln \left (x +{\mathrm e}^{\frac {2 \ln \left (x \right )-3}{\ln \left (x \right )}}-4\right )+\ln \left (x \right )\) | \(36\) |
risch | \(\ln \left (x \right )+\frac {3}{\ln \left (x \right )}-\ln \left (\ln \left (x \right )^{2}-\frac {1}{15}\right )+2 \ln \left (\ln \left (x \right )\right )+\frac {2 \ln \left (x \right )-3}{\ln \left (x \right )}-\ln \left (x +{\mathrm e}^{\frac {2 \ln \left (x \right )-3}{\ln \left (x \right )}}-4\right )\) | \(53\) |
int(((15*ln(x)^4-46*ln(x)^2-2*ln(x)+3)*exp((2*ln(x)-3)/ln(x))-60*ln(x)^4+4 *ln(x)^2+(-2*x+8)*ln(x))/((15*x*ln(x)^4-x*ln(x)^2)*exp((2*ln(x)-3)/ln(x))+ (15*x^2-60*x)*ln(x)^4+(-x^2+4*x)*ln(x)^2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=-\log \left (15 \, \log \left (x\right )^{2} - 1\right ) - \log \left (x + e^{\left (\frac {2 \, \log \left (x\right ) - 3}{\log \left (x\right )}\right )} - 4\right ) + \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]
integrate(((15*log(x)^4-46*log(x)^2-2*log(x)+3)*exp((2*log(x)-3)/log(x))-6 0*log(x)^4+4*log(x)^2+(-2*x+8)*log(x))/((15*x*log(x)^4-x*log(x)^2)*exp((2* log(x)-3)/log(x))+(15*x^2-60*x)*log(x)^4+(-x^2+4*x)*log(x)^2),x, algorithm =\
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\log {\left (x \right )} - \log {\left (\log {\left (x \right )}^{2} - \frac {1}{15} \right )} - \log {\left (x + e^{\frac {2 \log {\left (x \right )} - 3}{\log {\left (x \right )}}} - 4 \right )} + 2 \log {\left (\log {\left (x \right )} \right )} \]
integrate(((15*ln(x)**4-46*ln(x)**2-2*ln(x)+3)*exp((2*ln(x)-3)/ln(x))-60*l n(x)**4+4*ln(x)**2+(-2*x+8)*ln(x))/((15*x*ln(x)**4-x*ln(x)**2)*exp((2*ln(x )-3)/ln(x))+(15*x**2-60*x)*ln(x)**4+(-x**2+4*x)*ln(x)**2),x)
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\frac {3}{\log \left (x\right )} - \log \left (\log \left (x\right )^{2} - \frac {1}{15}\right ) - \log \left (x - 4\right ) + \log \left (x\right ) - \log \left (\frac {{\left (x - 4\right )} e^{\frac {3}{\log \left (x\right )}} + e^{2}}{x - 4}\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]
integrate(((15*log(x)^4-46*log(x)^2-2*log(x)+3)*exp((2*log(x)-3)/log(x))-6 0*log(x)^4+4*log(x)^2+(-2*x+8)*log(x))/((15*x*log(x)^4-x*log(x)^2)*exp((2* log(x)-3)/log(x))+(15*x^2-60*x)*log(x)^4+(-x^2+4*x)*log(x)^2),x, algorithm =\
3/log(x) - log(log(x)^2 - 1/15) - log(x - 4) + log(x) - log(((x - 4)*e^(3/ log(x)) + e^2)/(x - 4)) + 2*log(log(x))
Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=-\log \left (15 \, \log \left (x\right )^{2} - 1\right ) - \log \left (x + e^{\left (\frac {2 \, \log \left (x\right ) - 3}{\log \left (x\right )}\right )} - 4\right ) + \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]
integrate(((15*log(x)^4-46*log(x)^2-2*log(x)+3)*exp((2*log(x)-3)/log(x))-6 0*log(x)^4+4*log(x)^2+(-2*x+8)*log(x))/((15*x*log(x)^4-x*log(x)^2)*exp((2* log(x)-3)/log(x))+(15*x^2-60*x)*log(x)^4+(-x^2+4*x)*log(x)^2),x, algorithm =\
Time = 10.88 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=2\,\ln \left (\frac {1}{x^2}\right )-\ln \left (x+{\mathrm {e}}^{-\frac {3}{\ln \left (x\right )}}\,{\mathrm {e}}^2-4\right )+2\,\ln \left (\ln \left (x\right )\right )-\ln \left (\frac {15\,{\ln \left (x\right )}^2-1}{x}\right )+4\,\ln \left (x\right ) \]
int((60*log(x)^4 - 4*log(x)^2 + log(x)*(2*x - 8) + exp((2*log(x) - 3)/log( x))*(2*log(x) + 46*log(x)^2 - 15*log(x)^4 - 3))/(log(x)^4*(60*x - 15*x^2) - log(x)^2*(4*x - x^2) + exp((2*log(x) - 3)/log(x))*(x*log(x)^2 - 15*x*log (x)^4)),x)