Integrand size = 57, antiderivative size = 27 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {1}{5} \log \left (x^2 \left (25+4 \left (1+5 e^{\frac {1}{\log (x)}}-x\right )+x\right )\right ) \] Output:
1/5*ln(x^2*(-3*x+20*exp(1/ln(x))+29))
Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {1}{5} \left (\log \left (29+20 e^{\frac {1}{\log (x)}}-3 x\right )+2 \log (x)\right ) \] Input:
Integrate[((58 - 9*x)*Log[x]^2 + E^Log[x]^(-1)*(-20 + 40*Log[x]^2))/(100*E ^Log[x]^(-1)*x*Log[x]^2 + (145*x - 15*x^2)*Log[x]^2),x]
Output:
(Log[29 + 20*E^Log[x]^(-1) - 3*x] + 2*Log[x])/5
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.
\(\displaystyle \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (40 \log ^2(x)-20\right )}{\left (145 x-15 x^2\right ) \log ^2(x)+100 x e^{\frac {1}{\log (x)}} \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (40 \log ^2(x)-20\right )}{5 x \left (-3 x+20 e^{\frac {1}{\log (x)}}+29\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {(58-9 x) \log ^2(x)-20 e^{\frac {1}{\log (x)}} \left (1-2 \log ^2(x)\right )}{\left (-3 x+20 e^{\frac {1}{\log (x)}}+29\right ) x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {2 \log ^2(x)-1}{x \log ^2(x)}-\frac {3 x \log ^2(x)+3 x-29}{\left (-3 x+20 e^{\frac {1}{\log (x)}}+29\right ) x \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-3 \int \frac {1}{\left (-3 x+20 e^{\frac {1}{\log (x)}}+29\right ) \log ^2(x)}dx+29 \int \frac {1}{\left (-3 x+20 e^{\frac {1}{\log (x)}}+29\right ) x \log ^2(x)}dx-3 \int \frac {1}{-3 x+20 e^{\frac {1}{\log (x)}}+29}dx+2 \log (x)+\frac {1}{\log (x)}\right )\) |
Input:
Int[((58 - 9*x)*Log[x]^2 + E^Log[x]^(-1)*(-20 + 40*Log[x]^2))/(100*E^Log[x ]^(-1)*x*Log[x]^2 + (145*x - 15*x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {2 \ln \left (x \right )}{5}+\frac {\ln \left (-\frac {3 x}{20}+{\mathrm e}^{\frac {1}{\ln \left (x \right )}}+\frac {29}{20}\right )}{5}\) | \(19\) |
parallelrisch | \(\frac {2 \ln \left (x \right )}{5}+\frac {\ln \left (x -\frac {20 \,{\mathrm e}^{\frac {1}{\ln \left (x \right )}}}{3}-\frac {29}{3}\right )}{5}\) | \(19\) |
norman | \(\frac {2 \ln \left (x \right )}{5}+\frac {\ln \left (3 x -20 \,{\mathrm e}^{\frac {1}{\ln \left (x \right )}}-29\right )}{5}\) | \(21\) |
Input:
int(((40*ln(x)^2-20)*exp(1/ln(x))+(-9*x+58)*ln(x)^2)/(100*x*ln(x)^2*exp(1/ ln(x))+(-15*x^2+145*x)*ln(x)^2),x,method=_RETURNVERBOSE)
Output:
2/5*ln(x)+1/5*ln(-3/20*x+exp(1/ln(x))+29/20)
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (-3 \, x + 20 \, e^{\frac {1}{\log \left (x\right )}} + 29\right ) \] Input:
integrate(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x )^2*exp(1/log(x))+(-15*x^2+145*x)*log(x)^2),x, algorithm="fricas")
Output:
2/5*log(x) + 1/5*log(-3*x + 20*e^(1/log(x)) + 29)
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2 \log {\left (x \right )}}{5} + \frac {\log {\left (- \frac {3 x}{20} + e^{\frac {1}{\log {\left (x \right )}}} + \frac {29}{20} \right )}}{5} \] Input:
integrate(((40*ln(x)**2-20)*exp(1/ln(x))+(-9*x+58)*ln(x)**2)/(100*x*ln(x)* *2*exp(1/ln(x))+(-15*x**2+145*x)*ln(x)**2),x)
Output:
2*log(x)/5 + log(-3*x/20 + exp(1/log(x)) + 29/20)/5
Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (-\frac {3}{20} \, x + e^{\frac {1}{\log \left (x\right )}} + \frac {29}{20}\right ) \] Input:
integrate(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x )^2*exp(1/log(x))+(-15*x^2+145*x)*log(x)^2),x, algorithm="maxima")
Output:
2/5*log(x) + 1/5*log(-3/20*x + e^(1/log(x)) + 29/20)
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (-3 \, x + 20 \, e^{\frac {1}{\log \left (x\right )}} + 29\right ) \] Input:
integrate(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x )^2*exp(1/log(x))+(-15*x^2+145*x)*log(x)^2),x, algorithm="giac")
Output:
2/5*log(x) + 1/5*log(-3*x + 20*e^(1/log(x)) + 29)
Time = 3.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{\frac {1}{\ln \left (x\right )}}-\frac {3\,x}{20}+\frac {29}{20}\right )}{5}+\frac {2\,\ln \left (x\right )}{5} \] Input:
int((exp(1/log(x))*(40*log(x)^2 - 20) - log(x)^2*(9*x - 58))/(log(x)^2*(14 5*x - 15*x^2) + 100*x*exp(1/log(x))*log(x)^2),x)
Output:
log(exp(1/log(x)) - (3*x)/20 + 29/20)/5 + (2*log(x))/5
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {\mathrm {log}\left (20 e^{\frac {1}{\mathrm {log}\left (x \right )}}-3 x +29\right )}{5}+\frac {2 \,\mathrm {log}\left (x \right )}{5} \] Input:
int(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x)^2*ex p(1/log(x))+(-15*x^2+145*x)*log(x)^2),x)
Output:
(log(20*e**(1/log(x)) - 3*x + 29) + 2*log(x))/5