Integrand size = 68, antiderivative size = 36 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=\log \left (\frac {3}{4 \left (3+e^x+x-\frac {\frac {1}{5} \left (4-\frac {x}{\log (5)}\right )-4 \log (x)}{x}\right )}\right ) \] Output:
ln(3/4/(3-(4/5-1/5*x/ln(5)-4*ln(x))/x+exp(x)+x))
Time = 0.84 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.42 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log (5) \left (-\frac {\log (x)}{\log (5)}+\frac {\log \left (x-4 \log (5)+15 x \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+20 \log (5) \log (x)\right )}{\log (5)}\right ) \] Input:
Integrate[(-5*E^x*x^2*Log[5] + (-24 - 5*x^2)*Log[5] + 20*Log[5]*Log[x])/(x ^2 + 5*E^x*x^2*Log[5] + (-4*x + 15*x^2 + 5*x^3)*Log[5] + 20*x*Log[5]*Log[x ]),x]
Output:
-(Log[5]*(-(Log[x]/Log[5]) + Log[x - 4*Log[5] + 15*x*Log[5] + 5*E^x*x*Log[ 5] + 5*x^2*Log[5] + 20*Log[5]*Log[x]]/Log[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.
\(\displaystyle \int \frac {-5 e^x x^2 \log (5)+\left (-5 x^2-24\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (5 x^3+15 x^2-4 x\right ) \log (5)+20 x \log (5) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\log (5) \left (-5 e^x x^2-5 x^2+20 \log (x)-24\right )}{x^2+5 e^x x^2 \log (5)+\left (5 x^3+15 x^2-4 x\right ) \log (5)+20 x \log (5) \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (5) \int -\frac {5 e^x x^2+5 x^2-20 \log (x)+24}{5 e^x \log (5) x^2+x^2+20 \log (5) \log (x) x-\left (-5 x^3-15 x^2+4 x\right ) \log (5)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (5) \int \frac {5 e^x x^2+5 x^2-20 \log (x)+24}{5 e^x \log (5) x^2+x^2+20 \log (5) \log (x) x-\left (-5 x^3-15 x^2+4 x\right ) \log (5)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\log (5) \int \left (\frac {5 \log (5) x^3+(1+10 \log (5)) x^2+20 \log (5) \log (x) x-4 \log (5) x+20 \log (5) \log (x)-24 \log (5)}{x \log (5) \left (-5 \log (5) x^2-(1+15 \log (5)) x-5 e^x \log (5) x-20 \log (5) \log (x)+4 \log (5)\right )}+\frac {1}{\log (5)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log (5) \left (\frac {(1+10 \log (5)) \int \frac {x}{-5 \log (5) x^2-(1+15 \log (5)) x-5 e^x \log (5) x-20 \log (5) \log (x)+4 \log (5)}dx}{\log (5)}+5 \int \frac {x^2}{-5 \log (5) x^2-(1+15 \log (5)) x-5 e^x \log (5) x-20 \log (5) \log (x)+4 \log (5)}dx+20 \int \frac {\log (x)}{-5 \log (5) x^2-(1+15 \log (5)) x-5 e^x \log (5) x-20 \log (5) \log (x)+4 \log (5)}dx+20 \int \frac {\log (x)}{x \left (-5 \log (5) x^2-(1+15 \log (5)) x-5 e^x \log (5) x-20 \log (5) \log (x)+4 \log (5)\right )}dx+4 \int \frac {1}{5 \log (5) x^2+(1+15 \log (5)) x+5 e^x \log (5) x+20 \log (5) \log (x)-4 \log (5)}dx+24 \int \frac {1}{x \left (5 \log (5) x^2+(1+15 \log (5)) x+5 e^x \log (5) x+20 \log (5) \log (x)-4 \log (5)\right )}dx+\frac {x}{\log (5)}\right )\) |
Input:
Int[(-5*E^x*x^2*Log[5] + (-24 - 5*x^2)*Log[5] + 20*Log[5]*Log[x])/(x^2 + 5 *E^x*x^2*Log[5] + (-4*x + 15*x^2 + 5*x^3)*Log[5] + 20*x*Log[5]*Log[x]),x]
Output:
$Aborted
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06
method | result | size |
norman | \(\ln \left (x \right )-\ln \left (5 x \,{\mathrm e}^{x} \ln \left (5\right )+5 x^{2} \ln \left (5\right )+20 \ln \left (5\right ) \ln \left (x \right )+15 x \ln \left (5\right )-4 \ln \left (5\right )+x \right )\) | \(38\) |
risch | \(\ln \left (x \right )-\ln \left (\ln \left (x \right )+\frac {5 x^{2} \ln \left (5\right )+5 x \,{\mathrm e}^{x} \ln \left (5\right )+15 x \ln \left (5\right )-4 \ln \left (5\right )+x}{20 \ln \left (5\right )}\right )\) | \(41\) |
parallelrisch | \(-\ln \left (\frac {5 x \,{\mathrm e}^{x} \ln \left (5\right )+5 x^{2} \ln \left (5\right )+20 \ln \left (5\right ) \ln \left (x \right )+15 x \ln \left (5\right )-4 \ln \left (5\right )+x}{5 \ln \left (5\right )}\right )+\ln \left (x \right )\) | \(44\) |
Input:
int((20*ln(5)*ln(x)-5*x^2*ln(5)*exp(x)+(-5*x^2-24)*ln(5))/(20*x*ln(5)*ln(x )+5*x^2*ln(5)*exp(x)+(5*x^3+15*x^2-4*x)*ln(5)+x^2),x,method=_RETURNVERBOSE )
Output:
ln(x)-ln(5*x*exp(x)*ln(5)+5*x^2*ln(5)+20*ln(5)*ln(x)+15*x*ln(5)-4*ln(5)+x)
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log \left (5 \, x e^{x} \log \left (5\right ) + {\left (5 \, x^{2} + 15 \, x - 4\right )} \log \left (5\right ) + 20 \, \log \left (5\right ) \log \left (x\right ) + x\right ) + \log \left (x\right ) \] Input:
integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x* log(5)*log(x)+5*x^2*log(5)*exp(x)+(5*x^3+15*x^2-4*x)*log(5)+x^2),x, algori thm="fricas")
Output:
-log(5*x*e^x*log(5) + (5*x^2 + 15*x - 4)*log(5) + 20*log(5)*log(x) + x) + log(x)
Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=- \log {\left (e^{x} + \frac {5 x^{2} \log {\left (5 \right )} + x + 15 x \log {\left (5 \right )} + 20 \log {\left (5 \right )} \log {\left (x \right )} - 4 \log {\left (5 \right )}}{5 x \log {\left (5 \right )}} \right )} \] Input:
integrate((20*ln(5)*ln(x)-5*x**2*ln(5)*exp(x)+(-5*x**2-24)*ln(5))/(20*x*ln (5)*ln(x)+5*x**2*ln(5)*exp(x)+(5*x**3+15*x**2-4*x)*ln(5)+x**2),x)
Output:
-log(exp(x) + (5*x**2*log(5) + x + 15*x*log(5) + 20*log(5)*log(x) - 4*log( 5))/(5*x*log(5)))
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log \left (\frac {5 \, x^{2} \log \left (5\right ) + 5 \, x e^{x} \log \left (5\right ) + x {\left (15 \, \log \left (5\right ) + 1\right )} + 20 \, \log \left (5\right ) \log \left (x\right ) - 4 \, \log \left (5\right )}{5 \, x \log \left (5\right )}\right ) \] Input:
integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x* log(5)*log(x)+5*x^2*log(5)*exp(x)+(5*x^3+15*x^2-4*x)*log(5)+x^2),x, algori thm="maxima")
Output:
-log(1/5*(5*x^2*log(5) + 5*x*e^x*log(5) + x*(15*log(5) + 1) + 20*log(5)*lo g(x) - 4*log(5))/(x*log(5)))
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log \left (5 \, x^{2} \log \left (5\right ) + 5 \, x e^{x} \log \left (5\right ) + 15 \, x \log \left (5\right ) + 20 \, \log \left (5\right ) \log \left (x\right ) + x - 4 \, \log \left (5\right )\right ) + \log \left (x\right ) \] Input:
integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x* log(5)*log(x)+5*x^2*log(5)*exp(x)+(5*x^3+15*x^2-4*x)*log(5)+x^2),x, algori thm="giac")
Output:
-log(5*x^2*log(5) + 5*x*e^x*log(5) + 15*x*log(5) + 20*log(5)*log(x) + x - 4*log(5)) + log(x)
Timed out. \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=\int -\frac {\ln \left (5\right )\,\left (5\,x^2+24\right )-20\,\ln \left (5\right )\,\ln \left (x\right )+5\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )}{\ln \left (5\right )\,\left (5\,x^3+15\,x^2-4\,x\right )+x^2+20\,x\,\ln \left (5\right )\,\ln \left (x\right )+5\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )} \,d x \] Input:
int(-(log(5)*(5*x^2 + 24) - 20*log(5)*log(x) + 5*x^2*exp(x)*log(5))/(log(5 )*(15*x^2 - 4*x + 5*x^3) + x^2 + 20*x*log(5)*log(x) + 5*x^2*exp(x)*log(5)) ,x)
Output:
int(-(log(5)*(5*x^2 + 24) - 20*log(5)*log(x) + 5*x^2*exp(x)*log(5))/(log(5 )*(15*x^2 - 4*x + 5*x^3) + x^2 + 20*x*log(5)*log(x) + 5*x^2*exp(x)*log(5)) , x)
Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\mathrm {log}\left (5 e^{x} \mathrm {log}\left (5\right ) x +20 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (5\right )+5 \,\mathrm {log}\left (5\right ) x^{2}+15 \,\mathrm {log}\left (5\right ) x -4 \,\mathrm {log}\left (5\right )+x \right )+\mathrm {log}\left (x \right ) \] Input:
int((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5) *log(x)+5*x^2*log(5)*exp(x)+(5*x^3+15*x^2-4*x)*log(5)+x^2),x)
Output:
- log(5*e**x*log(5)*x + 20*log(x)*log(5) + 5*log(5)*x**2 + 15*log(5)*x - 4*log(5) + x) + log(x)