Integrand size = 73, antiderivative size = 25 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {4+\frac {25}{x}-x}{-6+\frac {5}{\log \left (e^6 x\right )}} \] Output:
(25/x-x+4)/(5/ln(x*exp(6))-6)
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {-25+x^2+\frac {5 \left (-25-4 x+x^2\right )}{31+6 \log (x)}}{6 x} \] Input:
Integrate[(125 + 20*x - 5*x^2 + (-125 - 5*x^2)*Log[E^6*x] + (150 + 6*x^2)* Log[E^6*x]^2)/(25*x^2 - 60*x^2*Log[E^6*x] + 36*x^2*Log[E^6*x]^2),x]
Output:
(-25 + x^2 + (5*(-25 - 4*x + x^2))/(31 + 6*Log[x]))/(6*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.
\(\displaystyle \int \frac {-5 x^2+\left (6 x^2+150\right ) \log ^2\left (e^6 x\right )+\left (-5 x^2-125\right ) \log \left (e^6 x\right )+20 x+125}{25 x^2+36 x^2 \log ^2\left (e^6 x\right )-60 x^2 \log \left (e^6 x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {181 x^2+6 \left (x^2+25\right ) \log ^2(x)+67 \left (x^2+25\right ) \log (x)+20 x+4775}{x^2 (6 \log (x)+31)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2+25}{6 x^2}+\frac {5 \left (x^2+25\right )}{6 x^2 (6 \log (x)+31)}-\frac {5 \left (x^2-4 x-25\right )}{x^2 (6 \log (x)+31)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int \frac {x^2-4 x-25}{x^2 (6 \log (x)+31)^2}dx+\frac {125}{36} e^{31/6} \operatorname {ExpIntegralEi}\left (\frac {1}{6} (-6 \log (x)-31)\right )+\frac {5 \operatorname {ExpIntegralEi}\left (\frac {1}{6} (6 \log (x)+31)\right )}{36 e^{31/6}}+\frac {x}{6}-\frac {25}{6 x}\) |
Input:
Int[(125 + 20*x - 5*x^2 + (-125 - 5*x^2)*Log[E^6*x] + (150 + 6*x^2)*Log[E^ 6*x]^2)/(25*x^2 - 60*x^2*Log[E^6*x] + 36*x^2*Log[E^6*x]^2),x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
method | result | size |
norman | \(\frac {x^{2} \ln \left (x \,{\mathrm e}^{6}\right )-\frac {10 x}{3}-25 \ln \left (x \,{\mathrm e}^{6}\right )}{x \left (-5+6 \ln \left (x \,{\mathrm e}^{6}\right )\right )}\) | \(36\) |
risch | \(\frac {x^{2}-25}{6 x}+\frac {\frac {5}{6} x^{2}-\frac {10}{3} x -\frac {125}{6}}{x \left (-5+6 \ln \left (x \,{\mathrm e}^{6}\right )\right )}\) | \(36\) |
parallelrisch | \(\frac {6 x^{2} \ln \left (x \,{\mathrm e}^{6}\right )-20 x -150 \ln \left (x \,{\mathrm e}^{6}\right )}{6 x \left (-5+6 \ln \left (x \,{\mathrm e}^{6}\right )\right )}\) | \(38\) |
default | \(-\frac {10}{3 \left (31+6 \ln \left (x \right )\right )}-\frac {125}{6 x \left (31+6 \ln \left (x \right )\right )}+\frac {5 x}{36 \left (\ln \left (x \right )+\frac {31}{6}\right )}-\frac {25}{6 x}+\frac {x}{6}\) | \(42\) |
Input:
int(((6*x^2+150)*ln(x*exp(6))^2+(-5*x^2-125)*ln(x*exp(6))-5*x^2+20*x+125)/ (36*x^2*ln(x*exp(6))^2-60*x^2*ln(x*exp(6))+25*x^2),x,method=_RETURNVERBOSE )
Output:
(x^2*ln(x*exp(6))-10/3*x-25*ln(x*exp(6)))/x/(-5+6*ln(x*exp(6)))
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {3 \, {\left (x^{2} - 25\right )} \log \left (x e^{6}\right ) - 10 \, x}{3 \, {\left (6 \, x \log \left (x e^{6}\right ) - 5 \, x\right )}} \] Input:
integrate(((6*x^2+150)*log(x*exp(6))^2+(-5*x^2-125)*log(x*exp(6))-5*x^2+20 *x+125)/(36*x^2*log(x*exp(6))^2-60*x^2*log(x*exp(6))+25*x^2),x, algorithm= "fricas")
Output:
1/3*(3*(x^2 - 25)*log(x*e^6) - 10*x)/(6*x*log(x*e^6) - 5*x)
Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (15) = 30\).
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {x}{6} + \frac {5 x^{2} - 20 x - 125}{36 x \log {\left (x e^{6} \right )} - 30 x} - \frac {25}{6 x} \] Input:
integrate(((6*x**2+150)*ln(x*exp(6))**2+(-5*x**2-125)*ln(x*exp(6))-5*x**2+ 20*x+125)/(36*x**2*ln(x*exp(6))**2-60*x**2*ln(x*exp(6))+25*x**2),x)
Output:
x/6 + (5*x**2 - 20*x - 125)/(36*x*log(x*exp(6)) - 30*x) - 25/(6*x)
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {18 \, x^{2} + 3 \, {\left (x^{2} - 25\right )} \log \left (x\right ) - 10 \, x - 450}{3 \, {\left (6 \, x \log \left (x\right ) + 31 \, x\right )}} \] Input:
integrate(((6*x^2+150)*log(x*exp(6))^2+(-5*x^2-125)*log(x*exp(6))-5*x^2+20 *x+125)/(36*x^2*log(x*exp(6))^2-60*x^2*log(x*exp(6))+25*x^2),x, algorithm= "maxima")
Output:
1/3*(18*x^2 + 3*(x^2 - 25)*log(x) - 10*x - 450)/(6*x*log(x) + 31*x)
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {1}{6} \, x + \frac {5 \, {\left (x^{2} - 4 \, x - 25\right )}}{6 \, {\left (6 \, x \log \left (x\right ) + 31 \, x\right )}} - \frac {25}{6 \, x} \] Input:
integrate(((6*x^2+150)*log(x*exp(6))^2+(-5*x^2-125)*log(x*exp(6))-5*x^2+20 *x+125)/(36*x^2*log(x*exp(6))^2-60*x^2*log(x*exp(6))+25*x^2),x, algorithm= "giac")
Output:
1/6*x + 5/6*(x^2 - 4*x - 25)/(6*x*log(x) + 31*x) - 25/6/x
Time = 0.74 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=-\frac {\ln \left (x\,{\mathrm {e}}^6\right )\,\left (-x^2+4\,x+25\right )}{x\,\left (6\,\ln \left (x\,{\mathrm {e}}^6\right )-5\right )} \] Input:
int((20*x - log(x*exp(6))*(5*x^2 + 125) - 5*x^2 + log(x*exp(6))^2*(6*x^2 + 150) + 125)/(36*x^2*log(x*exp(6))^2 + 25*x^2 - 60*x^2*log(x*exp(6))),x)
Output:
-(log(x*exp(6))*(4*x - x^2 + 25))/(x*(6*log(x*exp(6)) - 5))
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {125+20 x-5 x^2+\left (-125-5 x^2\right ) \log \left (e^6 x\right )+\left (150+6 x^2\right ) \log ^2\left (e^6 x\right )}{25 x^2-60 x^2 \log \left (e^6 x\right )+36 x^2 \log ^2\left (e^6 x\right )} \, dx=\frac {\mathrm {log}\left (e^{6} x \right ) \left (x^{2}-4 x -25\right )}{x \left (6 \,\mathrm {log}\left (e^{6} x \right )-5\right )} \] Input:
int(((6*x^2+150)*log(x*exp(6))^2+(-5*x^2-125)*log(x*exp(6))-5*x^2+20*x+125 )/(36*x^2*log(x*exp(6))^2-60*x^2*log(x*exp(6))+25*x^2),x)
Output:
(log(e**6*x)*(x**2 - 4*x - 25))/(x*(6*log(e**6*x) - 5))