Integrand size = 108, antiderivative size = 26 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=e-\frac {9 \left (e^{10} \log (4)-\frac {\log (5)}{x+\log (x)}\right )}{3+x} \] Output:
exp(1)-9*(2*exp(5)^2*ln(2)-ln(5)/(x+ln(x)))/(3+x)
Time = 5.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {9 \left (-e^{10} x \log (4)+\log (5)-e^{10} \log (4) \log (x)\right )}{(3+x) (x+\log (x))} \] Input:
Integrate[(9*E^10*x^3*Log[4] + (-27 - 36*x - 18*x^2)*Log[5] + (18*E^10*x^2 *Log[4] - 9*x*Log[5])*Log[x] + 9*E^10*x*Log[4]*Log[x]^2)/(9*x^3 + 6*x^4 + x^5 + (18*x^2 + 12*x^3 + 2*x^4)*Log[x] + (9*x + 6*x^2 + x^3)*Log[x]^2),x]
Output:
(9*(-(E^10*x*Log[4]) + Log[5] - E^10*Log[4]*Log[x]))/((3 + x)*(x + 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.
\(\displaystyle \int \frac {9 e^{10} x^3 \log (4)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+\left (-18 x^2-36 x-27\right ) \log (5)+9 e^{10} x \log (4) \log ^2(x)}{x^5+6 x^4+9 x^3+\left (x^3+6 x^2+9 x\right ) \log ^2(x)+\left (2 x^4+12 x^3+18 x^2\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {9 \left (e^{10} x^3 \log (4)+e^{10} x^2 \log (16) \log (x)-2 x^2 \log (5)+e^{10} x \log (4) \log ^2(x)-x \log (5) \log (x)-4 x \log (5)-3 \log (5)\right )}{x (x+3)^2 (x+\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9 \int \frac {e^{10} \log (4) x^3+e^{10} \log (16) \log (x) x^2-2 \log (5) x^2+e^{10} \log (4) \log ^2(x) x-\log (5) \log (x) x-4 \log (5) x-3 \log (5)}{x (x+3)^2 (x+\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 9 \int \left (-\frac {\log (5) (x+1)}{x (x+3) (x+\log (x))^2}-\frac {\log (5)}{(x+3)^2 (x+\log (x))}+\frac {e^{10} \log (4)}{(x+3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 9 \left (-\frac {1}{3} \log (5) \int \frac {1}{x (x+\log (x))^2}dx-\frac {2}{3} \log (5) \int \frac {1}{(x+3) (x+\log (x))^2}dx-\log (5) \int \frac {1}{(x+3)^2 (x+\log (x))}dx-\frac {e^{10} \log (4)}{x+3}\right )\) |
Input:
Int[(9*E^10*x^3*Log[4] + (-27 - 36*x - 18*x^2)*Log[5] + (18*E^10*x^2*Log[4 ] - 9*x*Log[5])*Log[x] + 9*E^10*x*Log[4]*Log[x]^2)/(9*x^3 + 6*x^4 + x^5 + (18*x^2 + 12*x^3 + 2*x^4)*Log[x] + (9*x + 6*x^2 + x^3)*Log[x]^2),x]
Output:
$Aborted
Time = 0.95 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-\frac {18 \,{\mathrm e}^{10} \ln \left (2\right )}{3+x}+\frac {9 \ln \left (5\right )}{\left (3+x \right ) \left (x +\ln \left (x \right )\right )}\) | \(28\) |
default | \(-\frac {18 \,{\mathrm e}^{10} \ln \left (2\right )}{3+x}+\frac {9 \ln \left (5\right )}{\left (3+x \right ) \left (x +\ln \left (x \right )\right )}\) | \(30\) |
norman | \(\frac {-18 \,{\mathrm e}^{10} \ln \left (2\right ) x -18 \,{\mathrm e}^{10} \ln \left (2\right ) \ln \left (x \right )+9 \ln \left (5\right )}{x \ln \left (x \right )+x^{2}+3 \ln \left (x \right )+3 x}\) | \(43\) |
parallelrisch | \(-\frac {18 \,{\mathrm e}^{10} \ln \left (2\right ) x +18 \,{\mathrm e}^{10} \ln \left (2\right ) \ln \left (x \right )-9 \ln \left (5\right )}{x \ln \left (x \right )+x^{2}+3 \ln \left (x \right )+3 x}\) | \(44\) |
Input:
int((18*x*exp(5)^2*ln(2)*ln(x)^2+(-9*x*ln(5)+36*x^2*exp(5)^2*ln(2))*ln(x)+ (-18*x^2-36*x-27)*ln(5)+18*x^3*exp(5)^2*ln(2))/((x^3+6*x^2+9*x)*ln(x)^2+(2 *x^4+12*x^3+18*x^2)*ln(x)+x^5+6*x^4+9*x^3),x,method=_RETURNVERBOSE)
Output:
-18*exp(10)*ln(2)/(3+x)+9*ln(5)/(3+x)/(x+ln(x))
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {9 \, {\left (2 \, x e^{10} \log \left (2\right ) + 2 \, e^{10} \log \left (2\right ) \log \left (x\right ) - \log \left (5\right )\right )}}{x^{2} + {\left (x + 3\right )} \log \left (x\right ) + 3 \, x} \] Input:
integrate((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log( 2))*log(x)+(-18*x^2-36*x-27)*log(5)+18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9* x)*log(x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x, algorithm="fr icas")
Output:
-9*(2*x*e^10*log(2) + 2*e^10*log(2)*log(x) - log(5))/(x^2 + (x + 3)*log(x) + 3*x)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {9 \log {\left (5 \right )}}{x^{2} + 3 x + \left (x + 3\right ) \log {\left (x \right )}} - \frac {18 e^{10} \log {\left (2 \right )}}{x + 3} \] Input:
integrate((18*x*exp(5)**2*ln(2)*ln(x)**2+(-9*x*ln(5)+36*x**2*exp(5)**2*ln( 2))*ln(x)+(-18*x**2-36*x-27)*ln(5)+18*x**3*exp(5)**2*ln(2))/((x**3+6*x**2+ 9*x)*ln(x)**2+(2*x**4+12*x**3+18*x**2)*ln(x)+x**5+6*x**4+9*x**3),x)
Output:
9*log(5)/(x**2 + 3*x + (x + 3)*log(x)) - 18*exp(10)*log(2)/(x + 3)
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {9 \, {\left (2 \, x e^{10} \log \left (2\right ) + 2 \, e^{10} \log \left (2\right ) \log \left (x\right ) - \log \left (5\right )\right )}}{x^{2} + {\left (x + 3\right )} \log \left (x\right ) + 3 \, x} \] Input:
integrate((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log( 2))*log(x)+(-18*x^2-36*x-27)*log(5)+18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9* x)*log(x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x, algorithm="ma xima")
Output:
-9*(2*x*e^10*log(2) + 2*e^10*log(2)*log(x) - log(5))/(x^2 + (x + 3)*log(x) + 3*x)
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {9 \, {\left (2 \, x e^{10} \log \left (2\right ) + 2 \, e^{10} \log \left (2\right ) \log \left (x\right ) - \log \left (5\right )\right )}}{x^{2} + x \log \left (x\right ) + 3 \, x + 3 \, \log \left (x\right )} \] Input:
integrate((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log( 2))*log(x)+(-18*x^2-36*x-27)*log(5)+18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9* x)*log(x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x, algorithm="gi ac")
Output:
-9*(2*x*e^10*log(2) + 2*e^10*log(2)*log(x) - log(5))/(x^2 + x*log(x) + 3*x + 3*log(x))
Time = 7.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {9\,\ln \left (5\right )}{\left (x+\ln \left (x\right )\right )\,\left (x+3\right )}-\frac {18\,{\mathrm {e}}^{10}\,\ln \left (2\right )}{x+3} \] Input:
int(-(log(5)*(36*x + 18*x^2 + 27) + log(x)*(9*x*log(5) - 36*x^2*exp(10)*lo g(2)) - 18*x^3*exp(10)*log(2) - 18*x*exp(10)*log(2)*log(x)^2)/(log(x)*(18* x^2 + 12*x^3 + 2*x^4) + 9*x^3 + 6*x^4 + x^5 + log(x)^2*(9*x + 6*x^2 + x^3) ),x)
Output:
(9*log(5))/((x + log(x))*(x + 3)) - (18*exp(10)*log(2))/(x + 3)
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {-18 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) e^{10}+9 \,\mathrm {log}\left (5\right )-18 \,\mathrm {log}\left (2\right ) e^{10} x}{\mathrm {log}\left (x \right ) x +3 \,\mathrm {log}\left (x \right )+x^{2}+3 x} \] Input:
int((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log(2))*lo g(x)+(-18*x^2-36*x-27)*log(5)+18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9*x)*log (x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x)
Output:
(9*( - 2*log(x)*log(2)*e**10 + log(5) - 2*log(2)*e**10*x))/(log(x)*x + 3*l og(x) + x**2 + 3*x)