Integrand size = 95, antiderivative size = 26 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {\log (x)}{1+\frac {8}{1-e^x+2 \log \left (\frac {3 x}{2}\right )}} \]
Time = 0.87 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\log (x)+\frac {4 \left (-9+e^x-\log \left (\frac {9}{4}\right )\right )}{-9+e^x-2 \log \left (\frac {3 x}{2}\right )} \]
Integrate[(9 - 10*E^x + E^(2*x) + (16 - 8*E^x*x)*Log[x] + (20 - 4*E^x)*Log [(3*x)/2] + 4*Log[(3*x)/2]^2)/(81*x - 18*E^x*x + E^(2*x)*x + (36*x - 4*E^x *x)*Log[(3*x)/2] + 4*x*Log[(3*x)/2]^2),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.
\(\displaystyle \int \frac {-10 e^x+e^{2 x}+4 \log ^2\left (\frac {3 x}{2}\right )+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+\left (16-8 e^x x\right ) \log (x)+9}{-18 e^x x+e^{2 x} x+81 x+4 x \log ^2\left (\frac {3 x}{2}\right )+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-10 e^x+e^{2 x}+4 \log ^2\left (\frac {3 x}{2}\right )+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+\left (16-8 e^x x\right ) \log (x)+9}{x \left (-e^x+2 \log \left (\frac {3 x}{2}\right )+9\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{x}-\frac {8 (x \log (x)-1)}{x \left (e^x-2 \log \left (\frac {3 x}{2}\right )-9\right )}-\frac {8 \log (x) \left (9 x+2 x \log \left (\frac {3 x}{2}\right )-2\right )}{x \left (e^x-2 \log \left (\frac {3 x}{2}\right )-9\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -144 \text {Subst}\left (\int \frac {\log (2 x)}{\left (-2 \log (3 x)+e^{2 x}-9\right )^2}dx,x,\frac {x}{2}\right )-16 \text {Subst}\left (\int \frac {\log (2 x)}{-2 \log (3 x)+e^{2 x}-9}dx,x,\frac {x}{2}\right )-32 \text {Subst}\left (\int \frac {\log (2 x) \log (3 x)}{\left (-2 \log (3 x)+e^{2 x}-9\right )^2}dx,x,\frac {x}{2}\right )-8 \int \frac {1}{x \left (2 \log (x)-e^x+9 \left (1+\frac {2}{9} \log \left (\frac {3}{2}\right )\right )\right )}dx+16 \int \frac {\log (x)}{x \left (-2 \log \left (\frac {3 x}{2}\right )+e^x-9\right )^2}dx+\log (x)\) |
Int[(9 - 10*E^x + E^(2*x) + (16 - 8*E^x*x)*Log[x] + (20 - 4*E^x)*Log[(3*x) /2] + 4*Log[(3*x)/2]^2)/(81*x - 18*E^x*x + E^(2*x)*x + (36*x - 4*E^x*x)*Lo g[(3*x)/2] + 4*x*Log[(3*x)/2]^2),x]
3.3.39.3.1 Defintions of rubi rules used
Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
parallelrisch | \(\frac {8 \,{\mathrm e}^{x} \ln \left (x \right )-8 \ln \left (x \right )-16 \ln \left (x \right ) \ln \left (\frac {3 x}{2}\right )}{-72+8 \,{\mathrm e}^{x}-16 \ln \left (\frac {3 x}{2}\right )}\) | \(34\) |
risch | \(\ln \left (x \right )+\frac {-36+8 \ln \left (2\right )-8 \ln \left (3\right )+4 \,{\mathrm e}^{x}}{-9+2 \ln \left (2\right )-2 \ln \left (3\right )+{\mathrm e}^{x}-2 \ln \left (x \right )}\) | \(36\) |
int((4*ln(3/2*x)^2+(-4*exp(x)+20)*ln(3/2*x)+(-8*exp(x)*x+16)*ln(x)+exp(x)^ 2-10*exp(x)+9)/(4*x*ln(3/2*x)^2+(-4*exp(x)*x+36*x)*ln(3/2*x)+x*exp(x)^2-18 *exp(x)*x+81*x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {{\left (e^{x} - 2 \, \log \left (\frac {3}{2}\right ) - 9\right )} \log \left (x\right ) - 2 \, \log \left (x\right )^{2} + 4 \, e^{x} - 8 \, \log \left (\frac {3}{2}\right ) - 36}{e^{x} - 2 \, \log \left (\frac {3}{2}\right ) - 2 \, \log \left (x\right ) - 9} \]
integrate((4*log(3/2*x)^2+(-4*exp(x)+20)*log(3/2*x)+(-8*exp(x)*x+16)*log(x )+exp(x)^2-10*exp(x)+9)/(4*x*log(3/2*x)^2+(-4*exp(x)*x+36*x)*log(3/2*x)+x* exp(x)^2-18*exp(x)*x+81*x),x, algorithm=\
((e^x - 2*log(3/2) - 9)*log(x) - 2*log(x)^2 + 4*e^x - 8*log(3/2) - 36)/(e^ x - 2*log(3/2) - 2*log(x) - 9)
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\log {\left (x \right )} + \frac {8 \log {\left (x \right )}}{e^{x} - 2 \log {\left (x \right )} - 9 - 2 \log {\left (3 \right )} + 2 \log {\left (2 \right )}} \]
integrate((4*ln(3/2*x)**2+(-4*exp(x)+20)*ln(3/2*x)+(-8*exp(x)*x+16)*ln(x)+ exp(x)**2-10*exp(x)+9)/(4*x*ln(3/2*x)**2+(-4*exp(x)*x+36*x)*ln(3/2*x)+x*ex p(x)**2-18*exp(x)*x+81*x),x)
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {8 \, \log \left (x\right )}{e^{x} - 2 \, \log \left (3\right ) + 2 \, \log \left (2\right ) - 2 \, \log \left (x\right ) - 9} + \log \left (x\right ) \]
integrate((4*log(3/2*x)^2+(-4*exp(x)+20)*log(3/2*x)+(-8*exp(x)*x+16)*log(x )+exp(x)^2-10*exp(x)+9)/(4*x*log(3/2*x)^2+(-4*exp(x)*x+36*x)*log(3/2*x)+x* exp(x)^2-18*exp(x)*x+81*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {e^{x} \log \left (\frac {1}{2} \, x\right ) - 2 \, \log \left (3\right ) \log \left (\frac {1}{2} \, x\right ) - 2 \, \log \left (\frac {1}{2} \, x\right )^{2} + 8 \, \log \left (2\right ) - \log \left (\frac {1}{2} \, x\right )}{e^{x} - 2 \, \log \left (3\right ) - 2 \, \log \left (\frac {1}{2} \, x\right ) - 9} \]
integrate((4*log(3/2*x)^2+(-4*exp(x)+20)*log(3/2*x)+(-8*exp(x)*x+16)*log(x )+exp(x)^2-10*exp(x)+9)/(4*x*log(3/2*x)^2+(-4*exp(x)*x+36*x)*log(3/2*x)+x* exp(x)^2-18*exp(x)*x+81*x),x, algorithm=\
(e^x*log(1/2*x) - 2*log(3)*log(1/2*x) - 2*log(1/2*x)^2 + 8*log(2) - log(1/ 2*x))/(e^x - 2*log(3) - 2*log(1/2*x) - 9)
Time = 12.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {9-10 e^x+e^{2 x}+\left (16-8 e^x x\right ) \log (x)+\left (20-4 e^x\right ) \log \left (\frac {3 x}{2}\right )+4 \log ^2\left (\frac {3 x}{2}\right )}{81 x-18 e^x x+e^{2 x} x+\left (36 x-4 e^x x\right ) \log \left (\frac {3 x}{2}\right )+4 x \log ^2\left (\frac {3 x}{2}\right )} \, dx=\frac {\ln \left (x\right )\,\left (2\,\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^x+1\right )}{2\,\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^x+9} \]