Integrand size = 66, antiderivative size = 25 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=3-\frac {\log (3)}{6 \left (3-e^x\right ) \log ^4(x)}-\log (x) \] Output:
-1/6*ln(3)/ln(x)^4/(-exp(x)+3)-ln(x)+3
Time = 0.61 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=\frac {\log (9)}{12 \left (-3+e^x\right ) \log ^4(x)}-\log (x) \] Input:
Integrate[(12*Log[3] - 4*E^x*Log[3] - E^x*x*Log[3]*Log[x] + (-54 + 36*E^x - 6*E^(2*x))*Log[x]^5)/((54*x - 36*E^x*x + 6*E^(2*x)*x)*Log[x]^5),x]
Output:
Log[9]/(12*(-3 + E^x)*Log[x]^4) - 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.
\(\displaystyle \int \frac {\left (36 e^x-6 e^{2 x}-54\right ) \log ^5(x)-e^x x \log (3) \log (x)-4 e^x \log (3)+12 \log (3)}{\left (-36 e^x x+6 e^{2 x} x+54 x\right ) \log ^5(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (36 e^x-6 e^{2 x}-54\right ) \log ^5(x)-e^x x \log (3) \log (x)-4 e^x \log (3)+12 \log (3)}{6 \left (3-e^x\right )^2 x \log ^5(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {-6 \left (9-6 e^x+e^{2 x}\right ) \log ^5(x)-e^x x \log (3) \log (x)-4 e^x \log (3)+12 \log (3)}{\left (3-e^x\right )^2 x \log ^5(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{6} \int \left (-\frac {\log (3) (x \log (x)+4)}{\left (-3+e^x\right ) x \log ^5(x)}-\frac {6}{x}-\frac {3 \log (3)}{\left (-3+e^x\right )^2 \log ^4(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (-4 \log (3) \int \frac {1}{\left (-3+e^x\right ) x \log ^5(x)}dx-3 \log (3) \int \frac {1}{\left (-3+e^x\right )^2 \log ^4(x)}dx-\log (3) \int \frac {1}{\left (-3+e^x\right ) \log ^4(x)}dx-6 \log (x)\right )\) |
Input:
Int[(12*Log[3] - 4*E^x*Log[3] - E^x*x*Log[3]*Log[x] + (-54 + 36*E^x - 6*E^ (2*x))*Log[x]^5)/((54*x - 36*E^x*x + 6*E^(2*x)*x)*Log[x]^5),x]
Output:
$Aborted
Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
\[-\ln \left (x \right )+\frac {\ln \left (3\right )}{6 \left ({\mathrm e}^{x}-3\right ) \ln \left (x \right )^{4}}\]
Input:
int(((-6*exp(x)^2+36*exp(x)-54)*ln(x)^5-x*ln(3)*exp(x)*ln(x)-4*ln(3)*exp(x )+12*ln(3))/(6*x*exp(x)^2-36*exp(x)*x+54*x)/ln(x)^5,x)
Output:
-ln(x)+1/6*ln(3)/(exp(x)-3)/ln(x)^4
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=-\frac {6 \, {\left (e^{x} - 3\right )} \log \left (x\right )^{5} - \log \left (3\right )}{6 \, {\left (e^{x} - 3\right )} \log \left (x\right )^{4}} \] Input:
integrate(((-6*exp(x)^2+36*exp(x)-54)*log(x)^5-x*log(3)*exp(x)*log(x)-4*lo g(3)*exp(x)+12*log(3))/(6*x*exp(x)^2-36*exp(x)*x+54*x)/log(x)^5,x, algorit hm="fricas")
Output:
-1/6*(6*(e^x - 3)*log(x)^5 - log(3))/((e^x - 3)*log(x)^4)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=- \log {\left (x \right )} + \frac {\log {\left (3 \right )}}{6 e^{x} \log {\left (x \right )}^{4} - 18 \log {\left (x \right )}^{4}} \] Input:
integrate(((-6*exp(x)**2+36*exp(x)-54)*ln(x)**5-x*ln(3)*exp(x)*ln(x)-4*ln( 3)*exp(x)+12*ln(3))/(6*x*exp(x)**2-36*exp(x)*x+54*x)/ln(x)**5,x)
Output:
-log(x) + log(3)/(6*exp(x)*log(x)**4 - 18*log(x)**4)
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=\frac {\log \left (3\right )}{6 \, {\left (e^{x} \log \left (x\right )^{4} - 3 \, \log \left (x\right )^{4}\right )}} - \log \left (x\right ) \] Input:
integrate(((-6*exp(x)^2+36*exp(x)-54)*log(x)^5-x*log(3)*exp(x)*log(x)-4*lo g(3)*exp(x)+12*log(3))/(6*x*exp(x)^2-36*exp(x)*x+54*x)/log(x)^5,x, algorit hm="maxima")
Output:
1/6*log(3)/(e^x*log(x)^4 - 3*log(x)^4) - log(x)
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=-\frac {6 \, e^{x} \log \left (x\right )^{5} - 18 \, \log \left (x\right )^{5} - \log \left (3\right )}{6 \, {\left (e^{x} \log \left (x\right )^{4} - 3 \, \log \left (x\right )^{4}\right )}} \] Input:
integrate(((-6*exp(x)^2+36*exp(x)-54)*log(x)^5-x*log(3)*exp(x)*log(x)-4*lo g(3)*exp(x)+12*log(3))/(6*x*exp(x)^2-36*exp(x)*x+54*x)/log(x)^5,x, algorit hm="giac")
Output:
-1/6*(6*e^x*log(x)^5 - 18*log(x)^5 - log(3))/(e^x*log(x)^4 - 3*log(x)^4)
Time = 3.13 (sec) , antiderivative size = 800, normalized size of antiderivative = 32.00 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=\text {Too large to display} \] Input:
int(-(4*exp(x)*log(3) - 12*log(3) + log(x)^5*(6*exp(2*x) - 36*exp(x) + 54) + x*exp(x)*log(3)*log(x))/(log(x)^5*(54*x + 6*x*exp(2*x) - 36*x*exp(x))), x)
Output:
(log(3)/(6*(exp(x) - 3)) + (x*exp(x)*log(3)*log(x))/(24*(exp(x) - 3)^2))/l og(x)^4 - log(x) + ((5*x*log(3))/48 - (3*x^2*log(3))/16 - (7*x^3*log(3))/2 4 + (3*x^4*log(3))/16)/(exp(2*x) - 6*exp(x) + 9) + (x^2*log(3) - (5*x^4*lo g(3))/4)/(9*exp(2*x) - exp(3*x) - 27*exp(x) + 27) + ((x*log(3))/72 - log(3 )/720 + (x^2*log(3))/72 - (x^3*log(3))/36 + (x^4*log(3))/180)/(exp(x) - 3) + ((9*x^3*log(3))/4 + (9*x^4*log(3))/4)/(54*exp(2*x) - 12*exp(3*x) + exp( 4*x) - 108*exp(x) + 81) + ((x*(3*exp(x)*log(3) - exp(2*x)*log(3) + 3*x*exp (x)*log(3) + x*exp(2*x)*log(3)))/(72*(exp(x) - 3)^3) + (x*log(x)*(exp(3*x) *log(3) - 6*exp(2*x)*log(3) + 9*exp(x)*log(3) + 27*x*exp(x)*log(3) - 3*x*e xp(3*x)*log(3) + 9*x^2*exp(x)*log(3) + 12*x^2*exp(2*x)*log(3) + x^2*exp(3* x)*log(3)))/(144*(exp(x) - 3)^4))/log(x)^2 - ((x*exp(x)*log(3))/(24*(exp(x ) - 3)^2) + (x*exp(x)*log(3)*log(x)*(3*x - exp(x) + x*exp(x) + 3))/(72*(ex p(x) - 3)^3))/log(x)^3 - ((9*log(3))/80 + exp(2*x)*((3*log(3))/40 - (15*x^ 2*log(3))/8 + (33*x^4*log(3))/40) + (27*x*log(3))/16 + exp(x)*((15*x^2*log (3))/8 - (9*x*log(3))/8 - (3*log(3))/20 + (15*x^3*log(3))/4 + (39*x^4*log( 3))/40) + (45*x^2*log(3))/16 + (9*x^3*log(3))/8 + (9*x^4*log(3))/80 + exp( 3*x)*((x*log(3))/8 - log(3)/60 + (5*x^2*log(3))/24 - (5*x^3*log(3))/12 + ( 13*x^4*log(3))/120) + exp(4*x)*(log(3)/720 - (x*log(3))/48 + (5*x^2*log(3) )/144 - (x^3*log(3))/72 + (x^4*log(3))/720))/(270*exp(2*x) - 90*exp(3*x) + 15*exp(4*x) - exp(5*x) - 405*exp(x) + 243) - ((x*(exp(3*x)*log(3) - 6*...
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {12 \log (3)-4 e^x \log (3)-e^x x \log (3) \log (x)+\left (-54+36 e^x-6 e^{2 x}\right ) \log ^5(x)}{\left (54 x-36 e^x x+6 e^{2 x} x\right ) \log ^5(x)} \, dx=\frac {-6 e^{x} \mathrm {log}\left (x \right )^{5}+18 \mathrm {log}\left (x \right )^{5}+\mathrm {log}\left (3\right )}{6 \mathrm {log}\left (x \right )^{4} \left (e^{x}-3\right )} \] Input:
int(((-6*exp(x)^2+36*exp(x)-54)*log(x)^5-x*log(3)*exp(x)*log(x)-4*log(3)*e xp(x)+12*log(3))/(6*x*exp(x)^2-36*exp(x)*x+54*x)/log(x)^5,x)
Output:
( - 6*e**x*log(x)**5 + 18*log(x)**5 + log(3))/(6*log(x)**4*(e**x - 3))