Integrand size = 61, antiderivative size = 23 \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=1+3 \log \left (-5+2 e^x (5-x-\log (3)+\log (x))\right ) \] Output:
3*ln(2*(5-ln(3)+ln(x)-x)*exp(x)-5)+1
\[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=\int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx \] Input:
Integrate[(E^x*(6 + 24*x - 6*x^2 - 6*x*Log[3]) + 6*E^x*x*Log[x])/(-5*x + E ^x*(10*x - 2*x^2 - 2*x*Log[3]) + 2*E^x*x*Log[x]),x]
Output:
Integrate[(E^x*(6 + 24*x - 6*x^2 - 6*x*Log[3]) + 6*E^x*x*Log[x])/(-5*x + E ^x*(10*x - 2*x^2 - 2*x*Log[3]) + 2*E^x*x*Log[x]), 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.
\(\displaystyle \int \frac {e^x \left (-6 x^2+24 x-6 x \log (3)+6\right )+6 e^x x \log (x)}{e^x \left (-2 x^2+10 x-2 x \log (3)\right )-5 x+2 e^x x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {6 e^x \left (x^2-x \log (x)-4 x \left (1-\frac {\log (3)}{4}\right )-1\right )}{-e^x \left (-2 x^2+10 x-2 x \log (3)\right )+5 x-2 e^x x \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 6 \int -\frac {e^x \left (-x^2+\log (x) x+(4-\log (3)) x+1\right )}{-2 e^x \log (x) x+5 x-2 e^x \left (-x^2-\log (3) x+5 x\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -6 \int \frac {e^x \left (-x^2+\log (x) x+(4-\log (3)) x+1\right )}{-2 e^x \log (x) x+5 x-2 e^x \left (-x^2-\log (3) x+5 x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -6 \int \left (\frac {e^x x}{-2 e^x x+2 e^x \log (x)+10 e^x \left (1-\frac {\log (3)}{5}\right )-5}+\frac {e^x \log (x)}{2 e^x x-2 e^x \log (x)-10 e^x \left (1-\frac {\log (3)}{5}\right )+5}+\frac {e^x (4-\log (3))}{2 e^x x-2 e^x \log (x)-10 e^x \left (1-\frac {\log (3)}{5}\right )+5}+\frac {e^x}{\left (2 e^x x-2 e^x \log (x)-10 e^x \left (1-\frac {\log (3)}{5}\right )+5\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -6 \left ((4-\log (3)) \int \frac {e^x}{2 e^x x-2 e^x \log (x)-10 e^x \left (1-\frac {\log (3)}{5}\right )+5}dx+\int \frac {e^x}{x \left (2 e^x x-2 e^x \log (x)-10 e^x \left (1-\frac {\log (3)}{5}\right )+5\right )}dx+\int \frac {e^x \log (x)}{2 e^x x-2 e^x \log (x)-10 e^x \left (1-\frac {\log (3)}{5}\right )+5}dx+\int \frac {e^x x}{-2 e^x x+2 e^x \log (x)+10 e^x \left (1-\frac {\log (3)}{5}\right )-5}dx\right )\) |
Input:
Int[(E^x*(6 + 24*x - 6*x^2 - 6*x*Log[3]) + 6*E^x*x*Log[x])/(-5*x + E^x*(10 *x - 2*x^2 - 2*x*Log[3]) + 2*E^x*x*Log[x]),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
risch | \(3 x +3 \ln \left (\ln \left (x \right )-\ln \left (3\right )-x +5-\frac {5 \,{\mathrm e}^{-x}}{2}\right )\) | \(25\) |
parallelrisch | \(3 \ln \left (\ln \left (3\right ) {\mathrm e}^{x}+{\mathrm e}^{x} x -{\mathrm e}^{x} \ln \left (x \right )-5 \,{\mathrm e}^{x}+\frac {5}{2}\right )\) | \(25\) |
norman | \(3 \ln \left (2 \ln \left (3\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x} \ln \left (x \right )-10 \,{\mathrm e}^{x}+5\right )\) | \(27\) |
Input:
int((6*x*exp(x)*ln(x)+(-6*x*ln(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x)*ln(x)+ (-2*x*ln(3)-2*x^2+10*x)*exp(x)-5*x),x,method=_RETURNVERBOSE)
Output:
3*x+3*ln(ln(x)-ln(3)-x+5-5/2*exp(-x))
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=3 \, x + 3 \, \log \left (-{\left (2 \, {\left (x + \log \left (3\right ) - 5\right )} e^{x} - 2 \, e^{x} \log \left (x\right ) + 5\right )} e^{\left (-x\right )}\right ) \] Input:
integrate((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x )*log(x)+(-2*x*log(3)-2*x^2+10*x)*exp(x)-5*x),x, algorithm="fricas")
Output:
3*x + 3*log(-(2*(x + log(3) - 5)*e^x - 2*e^x*log(x) + 5)*e^(-x))
Time = 0.95 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=3 \log {\left (e^{x} + \frac {5}{2 x - 2 \log {\left (x \right )} - 10 + 2 \log {\left (3 \right )}} \right )} + 3 \log {\left (- x + \log {\left (x \right )} - \log {\left (3 \right )} + 5 \right )} \] Input:
integrate((6*x*exp(x)*ln(x)+(-6*x*ln(3)-6*x**2+24*x+6)*exp(x))/(2*x*exp(x) *ln(x)+(-2*x*ln(3)-2*x**2+10*x)*exp(x)-5*x),x)
Output:
3*log(exp(x) + 5/(2*x - 2*log(x) - 10 + 2*log(3))) + 3*log(-x + log(x) - l og(3) + 5)
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=3 \, \log \left (-x - \log \left (3\right ) + \log \left (x\right ) + 5\right ) + 3 \, \log \left (\frac {2 \, {\left (x + \log \left (3\right ) - \log \left (x\right ) - 5\right )} e^{x} + 5}{2 \, {\left (x + \log \left (3\right ) - \log \left (x\right ) - 5\right )}}\right ) \] Input:
integrate((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x )*log(x)+(-2*x*log(3)-2*x^2+10*x)*exp(x)-5*x),x, algorithm="maxima")
Output:
3*log(-x - log(3) + log(x) + 5) + 3*log(1/2*(2*(x + log(3) - log(x) - 5)*e ^x + 5)/(x + log(3) - log(x) - 5))
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=3 \, \log \left (2 \, x e^{x} + 2 \, e^{x} \log \left (3\right ) - 2 \, e^{x} \log \left (x\right ) - 10 \, e^{x} + 5\right ) \] Input:
integrate((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x )*log(x)+(-2*x*log(3)-2*x^2+10*x)*exp(x)-5*x),x, algorithm="giac")
Output:
3*log(2*x*e^x + 2*e^x*log(3) - 2*e^x*log(x) - 10*e^x + 5)
Timed out. \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=\int -\frac {{\mathrm {e}}^x\,\left (24\,x-6\,x\,\ln \left (3\right )-6\,x^2+6\right )+6\,x\,{\mathrm {e}}^x\,\ln \left (x\right )}{5\,x+{\mathrm {e}}^x\,\left (2\,x\,\ln \left (3\right )-10\,x+2\,x^2\right )-2\,x\,{\mathrm {e}}^x\,\ln \left (x\right )} \,d x \] Input:
int(-(exp(x)*(24*x - 6*x*log(3) - 6*x^2 + 6) + 6*x*exp(x)*log(x))/(5*x + e xp(x)*(2*x*log(3) - 10*x + 2*x^2) - 2*x*exp(x)*log(x)),x)
Output:
int(-(exp(x)*(24*x - 6*x*log(3) - 6*x^2 + 6) + 6*x*exp(x)*log(x))/(5*x + e xp(x)*(2*x*log(3) - 10*x + 2*x^2) - 2*x*exp(x)*log(x)), x)
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^x \left (6+24 x-6 x^2-6 x \log (3)\right )+6 e^x x \log (x)}{-5 x+e^x \left (10 x-2 x^2-2 x \log (3)\right )+2 e^x x \log (x)} \, dx=3 \,\mathrm {log}\left (2 e^{x} \mathrm {log}\left (x \right )-2 e^{x} \mathrm {log}\left (3\right )-2 e^{x} x +10 e^{x}-5\right ) \] Input:
int((6*x*exp(x)*log(x)+(-6*x*log(3)-6*x^2+24*x+6)*exp(x))/(2*x*exp(x)*log( x)+(-2*x*log(3)-2*x^2+10*x)*exp(x)-5*x),x)
Output:
3*log(2*e**x*log(x) - 2*e**x*log(3) - 2*e**x*x + 10*e**x - 5)