Integrand size = 96, antiderivative size = 23 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log ^2\left (x \left (-e^x+\frac {\log (2 x)}{5-2 x}\right )\right ) \] Output:
ln((ln(2*x)/(5-2*x)-exp(x))*x)^2
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log ^2\left (-e^x x-\frac {x \log (2 x)}{-5+2 x}\right ) \] Input:
Integrate[((-10 + 4*x + E^x*(50 + 10*x - 32*x^2 + 8*x^3) - 10*Log[2*x])*Lo g[(E^x*(5*x - 2*x^2) - x*Log[2*x])/(-5 + 2*x)])/(E^x*(25*x - 20*x^2 + 4*x^ 3) + (-5*x + 2*x^2)*Log[2*x]),x]
Output:
Log[-(E^x*x) - (x*Log[2*x])/(-5 + 2*x)]^2
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.
| \(\displaystyle \int \frac {\left (e^x \left (8 x^3-32 x^2+10 x+50\right )+4 x-10 \log (2 x)-10\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{2 x-5}\right )}{\left (2 x^2-5 x\right ) \log (2 x)+e^x \left (4 x^3-20 x^2+25 x\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (e^x \left (8 x^3-32 x^2+10 x+50\right )+4 x-10 \log (2 x)-10\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{2 x-5}\right )}{(5-2 x) x \left (-2 e^x x+5 e^x-\log (2 x)\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 (x+1) \log \left (-e^x x-\frac {x \log (2 x)}{2 x-5}\right )}{x}-\frac {2 \left (2 x^2 \log (2 x)-2 x-3 x \log (2 x)+5\right ) \log \left (-e^x x-\frac {x \log (2 x)}{2 x-5}\right )}{x (2 x-5) \left (2 e^x x-5 e^x+\log (2 x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int \log \left (-e^x x-\frac {\log (2 x) x}{2 x-5}\right )dx+2 \int \frac {\log \left (-e^x x-\frac {\log (2 x) x}{2 x-5}\right )}{x}dx+2 \int \frac {\log \left (-e^x x-\frac {\log (2 x) x}{2 x-5}\right )}{x \left (2 e^x x-5 e^x+\log (2 x)\right )}dx-2 \int \frac {\log (2 x) \log \left (-e^x x-\frac {\log (2 x) x}{2 x-5}\right )}{2 e^x x-5 e^x+\log (2 x)}dx-4 \int \frac {\log (2 x) \log \left (-e^x x-\frac {\log (2 x) x}{2 x-5}\right )}{(2 x-5) \left (2 e^x x-5 e^x+\log (2 x)\right )}dx\) |
Input:
Int[((-10 + 4*x + E^x*(50 + 10*x - 32*x^2 + 8*x^3) - 10*Log[2*x])*Log[(E^x *(5*x - 2*x^2) - x*Log[2*x])/(-5 + 2*x)])/(E^x*(25*x - 20*x^2 + 4*x^3) + ( -5*x + 2*x^2)*Log[2*x]),x]
Output:
$Aborted
Time = 14.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39
| method | result | size |
| parallelrisch | \({\ln \left (\frac {-x \ln \left (2 x \right )+\left (-2 x^{2}+5 x \right ) {\mathrm e}^{x}}{-5+2 x}\right )}^{2}\) | \(32\) |
Input:
int((-10*ln(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*ln((-x*ln(2*x)+(-2* x^2+5*x)*exp(x))/(-5+2*x))/((2*x^2-5*x)*ln(2*x)+(4*x^3-20*x^2+25*x)*exp(x) ),x,method=_RETURNVERBOSE)
Output:
ln((-x*ln(2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))^2
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log \left (-\frac {{\left (2 \, x^{2} - 5 \, x\right )} e^{x} + x \log \left (2 \, x\right )}{2 \, x - 5}\right )^{2} \] Input:
integrate((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log( 2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25 *x)*exp(x)),x, algorithm="fricas")
Output:
log(-((2*x^2 - 5*x)*e^x + x*log(2*x))/(2*x - 5))^2
Time = 0.89 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\log {\left (\frac {- x \log {\left (2 x \right )} + \left (- 2 x^{2} + 5 x\right ) e^{x}}{2 x - 5} \right )}^{2} \] Input:
integrate((-10*ln(2*x)+(8*x**3-32*x**2+10*x+50)*exp(x)+4*x-10)*ln((-x*ln(2 *x)+(-2*x**2+5*x)*exp(x))/(-5+2*x))/((2*x**2-5*x)*ln(2*x)+(4*x**3-20*x**2+ 25*x)*exp(x)),x)
Output:
log((-x*log(2*x) + (-2*x**2 + 5*x)*exp(x))/(2*x - 5))**2
\[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} + 2 \, x - 5 \, \log \left (2 \, x\right ) - 5\right )} \log \left (-\frac {{\left (2 \, x^{2} - 5 \, x\right )} e^{x} + x \log \left (2 \, x\right )}{2 \, x - 5}\right )}{{\left (4 \, x^{3} - 20 \, x^{2} + 25 \, x\right )} e^{x} + {\left (2 \, x^{2} - 5 \, x\right )} \log \left (2 \, x\right )} \,d x } \] Input:
integrate((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log( 2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25 *x)*exp(x)),x, algorithm="maxima")
Output:
2*integrate(((4*x^3 - 16*x^2 + 5*x + 25)*e^x + 2*x - 5*log(2*x) - 5)*log(- ((2*x^2 - 5*x)*e^x + x*log(2*x))/(2*x - 5))/((4*x^3 - 20*x^2 + 25*x)*e^x + (2*x^2 - 5*x)*log(2*x)), x)
\[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} + 2 \, x - 5 \, \log \left (2 \, x\right ) - 5\right )} \log \left (-\frac {{\left (2 \, x^{2} - 5 \, x\right )} e^{x} + x \log \left (2 \, x\right )}{2 \, x - 5}\right )}{{\left (4 \, x^{3} - 20 \, x^{2} + 25 \, x\right )} e^{x} + {\left (2 \, x^{2} - 5 \, x\right )} \log \left (2 \, x\right )} \,d x } \] Input:
integrate((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log( 2*x)+(-2*x^2+5*x)*exp(x))/(-5+2*x))/((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25 *x)*exp(x)),x, algorithm="giac")
Output:
integrate(2*((4*x^3 - 16*x^2 + 5*x + 25)*e^x + 2*x - 5*log(2*x) - 5)*log(- ((2*x^2 - 5*x)*e^x + x*log(2*x))/(2*x - 5))/((4*x^3 - 20*x^2 + 25*x)*e^x + (2*x^2 - 5*x)*log(2*x)), x)
Time = 3.77 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx={\ln \left (-\frac {x\,\ln \left (2\,x\right )-{\mathrm {e}}^x\,\left (5\,x-2\,x^2\right )}{2\,x-5}\right )}^2 \] Input:
int(-(log(-(x*log(2*x) - exp(x)*(5*x - 2*x^2))/(2*x - 5))*(4*x - 10*log(2* x) + exp(x)*(10*x - 32*x^2 + 8*x^3 + 50) - 10))/(log(2*x)*(5*x - 2*x^2) - exp(x)*(25*x - 20*x^2 + 4*x^3)),x)
Output:
log(-(x*log(2*x) - exp(x)*(5*x - 2*x^2))/(2*x - 5))^2
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {\left (-10+4 x+e^x \left (50+10 x-32 x^2+8 x^3\right )-10 \log (2 x)\right ) \log \left (\frac {e^x \left (5 x-2 x^2\right )-x \log (2 x)}{-5+2 x}\right )}{e^x \left (25 x-20 x^2+4 x^3\right )+\left (-5 x+2 x^2\right ) \log (2 x)} \, dx=\mathrm {log}\left (\frac {-2 e^{x} x^{2}+5 e^{x} x -\mathrm {log}\left (2 x \right ) x}{2 x -5}\right )^{2} \] Input:
int((-10*log(2*x)+(8*x^3-32*x^2+10*x+50)*exp(x)+4*x-10)*log((-x*log(2*x)+( -2*x^2+5*x)*exp(x))/(-5+2*x))/((2*x^2-5*x)*log(2*x)+(4*x^3-20*x^2+25*x)*ex p(x)),x)
Output:
log(( - 2*e**x*x**2 + 5*e**x*x - log(2*x)*x)/(2*x - 5))**2