Integrand size = 109, antiderivative size = 29 \[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=x+\left (3+\frac {1}{x}\right ) \left (\log (2)-\log ^2\left (\frac {x}{-4 e^{2 x}+x}\right )\right ) \]
Time = 0.50 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=x+\frac {\log (2)}{x}+\frac {(-1-3 x) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{x} \]
Integrate[(-x^3 + E^(2*x)*(4*x^2 - 4*Log[2]) + x*Log[2] + E^(2*x)*(-8 - 8* x + 48*x^2)*Log[-(x/(4*E^(2*x) - x))] + (4*E^(2*x) - x)*Log[-(x/(4*E^(2*x) - x))]^2)/(4*E^(2*x)*x^2 - x^3),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.
\(\displaystyle \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+e^{2 x} \left (48 x^2-8 x-8\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )+x \log (2)}{4 e^{2 x} x^2-x^3} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2+12 x^2 \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )-2 x \log \left (-\frac {x}{4 e^{2 x}-x}\right )-2 \log \left (-\frac {x}{4 e^{2 x}-x}\right )-\log (2)}{x^2}+\frac {2 \left (6 x^2-x-1\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )}{\left (4 e^{2 x}-x\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {e^{2 x}}{\left (4 e^{2 x}-x\right ) x^2}dx-16 \int \frac {e^{2 x} \int \frac {1}{4 e^{2 x} x-x^2}dx}{4 e^{2 x}-x}dx+8 \int \frac {e^{2 x} \int \frac {1}{4 e^{2 x} x-x^2}dx}{\left (4 e^{2 x}-x\right ) x}dx+\int \frac {\log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{x^2}dx-48 \int \frac {e^{2 x}}{4 e^{2 x}-x}dx+16 \int \frac {e^{2 x}}{\left (4 e^{2 x}-x\right ) x}dx+96 \int \frac {e^{2 x} x}{4 e^{2 x}-x}dx-16 \int \frac {e^{2 x} \int \frac {1}{4 e^{2 x}-x}dx}{4 e^{2 x}-x}dx+8 \int \frac {e^{2 x} \int \frac {1}{4 e^{2 x}-x}dx}{\left (4 e^{2 x}-x\right ) x}dx+96 \int \frac {e^{2 x} \int \frac {x}{4 e^{2 x}-x}dx}{4 e^{2 x}-x}dx-48 \int \frac {e^{2 x} \int \frac {x}{4 e^{2 x}-x}dx}{\left (4 e^{2 x}-x\right ) x}dx-2 \log \left (-\frac {x}{4 e^{2 x}-x}\right ) \int \frac {1}{4 e^{2 x}-x}dx-2 \log \left (-\frac {x}{4 e^{2 x}-x}\right ) \int \frac {1}{\left (4 e^{2 x}-x\right ) x}dx+12 \log \left (-\frac {x}{4 e^{2 x}-x}\right ) \int \frac {x}{4 e^{2 x}-x}dx-2 \int \frac {\log \left (-\frac {x}{4 e^{2 x}-x}\right )}{x}dx+x+12 x \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\frac {2 \log \left (-\frac {x}{4 e^{2 x}-x}\right )}{x}+\frac {\log (2)}{x}\) |
Int[(-x^3 + E^(2*x)*(4*x^2 - 4*Log[2]) + x*Log[2] + E^(2*x)*(-8 - 8*x + 48 *x^2)*Log[-(x/(4*E^(2*x) - x))] + (4*E^(2*x) - x)*Log[-(x/(4*E^(2*x) - x)) ]^2)/(4*E^(2*x)*x^2 - x^3),x]
3.18.86.3.1 Defintions of rubi rules used
Time = 0.48 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79
method | result | size |
parallelrisch | \(\frac {-3 \ln \left (-\frac {x}{4 \,{\mathrm e}^{2 x}-x}\right )^{2} x +x^{2}-\ln \left (-\frac {x}{4 \,{\mathrm e}^{2 x}-x}\right )^{2}+\ln \left (2\right )}{x}\) | \(52\) |
risch | \(\text {Expression too large to display}\) | \(934\) |
int(((4*exp(2*x)-x)*ln(-x/(4*exp(2*x)-x))^2+(48*x^2-8*x-8)*exp(2*x)*ln(-x/ (4*exp(2*x)-x))+(-4*ln(2)+4*x^2)*exp(2*x)+x*ln(2)-x^3)/(4*exp(2*x)*x^2-x^3 ),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=-\frac {{\left (3 \, x + 1\right )} \log \left (\frac {x}{x - 4 \, e^{\left (2 \, x\right )}}\right )^{2} - x^{2} - \log \left (2\right )}{x} \]
integrate(((4*exp(2*x)-x)*log(-x/(4*exp(2*x)-x))^2+(48*x^2-8*x-8)*exp(2*x) *log(-x/(4*exp(2*x)-x))+(-4*log(2)+4*x^2)*exp(2*x)+x*log(2)-x^3)/(4*exp(2* x)*x^2-x^3),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=x + \frac {\left (- 3 x - 1\right ) \log {\left (- \frac {x}{- x + 4 e^{2 x}} \right )}^{2}}{x} + \frac {\log {\left (2 \right )}}{x} \]
integrate(((4*exp(2*x)-x)*ln(-x/(4*exp(2*x)-x))**2+(48*x**2-8*x-8)*exp(2*x )*ln(-x/(4*exp(2*x)-x))+(-4*ln(2)+4*x**2)*exp(2*x)+x*ln(2)-x**3)/(4*exp(2* x)*x**2-x**3),x)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (29) = 58\).
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.07 \[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=-\frac {{\left (3 \, x + 1\right )} \log \left (x - 4 \, e^{\left (2 \, x\right )}\right )^{2} - 2 \, {\left (3 \, x + 1\right )} \log \left (x - 4 \, e^{\left (2 \, x\right )}\right ) \log \left (x\right ) + {\left (3 \, x + 1\right )} \log \left (x\right )^{2} - x^{2} - \log \left (2\right )}{x} \]
integrate(((4*exp(2*x)-x)*log(-x/(4*exp(2*x)-x))^2+(48*x^2-8*x-8)*exp(2*x) *log(-x/(4*exp(2*x)-x))+(-4*log(2)+4*x^2)*exp(2*x)+x*log(2)-x^3)/(4*exp(2* x)*x^2-x^3),x, algorithm=\
-((3*x + 1)*log(x - 4*e^(2*x))^2 - 2*(3*x + 1)*log(x - 4*e^(2*x))*log(x) + (3*x + 1)*log(x)^2 - x^2 - log(2))/x
\[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=\int { \frac {x^{3} - 8 \, {\left (6 \, x^{2} - x - 1\right )} e^{\left (2 \, x\right )} \log \left (\frac {x}{x - 4 \, e^{\left (2 \, x\right )}}\right ) + {\left (x - 4 \, e^{\left (2 \, x\right )}\right )} \log \left (\frac {x}{x - 4 \, e^{\left (2 \, x\right )}}\right )^{2} - 4 \, {\left (x^{2} - \log \left (2\right )\right )} e^{\left (2 \, x\right )} - x \log \left (2\right )}{x^{3} - 4 \, x^{2} e^{\left (2 \, x\right )}} \,d x } \]
integrate(((4*exp(2*x)-x)*log(-x/(4*exp(2*x)-x))^2+(48*x^2-8*x-8)*exp(2*x) *log(-x/(4*exp(2*x)-x))+(-4*log(2)+4*x^2)*exp(2*x)+x*log(2)-x^3)/(4*exp(2* x)*x^2-x^3),x, algorithm=\
integrate((x^3 - 8*(6*x^2 - x - 1)*e^(2*x)*log(x/(x - 4*e^(2*x))) + (x - 4 *e^(2*x))*log(x/(x - 4*e^(2*x)))^2 - 4*(x^2 - log(2))*e^(2*x) - x*log(2))/ (x^3 - 4*x^2*e^(2*x)), x)
Time = 14.95 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-x^3+e^{2 x} \left (4 x^2-4 \log (2)\right )+x \log (2)+e^{2 x} \left (-8-8 x+48 x^2\right ) \log \left (-\frac {x}{4 e^{2 x}-x}\right )+\left (4 e^{2 x}-x\right ) \log ^2\left (-\frac {x}{4 e^{2 x}-x}\right )}{4 e^{2 x} x^2-x^3} \, dx=x-{\ln \left (\frac {x}{x-4\,{\mathrm {e}}^{2\,x}}\right )}^2\,\left (\frac {1}{x}+3\right )+\frac {\ln \left (2\right )}{x} \]