Integrand size = 137, antiderivative size = 24 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\frac {x \log (-8+x-x (4 x-\log (x)))}{e^2+x} \]
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\log \left (8-x+4 x^2-x \log (x)\right )-\frac {e^2 \log \left (-8+x-4 x^2+x \log (x)\right )}{e^2+x} \]
Integrate[(2*x^2 - 8*x^3 + E^2*(2*x - 8*x^2) + (E^2*x + x^2)*Log[x] + (E^2 *(-8 + x - 4*x^2) + E^2*x*Log[x])*Log[-8 + x - 4*x^2 + x*Log[x]])/(-8*x^2 + x^3 - 4*x^4 + E^4*(-8 + x - 4*x^2) + E^2*(-16*x + 2*x^2 - 8*x^3) + (E^4* x + 2*E^2*x^2 + x^3)*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.
\(\displaystyle \int \frac {-8 x^3+2 x^2+e^2 \left (2 x-8 x^2\right )+\left (x^2+e^2 x\right ) \log (x)+\left (e^2 \left (-4 x^2+x-8\right )+e^2 x \log (x)\right ) \log \left (-4 x^2+x+x \log (x)-8\right )}{-4 x^4+x^3-8 x^2+e^4 \left (-4 x^2+x-8\right )+e^2 \left (-8 x^3+2 x^2-16 x\right )+\left (x^3+2 e^2 x^2+e^4 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {8 x^3-2 x^2-e^2 \left (2 x-8 x^2\right )-\left (x^2+e^2 x\right ) \log (x)-\left (e^2 \left (-4 x^2+x-8\right )+e^2 x \log (x)\right ) \log \left (-4 x^2+x+x \log (x)-8\right )}{\left (x+e^2\right )^2 \left (4 x^2-x-x \log (x)+8\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 x^2}{\left (x+e^2\right )^2 \left (4 x^2-x-x \log (x)+8\right )}+\frac {2 e^2 (4 x-1) x}{\left (x+e^2\right )^2 \left (4 x^2-x-x \log (x)+8\right )}-\frac {x \log (x)}{\left (x+e^2\right ) \left (4 x^2-x-x \log (x)+8\right )}+\frac {e^2 \log \left (-4 x^2+x+x \log (x)-8\right )}{\left (x+e^2\right )^2}+\frac {8 x^3}{\left (x+e^2\right )^2 \left (4 x^2-x-x \log (x)+8\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (1+4 e^2\right ) \int \frac {1}{4 x^2-\log (x) x-x+8}dx-8 e^2 \int \frac {1}{4 x^2-\log (x) x-x+8}dx-2 \int \frac {1}{4 x^2-\log (x) x-x+8}dx+4 \int \frac {x}{4 x^2-\log (x) x-x+8}dx+2 e^4 \left (1+4 e^2\right ) \int \frac {1}{\left (x+e^2\right )^2 \left (4 x^2-\log (x) x-x+8\right )}dx-8 e^6 \int \frac {1}{\left (x+e^2\right )^2 \left (4 x^2-\log (x) x-x+8\right )}dx-2 e^4 \int \frac {1}{\left (x+e^2\right )^2 \left (4 x^2-\log (x) x-x+8\right )}dx-\left (8+e^2+4 e^4\right ) \int \frac {1}{\left (x+e^2\right ) \left (4 x^2-\log (x) x-x+8\right )}dx-2 e^2 \left (1+8 e^2\right ) \int \frac {1}{\left (x+e^2\right ) \left (4 x^2-\log (x) x-x+8\right )}dx+24 e^4 \int \frac {1}{\left (x+e^2\right ) \left (4 x^2-\log (x) x-x+8\right )}dx+4 e^2 \int \frac {1}{\left (x+e^2\right ) \left (4 x^2-\log (x) x-x+8\right )}dx+e^2 \int \frac {\log \left (-4 x^2+\log (x) x+x-8\right )}{\left (x+e^2\right )^2}dx+\log \left (x+e^2\right )\) |
Int[(2*x^2 - 8*x^3 + E^2*(2*x - 8*x^2) + (E^2*x + x^2)*Log[x] + (E^2*(-8 + x - 4*x^2) + E^2*x*Log[x])*Log[-8 + x - 4*x^2 + x*Log[x]])/(-8*x^2 + x^3 - 4*x^4 + E^4*(-8 + x - 4*x^2) + E^2*(-16*x + 2*x^2 - 8*x^3) + (E^4*x + 2* E^2*x^2 + x^3)*Log[x]),x]
3.16.93.3.1 Defintions of rubi rules used
Time = 23.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {\ln \left (x \ln \left (x \right )-4 x^{2}+x -8\right ) x}{x +{\mathrm e}^{2}}\) | \(22\) |
risch | \(-\frac {{\mathrm e}^{2} \ln \left (x \ln \left (x \right )-4 x^{2}+x -8\right )}{x +{\mathrm e}^{2}}+\ln \left (x \right )+\ln \left (\ln \left (x \right )-\frac {4 x^{2}-x +8}{x}\right )\) | \(46\) |
int(((x*exp(2)*ln(x)+(-4*x^2+x-8)*exp(2))*ln(x*ln(x)-4*x^2+x-8)+(exp(2)*x+ x^2)*ln(x)+(-8*x^2+2*x)*exp(2)-8*x^3+2*x^2)/((x*exp(2)^2+2*x^2*exp(2)+x^3) *ln(x)+(-4*x^2+x-8)*exp(2)^2+(-8*x^3+2*x^2-16*x)*exp(2)-4*x^4+x^3-8*x^2),x ,method=_RETURNVERBOSE)
Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\frac {x \log \left (-4 \, x^{2} + x \log \left (x\right ) + x - 8\right )}{x + e^{2}} \]
integrate(((x*exp(2)*log(x)+(-4*x^2+x-8)*exp(2))*log(x*log(x)-4*x^2+x-8)+( exp(2)*x+x^2)*log(x)+(-8*x^2+2*x)*exp(2)-8*x^3+2*x^2)/((x*exp(2)^2+2*x^2*e xp(2)+x^3)*log(x)+(-4*x^2+x-8)*exp(2)^2+(-8*x^3+2*x^2-16*x)*exp(2)-4*x^4+x ^3-8*x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x \right )} + \frac {- 4 x^{2} + x - 8}{x} \right )} - \frac {e^{2} \log {\left (- 4 x^{2} + x \log {\left (x \right )} + x - 8 \right )}}{x + e^{2}} \]
integrate(((x*exp(2)*ln(x)+(-4*x**2+x-8)*exp(2))*ln(x*ln(x)-4*x**2+x-8)+(e xp(2)*x+x**2)*ln(x)+(-8*x**2+2*x)*exp(2)-8*x**3+2*x**2)/((x*exp(2)**2+2*x* *2*exp(2)+x**3)*ln(x)+(-4*x**2+x-8)*exp(2)**2+(-8*x**3+2*x**2-16*x)*exp(2) -4*x**4+x**3-8*x**2),x)
log(x) + log(log(x) + (-4*x**2 + x - 8)/x) - exp(2)*log(-4*x**2 + x*log(x) + x - 8)/(x + exp(2))
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=-\frac {e^{2} \log \left (-4 \, x^{2} + x \log \left (x\right ) + x - 8\right )}{x + e^{2}} + \log \left (x\right ) + \log \left (-\frac {4 \, x^{2} - x \log \left (x\right ) - x + 8}{x}\right ) \]
integrate(((x*exp(2)*log(x)+(-4*x^2+x-8)*exp(2))*log(x*log(x)-4*x^2+x-8)+( exp(2)*x+x^2)*log(x)+(-8*x^2+2*x)*exp(2)-8*x^3+2*x^2)/((x*exp(2)^2+2*x^2*e xp(2)+x^3)*log(x)+(-4*x^2+x-8)*exp(2)^2+(-8*x^3+2*x^2-16*x)*exp(2)-4*x^4+x ^3-8*x^2),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\frac {x \log \left (-4 \, x^{2} + x \log \left (x\right ) + x - 8\right )}{x + e^{2}} \]
integrate(((x*exp(2)*log(x)+(-4*x^2+x-8)*exp(2))*log(x*log(x)-4*x^2+x-8)+( exp(2)*x+x^2)*log(x)+(-8*x^2+2*x)*exp(2)-8*x^3+2*x^2)/((x*exp(2)^2+2*x^2*e xp(2)+x^3)*log(x)+(-4*x^2+x-8)*exp(2)^2+(-8*x^3+2*x^2-16*x)*exp(2)-4*x^4+x ^3-8*x^2),x, algorithm=\
Time = 15.77 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\ln \left (\frac {x+x\,\ln \left (x\right )-4\,x^2-8}{x}\right )+\ln \left (x\right )-\frac {\ln \left (x+x\,\ln \left (x\right )-4\,x^2-8\right )\,{\mathrm {e}}^2}{x+{\mathrm {e}}^2} \]
int(-(exp(2)*(2*x - 8*x^2) - log(x + x*log(x) - 4*x^2 - 8)*(exp(2)*(4*x^2 - x + 8) - x*exp(2)*log(x)) + log(x)*(x*exp(2) + x^2) + 2*x^2 - 8*x^3)/(ex p(4)*(4*x^2 - x + 8) - log(x)*(x*exp(4) + 2*x^2*exp(2) + x^3) + exp(2)*(16 *x - 2*x^2 + 8*x^3) + 8*x^2 - x^3 + 4*x^4),x)