Integrand size = 99, antiderivative size = 26 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {\left (2+\frac {1}{x}\right )^2}{\log ^2\left (\frac {4 e^x x}{-e^x+x}\right )} \] Output:
(2+1/x)^2/ln(4*x*exp(x)/(x-exp(x)))^2
Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {(1+2 x)^2}{x^2 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )} \] Input:
Integrate[(2*x^2 + 8*x^3 + 8*x^4 + E^x*(-2 - 8*x - 8*x^2) + (E^x*(-2 - 4*x ) + 2*x + 4*x^2)*Log[(-4*E^x*x)/(E^x - x)])/((E^x*x^3 - x^4)*Log[(-4*E^x*x )/(E^x - x)]^3),x]
Output:
(1 + 2*x)^2/(x^2*Log[(-4*E^x*x)/(E^x - 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.
| \(\displaystyle \int \frac {8 x^4+8 x^3+2 x^2+e^x \left (-8 x^2-8 x-2\right )+\left (4 x^2+2 x+e^x (-4 x-2)\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 (2 x+1) \left (2 x^3+x^2-2 e^x x-e^x+x \log \left (-\frac {4 e^x x}{e^x-x}\right )-e^x \log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int -\frac {(2 x+1) \left (-2 x^3-x^2+2 e^x x-\log \left (-\frac {4 e^x x}{e^x-x}\right ) x+e^x+e^x \log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {(2 x+1) \left (-2 x^3-x^2+2 e^x x-\log \left (-\frac {4 e^x x}{e^x-x}\right ) x+e^x+e^x \log \left (-\frac {4 e^x x}{e^x-x}\right )\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 \int \left (\frac {(2 x+1) \left (2 x+\log \left (-\frac {4 e^x x}{e^x-x}\right )+1\right )}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}-\frac {(x-1) (2 x+1)^2}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \left (\int \frac {1}{x^3 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx+\int \frac {1}{x^3 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )}dx+4 \int \frac {1}{x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx+\int \frac {1}{\left (e^x-x\right ) x^2 \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx+2 \int \frac {1}{x^2 \log ^2\left (-\frac {4 e^x x}{e^x-x}\right )}dx+4 \int \frac {1}{x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx+3 \int \frac {1}{\left (e^x-x\right ) x \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx-4 \int \frac {x}{\left (e^x-x\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )}dx\right )\) |
Input:
Int[(2*x^2 + 8*x^3 + 8*x^4 + E^x*(-2 - 8*x - 8*x^2) + (E^x*(-2 - 4*x) + 2* x + 4*x^2)*Log[(-4*E^x*x)/(E^x - x)])/((E^x*x^3 - x^4)*Log[(-4*E^x*x)/(E^x - x)]^3),x]
Output:
$Aborted
Time = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23
| method | result | size |
| parallelrisch | \(-\frac {-8 x^{2}-8 x -2}{2 x^{2} \ln \left (-\frac {4 x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}}\) | \(32\) |
| risch | \(-\frac {4 \left (4 x^{2}+4 x +1\right )}{x^{2} {\left (\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{3}+\pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-x}\right )^{3}+2 i \ln \left (x \right )-2 i \ln \left (x -{\mathrm e}^{x}\right )+4 i \ln \left (2\right )+2 i \ln \left ({\mathrm e}^{x}\right )\right )}^{2}}\) | \(267\) |
Input:
int((((-4*x-2)*exp(x)+4*x^2+2*x)*ln(-4*x*exp(x)/(exp(x)-x))+(-8*x^2-8*x-2) *exp(x)+8*x^4+8*x^3+2*x^2)/(exp(x)*x^3-x^4)/ln(-4*x*exp(x)/(exp(x)-x))^3,x ,method=_RETURNVERBOSE)
Output:
-1/2*(-8*x^2-8*x-2)/x^2/ln(-4*x*exp(x)/(exp(x)-x))^2
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {4 \, x^{2} + 4 \, x + 1}{x^{2} \log \left (\frac {4 \, x e^{x}}{x - e^{x}}\right )^{2}} \] Input:
integrate((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2 -8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(exp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x) -x))^3,x, algorithm="fricas")
Output:
(4*x^2 + 4*x + 1)/(x^2*log(4*x*e^x/(x - e^x))^2)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {4 x^{2} + 4 x + 1}{x^{2} \log {\left (- \frac {4 x e^{x}}{- x + e^{x}} \right )}^{2}} \] Input:
integrate((((-4*x-2)*exp(x)+4*x**2+2*x)*ln(-4*x*exp(x)/(exp(x)-x))+(-8*x** 2-8*x-2)*exp(x)+8*x**4+8*x**3+2*x**2)/(exp(x)*x**3-x**4)/ln(-4*x*exp(x)/(e xp(x)-x))**3,x)
Output:
(4*x**2 + 4*x + 1)/(x**2*log(-4*x*exp(x)/(-x + exp(x)))**2)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (24) = 48\).
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.65 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {4 \, x^{2} + 4 \, x + 1}{x^{4} + 4 \, x^{3} \log \left (2\right ) + 4 \, x^{2} \log \left (2\right )^{2} + x^{2} \log \left (x - e^{x}\right )^{2} + x^{2} \log \left (x\right )^{2} - 2 \, {\left (x^{3} + 2 \, x^{2} \log \left (2\right ) + x^{2} \log \left (x\right )\right )} \log \left (x - e^{x}\right ) + 2 \, {\left (x^{3} + 2 \, x^{2} \log \left (2\right )\right )} \log \left (x\right )} \] Input:
integrate((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2 -8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(exp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x) -x))^3,x, algorithm="maxima")
Output:
(4*x^2 + 4*x + 1)/(x^4 + 4*x^3*log(2) + 4*x^2*log(2)^2 + x^2*log(x - e^x)^ 2 + x^2*log(x)^2 - 2*(x^3 + 2*x^2*log(2) + x^2*log(x))*log(x - e^x) + 2*(x ^3 + 2*x^2*log(2))*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {4 \, x^{2} + 4 \, x + 1}{x^{4} + 2 \, x^{3} \log \left (\frac {4 \, x}{x - e^{x}}\right ) + x^{2} \log \left (\frac {4 \, x}{x - e^{x}}\right )^{2}} \] Input:
integrate((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2 -8*x-2)*exp(x)+8*x^4+8*x^3+2*x^2)/(exp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x) -x))^3,x, algorithm="giac")
Output:
(4*x^2 + 4*x + 1)/(x^4 + 2*x^3*log(4*x/(x - e^x)) + x^2*log(4*x/(x - e^x)) ^2)
Time = 4.24 (sec) , antiderivative size = 377, normalized size of antiderivative = 14.50 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {\frac {\left (x-{\mathrm {e}}^x\right )\,\left (2\,x+1\right )}{x^2\,\left ({\mathrm {e}}^x-x^2\right )}+\frac {\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )\,\left (x-{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^{2\,x}-5\,x^2\,{\mathrm {e}}^x-7\,x^3\,{\mathrm {e}}^x+2\,x^4\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x+3\,x^3+4\,x^4\right )}{x^2\,{\left ({\mathrm {e}}^x-x^2\right )}^3}}{\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )}+\frac {\frac {{\left (2\,x+1\right )}^2}{x^2}-\frac {\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )\,\left (x-{\mathrm {e}}^x\right )\,\left (2\,x+1\right )}{x^2\,\left ({\mathrm {e}}^x-x^2\right )}}{{\ln \left (\frac {4\,x\,{\mathrm {e}}^x}{x-{\mathrm {e}}^x}\right )}^2}+\frac {2\,x+2}{x^2}+\frac {-4\,x^6+18\,x^5-23\,x^4+4\,x^3+3\,x^2+2\,x}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{2\,x}-2\,x^2\,{\mathrm {e}}^x+x^4\right )}-\frac {2\,x^4-5\,x^3+x^2-x+6}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^x-x^2\right )}-\frac {2\,x^8-11\,x^7+20\,x^6-11\,x^5-4\,x^4+4\,x^3}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{3\,x}+3\,x^4\,{\mathrm {e}}^x-3\,x^2\,{\mathrm {e}}^{2\,x}-x^6\right )} \] Input:
int((log((4*x*exp(x))/(x - exp(x)))*(2*x - exp(x)*(4*x + 2) + 4*x^2) - exp (x)*(8*x + 8*x^2 + 2) + 2*x^2 + 8*x^3 + 8*x^4)/(log((4*x*exp(x))/(x - exp( x)))^3*(x^3*exp(x) - x^4)),x)
Output:
(((x - exp(x))*(2*x + 1))/(x^2*(exp(x) - x^2)) + (log((4*x*exp(x))/(x - ex p(x)))*(x - exp(x))*(2*exp(2*x) + 2*x*exp(2*x) - 5*x^2*exp(x) - 7*x^3*exp( x) + 2*x^4*exp(x) - x*exp(x) + 3*x^3 + 4*x^4))/(x^2*(exp(x) - x^2)^3))/log ((4*x*exp(x))/(x - exp(x))) + ((2*x + 1)^2/x^2 - (log((4*x*exp(x))/(x - ex p(x)))*(x - exp(x))*(2*x + 1))/(x^2*(exp(x) - x^2)))/log((4*x*exp(x))/(x - exp(x)))^2 + (2*x + 2)/x^2 + (2*x + 3*x^2 + 4*x^3 - 23*x^4 + 18*x^5 - 4*x ^6)/((2*x - x^2)*(exp(2*x) - 2*x^2*exp(x) + x^4)) - (x^2 - x - 5*x^3 + 2*x ^4 + 6)/((2*x - x^2)*(exp(x) - x^2)) - (4*x^3 - 4*x^4 - 11*x^5 + 20*x^6 - 11*x^7 + 2*x^8)/((2*x - x^2)*(exp(3*x) + 3*x^4*exp(x) - 3*x^2*exp(2*x) - x ^6))
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {2 x^2+8 x^3+8 x^4+e^x \left (-2-8 x-8 x^2\right )+\left (e^x (-2-4 x)+2 x+4 x^2\right ) \log \left (-\frac {4 e^x x}{e^x-x}\right )}{\left (e^x x^3-x^4\right ) \log ^3\left (-\frac {4 e^x x}{e^x-x}\right )} \, dx=\frac {4 x^{2}+4 x +1}{\mathrm {log}\left (-\frac {4 e^{x} x}{e^{x}-x}\right )^{2} x^{2}} \] Input:
int((((-4*x-2)*exp(x)+4*x^2+2*x)*log(-4*x*exp(x)/(exp(x)-x))+(-8*x^2-8*x-2 )*exp(x)+8*x^4+8*x^3+2*x^2)/(exp(x)*x^3-x^4)/log(-4*x*exp(x)/(exp(x)-x))^3 ,x)
Output:
(4*x**2 + 4*x + 1)/(log(( - 4*e**x*x)/(e**x - x))**2*x**2)