Integrand size = 121, antiderivative size = 29 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2 \left (-1+2 x+\log \left (\left (1-e^x\right ) x \log (x)\right )\right )}{(2-x)^2 x^2} \] Output:
2*(ln(ln(x)*x*(1-exp(x)))+2*x-1)/x^2/(2-x)^2
Time = 0.78 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2 \left (-1+2 x+\log \left (-\left (\left (-1+e^x\right ) x \log (x)\right )\right )\right )}{(-2+x)^2 x^2} \] Input:
Integrate[(4 - 2*x + E^x*(-4 + 2*x) + (12 - 18*x + 12*x^2 + E^x*(-12 + 14* x - 10*x^2))*Log[x] + (-8 + E^x*(8 - 8*x) + 8*x)*Log[x]*Log[(x - E^x*x)*Lo g[x]])/((8*x^3 - 12*x^4 + 6*x^5 - x^6 + E^x*(-8*x^3 + 12*x^4 - 6*x^5 + x^6 ))*Log[x]),x]
Output:
(2*(-1 + 2*x + Log[-((-1 + E^x)*x*Log[x])]))/((-2 + x)^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.
| \(\displaystyle \int \frac {\left (12 x^2+e^x \left (-10 x^2+14 x-12\right )-18 x+12\right ) \log (x)-2 x+e^x (2 x-4)+\left (e^x (8-8 x)+8 x-8\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )+4}{\left (-x^6+6 x^5-12 x^4+8 x^3+e^x \left (x^6-6 x^5+12 x^4-8 x^3\right )\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (12 x^2+e^x \left (-10 x^2+14 x-12\right )-18 x+12\right ) \log (x)-2 x+e^x (2 x-4)+\left (e^x (8-8 x)+8 x-8\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )+4}{\left (1-e^x\right ) (2-x)^3 x^3 \log (x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2}{\left (e^x-1\right ) (x-2)^2 x^2}-\frac {2 \left (5 x^2 \log (x)-x-7 x \log (x)+4 x \log (x) \log \left (-\left (\left (e^x-1\right ) x \log (x)\right )\right )+6 \log (x)-4 \log (x) \log \left (-\left (\left (e^x-1\right ) x \log (x)\right )\right )+2\right )}{(x-2)^3 x^3 \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \int \frac {1}{(x-2)^3 x^3 \log (x)}dx-2 \int \frac {1}{(x-2)^2 x^3 \log (x)}dx+2 \int \frac {1}{(x-2)^3 x^2 \log (x)}dx-\frac {1}{2 x^2}+\frac {2 \log \left (\left (1-e^x\right ) x \log (x)\right )}{(2-x)^2 x^2}+\frac {1}{2 (2-x)}+\frac {1}{2 x}+\frac {3}{2 (2-x)^2}\) |
Input:
Int[(4 - 2*x + E^x*(-4 + 2*x) + (12 - 18*x + 12*x^2 + E^x*(-12 + 14*x - 10 *x^2))*Log[x] + (-8 + E^x*(8 - 8*x) + 8*x)*Log[x]*Log[(x - E^x*x)*Log[x]]) /((8*x^3 - 12*x^4 + 6*x^5 - x^6 + E^x*(-8*x^3 + 12*x^4 - 6*x^5 + x^6))*Log [x]),x]
Output:
$Aborted
Time = 51.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
| method | result | size |
| parallelrisch | \(\frac {-16+32 x +16 \ln \left (-x \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )\right )}{8 x^{2} \left (x^{2}-4 x +4\right )}\) | \(33\) |
| risch | \(\frac {2 \ln \left ({\mathrm e}^{x}-1\right )}{\left (x^{2}-4 x +4\right ) x^{2}}+\frac {i \pi {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{2} \operatorname {csgn}\left (i x \right )+2 i \pi -i \pi {\operatorname {csgn}\left (i \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{3}+i \pi {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{3}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-1\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )-i \pi \,\operatorname {csgn}\left (i \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right ) \operatorname {csgn}\left (i x \right )+i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-1\right )\right ) {\operatorname {csgn}\left (i \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right ) {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{2}-2 i \pi {\operatorname {csgn}\left (i x \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{2}+i \pi {\operatorname {csgn}\left (i \ln \left (x \right ) \left ({\mathrm e}^{x}-1\right )\right )}^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )-2+4 x +2 \ln \left (x \right )+2 \ln \left (\ln \left (x \right )\right )}{\left (x^{2}-4 x +4\right ) x^{2}}\) | \(257\) |
Input:
int((((-8*x+8)*exp(x)+8*x-8)*ln(x)*ln((x-exp(x)*x)*ln(x))+((-10*x^2+14*x-1 2)*exp(x)+12*x^2-18*x+12)*ln(x)+(2*x-4)*exp(x)+4-2*x)/((x^6-6*x^5+12*x^4-8 *x^3)*exp(x)-x^6+6*x^5-12*x^4+8*x^3)/ln(x),x,method=_RETURNVERBOSE)
Output:
1/8*(-16+32*x+16*ln(-x*(exp(x)-1)*ln(x)))/x^2/(x^2-4*x+4)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2 \, {\left (2 \, x + \log \left (-{\left (x e^{x} - x\right )} \log \left (x\right )\right ) - 1\right )}}{x^{4} - 4 \, x^{3} + 4 \, x^{2}} \] Input:
integrate((((-8*x+8)*exp(x)+8*x-8)*log(x)*log((x-exp(x)*x)*log(x))+((-10*x ^2+14*x-12)*exp(x)+12*x^2-18*x+12)*log(x)+(2*x-4)*exp(x)+4-2*x)/((x^6-6*x^ 5+12*x^4-8*x^3)*exp(x)-x^6+6*x^5-12*x^4+8*x^3)/log(x),x, algorithm="fricas ")
Output:
2*(2*x + log(-(x*e^x - x)*log(x)) - 1)/(x^4 - 4*x^3 + 4*x^2)
Time = 0.38 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=- \frac {2 - 4 x}{x^{4} - 4 x^{3} + 4 x^{2}} + \frac {2 \log {\left (\left (- x e^{x} + x\right ) \log {\left (x \right )} \right )}}{x^{4} - 4 x^{3} + 4 x^{2}} \] Input:
integrate((((-8*x+8)*exp(x)+8*x-8)*ln(x)*ln((x-exp(x)*x)*ln(x))+((-10*x**2 +14*x-12)*exp(x)+12*x**2-18*x+12)*ln(x)+(2*x-4)*exp(x)+4-2*x)/((x**6-6*x** 5+12*x**4-8*x**3)*exp(x)-x**6+6*x**5-12*x**4+8*x**3)/ln(x),x)
Output:
-(2 - 4*x)/(x**4 - 4*x**3 + 4*x**2) + 2*log((-x*exp(x) + x)*log(x))/(x**4 - 4*x**3 + 4*x**2)
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2 \, {\left (2 \, x + \log \left (x\right ) + \log \left (-e^{x} + 1\right ) + \log \left (\log \left (x\right )\right ) - 1\right )}}{x^{4} - 4 \, x^{3} + 4 \, x^{2}} \] Input:
integrate((((-8*x+8)*exp(x)+8*x-8)*log(x)*log((x-exp(x)*x)*log(x))+((-10*x ^2+14*x-12)*exp(x)+12*x^2-18*x+12)*log(x)+(2*x-4)*exp(x)+4-2*x)/((x^6-6*x^ 5+12*x^4-8*x^3)*exp(x)-x^6+6*x^5-12*x^4+8*x^3)/log(x),x, algorithm="maxima ")
Output:
2*(2*x + log(x) + log(-e^x + 1) + log(log(x)) - 1)/(x^4 - 4*x^3 + 4*x^2)
Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2 \, {\left (2 \, x + \log \left (-e^{x} \log \left (x\right ) + \log \left (x\right )\right ) + \log \left (x\right ) - 1\right )}}{x^{4} - 4 \, x^{3} + 4 \, x^{2}} \] Input:
integrate((((-8*x+8)*exp(x)+8*x-8)*log(x)*log((x-exp(x)*x)*log(x))+((-10*x ^2+14*x-12)*exp(x)+12*x^2-18*x+12)*log(x)+(2*x-4)*exp(x)+4-2*x)/((x^6-6*x^ 5+12*x^4-8*x^3)*exp(x)-x^6+6*x^5-12*x^4+8*x^3)/log(x),x, algorithm="giac")
Output:
2*(2*x + log(-e^x*log(x) + log(x)) + log(x) - 1)/(x^4 - 4*x^3 + 4*x^2)
Time = 4.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2\,\left (2\,x+\ln \left (\ln \left (x\right )\,\left (x-x\,{\mathrm {e}}^x\right )\right )-1\right )}{x^2\,{\left (x-2\right )}^2} \] Input:
int((2*x + log(x)*(18*x + exp(x)*(10*x^2 - 14*x + 12) - 12*x^2 - 12) - exp (x)*(2*x - 4) + log(log(x)*(x - x*exp(x)))*log(x)*(exp(x)*(8*x - 8) - 8*x + 8) - 4)/(log(x)*(exp(x)*(8*x^3 - 12*x^4 + 6*x^5 - x^6) - 8*x^3 + 12*x^4 - 6*x^5 + x^6)),x)
Output:
(2*(2*x + log(log(x)*(x - x*exp(x))) - 1))/(x^2*(x - 2)^2)
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {4-2 x+e^x (-4+2 x)+\left (12-18 x+12 x^2+e^x \left (-12+14 x-10 x^2\right )\right ) \log (x)+\left (-8+e^x (8-8 x)+8 x\right ) \log (x) \log \left (\left (x-e^x x\right ) \log (x)\right )}{\left (8 x^3-12 x^4+6 x^5-x^6+e^x \left (-8 x^3+12 x^4-6 x^5+x^6\right )\right ) \log (x)} \, dx=\frac {2 \,\mathrm {log}\left (-e^{x} \mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right ) x \right )+4 x -2}{x^{2} \left (x^{2}-4 x +4\right )} \] Input:
int((((-8*x+8)*exp(x)+8*x-8)*log(x)*log((x-exp(x)*x)*log(x))+((-10*x^2+14* x-12)*exp(x)+12*x^2-18*x+12)*log(x)+(2*x-4)*exp(x)+4-2*x)/((x^6-6*x^5+12*x ^4-8*x^3)*exp(x)-x^6+6*x^5-12*x^4+8*x^3)/log(x),x)
Output:
(2*(log( - e**x*log(x)*x + log(x)*x) + 2*x - 1))/(x**2*(x**2 - 4*x + 4))