Integrand size = 117, antiderivative size = 32 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {3 \left (5-\frac {x+\log \left (e^x x^2\right )}{1-e^x-\log (x)}\right )}{x} \] Output:
3*(5-(ln(exp(x)*x^2)+x)/(1-exp(x)-ln(x)))/x
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {3 \left (-5+5 e^x+x+5 \log (x)+\log \left (e^x x^2\right )\right )}{x \left (-1+e^x+\log (x)\right )} \] Input:
Integrate[(-21 - 15*E^(2*x) - 6*x + E^x*(36 + 3*x - 3*x^2) + (36 - 30*E^x + 3*x)*Log[x] - 15*Log[x]^2 + (E^x*(-3 - 3*x) - 3*Log[x])*Log[E^x*x^2])/(x ^2 - 2*E^x*x^2 + E^(2*x)*x^2 + (-2*x^2 + 2*E^x*x^2)*Log[x] + x^2*Log[x]^2) ,x]
Output:
(3*(-5 + 5*E^x + x + 5*Log[x] + Log[E^x*x^2]))/(x*(-1 + E^x + Log[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.
| \(\displaystyle \int \frac {e^x \left (-3 x^2+3 x+36\right )+\left (e^x (-3 x-3)-3 \log (x)\right ) \log \left (e^x x^2\right )-15 e^{2 x}-6 x-15 \log ^2(x)+\left (3 x-30 e^x+36\right ) \log (x)-21}{-2 e^x x^2+e^{2 x} x^2+x^2+x^2 \log ^2(x)+\left (2 e^x x^2-2 x^2\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^x \left (-3 x^2+3 x+36\right )+\left (e^x (-3 x-3)-3 \log (x)\right ) \log \left (e^x x^2\right )-15 e^{2 x}-6 x-15 \log ^2(x)+\left (3 x-30 e^x+36\right ) \log (x)-21}{x^2 \left (-e^x-\log (x)+1\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {15}{x^2}-\frac {3 (x+1) \left (\log \left (e^x x^2\right )+x-2\right )}{x^2 \left (e^x+\log (x)-1\right )}+\frac {3 (-x+x \log (x)-1) \left (\log \left (e^x x^2\right )+x\right )}{x^2 \left (e^x+\log (x)-1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 6 \int \frac {1}{x^2 \left (\log (x)+e^x-1\right )}dx-3 \int \frac {\log \left (e^x x^2\right )}{x^2 \left (\log (x)+e^x-1\right )^2}dx-3 \int \frac {\log \left (e^x x^2\right )}{x \left (\log (x)+e^x-1\right )^2}dx+3 \int \frac {\log (x) \log \left (e^x x^2\right )}{x \left (\log (x)+e^x-1\right )^2}dx-3 \int \frac {\log \left (e^x x^2\right )}{x^2 \left (\log (x)+e^x-1\right )}dx-3 \int \frac {\log \left (e^x x^2\right )}{x \left (\log (x)+e^x-1\right )}dx-3 \int \frac {1}{\left (\log (x)+e^x-1\right )^2}dx-3 \int \frac {1}{x \left (\log (x)+e^x-1\right )^2}dx+3 \int \frac {\log (x)}{\left (\log (x)+e^x-1\right )^2}dx-3 \int \frac {1}{\log (x)+e^x-1}dx+3 \int \frac {1}{x \left (\log (x)+e^x-1\right )}dx+\frac {15}{x}\) |
Input:
Int[(-21 - 15*E^(2*x) - 6*x + E^x*(36 + 3*x - 3*x^2) + (36 - 30*E^x + 3*x) *Log[x] - 15*Log[x]^2 + (E^x*(-3 - 3*x) - 3*Log[x])*Log[E^x*x^2])/(x^2 - 2 *E^x*x^2 + E^(2*x)*x^2 + (-2*x^2 + 2*E^x*x^2)*Log[x] + x^2*Log[x]^2),x]
Output:
$Aborted
Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12
| method | result | size |
| parallelrisch | \(\frac {-60+12 x +12 \ln \left ({\mathrm e}^{x} x^{2}\right )+60 \ln \left (x \right )+60 \,{\mathrm e}^{x}}{4 x \left (-1+\ln \left (x \right )+{\mathrm e}^{x}\right )}\) | \(36\) |
| risch | \(\frac {3 \ln \left ({\mathrm e}^{x}\right )}{x \left (-1+\ln \left (x \right )+{\mathrm e}^{x}\right )}+\frac {-\frac {3 i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right )^{3}}{2}+\frac {3 i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )}{2}+\frac {3 i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {3 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {3 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}+3 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-\frac {3 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+3 x +15 \,{\mathrm e}^{x}+21 \ln \left (x \right )-15}{x \left (-1+\ln \left (x \right )+{\mathrm e}^{x}\right )}\) | \(177\) |
Input:
int(((-3*ln(x)+(-3*x-3)*exp(x))*ln(exp(x)*x^2)-15*ln(x)^2+(-30*exp(x)+3*x+ 36)*ln(x)-15*exp(x)^2+(-3*x^2+3*x+36)*exp(x)-6*x-21)/(x^2*ln(x)^2+(2*exp(x )*x^2-2*x^2)*ln(x)+exp(x)^2*x^2-2*exp(x)*x^2+x^2),x,method=_RETURNVERBOSE)
Output:
1/4/x*(-60+12*x+12*ln(exp(x)*x^2)+60*ln(x)+60*exp(x))/(-1+ln(x)+exp(x))
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {3 \, {\left (2 \, x + 5 \, e^{x} + 7 \, \log \left (x\right ) - 5\right )}}{x e^{x} + x \log \left (x\right ) - x} \] Input:
integrate(((-3*log(x)+(-3*x-3)*exp(x))*log(exp(x)*x^2)-15*log(x)^2+(-30*ex p(x)+3*x+36)*log(x)-15*exp(x)^2+(-3*x^2+3*x+36)*exp(x)-6*x-21)/(x^2*log(x) ^2+(2*exp(x)*x^2-2*x^2)*log(x)+exp(x)^2*x^2-2*exp(x)*x^2+x^2),x, algorithm ="fricas")
Output:
3*(2*x + 5*e^x + 7*log(x) - 5)/(x*e^x + x*log(x) - x)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {6 x + 6 \log {\left (x \right )}}{x e^{x} + x \log {\left (x \right )} - x} + \frac {15}{x} \] Input:
integrate(((-3*ln(x)+(-3*x-3)*exp(x))*ln(exp(x)*x**2)-15*ln(x)**2+(-30*exp (x)+3*x+36)*ln(x)-15*exp(x)**2+(-3*x**2+3*x+36)*exp(x)-6*x-21)/(x**2*ln(x) **2+(2*exp(x)*x**2-2*x**2)*ln(x)+exp(x)**2*x**2-2*exp(x)*x**2+x**2),x)
Output:
(6*x + 6*log(x))/(x*exp(x) + x*log(x) - x) + 15/x
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {3 \, {\left (2 \, x + 5 \, e^{x} + 7 \, \log \left (x\right ) - 5\right )}}{x e^{x} + x \log \left (x\right ) - x} \] Input:
integrate(((-3*log(x)+(-3*x-3)*exp(x))*log(exp(x)*x^2)-15*log(x)^2+(-30*ex p(x)+3*x+36)*log(x)-15*exp(x)^2+(-3*x^2+3*x+36)*exp(x)-6*x-21)/(x^2*log(x) ^2+(2*exp(x)*x^2-2*x^2)*log(x)+exp(x)^2*x^2-2*exp(x)*x^2+x^2),x, algorithm ="maxima")
Output:
3*(2*x + 5*e^x + 7*log(x) - 5)/(x*e^x + x*log(x) - x)
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {3 \, {\left (2 \, x + 5 \, e^{x} + 7 \, \log \left (x\right ) - 5\right )}}{x e^{x} + x \log \left (x\right ) - x} \] Input:
integrate(((-3*log(x)+(-3*x-3)*exp(x))*log(exp(x)*x^2)-15*log(x)^2+(-30*ex p(x)+3*x+36)*log(x)-15*exp(x)^2+(-3*x^2+3*x+36)*exp(x)-6*x-21)/(x^2*log(x) ^2+(2*exp(x)*x^2-2*x^2)*log(x)+exp(x)^2*x^2-2*exp(x)*x^2+x^2),x, algorithm ="giac")
Output:
3*(2*x + 5*e^x + 7*log(x) - 5)/(x*e^x + x*log(x) - x)
Time = 1.68 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {3\,\left (2\,x+\ln \left (x^2\right )+5\,{\mathrm {e}}^x+5\,\ln \left (x\right )-5\right )}{x\,\left ({\mathrm {e}}^x+\ln \left (x\right )-1\right )} \] Input:
int(-(6*x + 15*exp(2*x) + 15*log(x)^2 - exp(x)*(3*x - 3*x^2 + 36) - log(x) *(3*x - 30*exp(x) + 36) + log(x^2*exp(x))*(3*log(x) + exp(x)*(3*x + 3)) + 21)/(x^2*exp(2*x) - 2*x^2*exp(x) + x^2*log(x)^2 + x^2 + log(x)*(2*x^2*exp( x) - 2*x^2)),x)
Output:
(3*(2*x + log(x^2) + 5*exp(x) + 5*log(x) - 5))/(x*(exp(x) + log(x) - 1))
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-21-15 e^{2 x}-6 x+e^x \left (36+3 x-3 x^2\right )+\left (36-30 e^x+3 x\right ) \log (x)-15 \log ^2(x)+\left (e^x (-3-3 x)-3 \log (x)\right ) \log \left (e^x x^2\right )}{x^2-2 e^x x^2+e^{2 x} x^2+\left (-2 x^2+2 e^x x^2\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {15 e^{x}+3 \,\mathrm {log}\left (e^{x} x^{2}\right )+15 \,\mathrm {log}\left (x \right )+3 x -15}{x \left (e^{x}+\mathrm {log}\left (x \right )-1\right )} \] Input:
int(((-3*log(x)+(-3*x-3)*exp(x))*log(exp(x)*x^2)-15*log(x)^2+(-30*exp(x)+3 *x+36)*log(x)-15*exp(x)^2+(-3*x^2+3*x+36)*exp(x)-6*x-21)/(x^2*log(x)^2+(2* exp(x)*x^2-2*x^2)*log(x)+exp(x)^2*x^2-2*exp(x)*x^2+x^2),x)
Output:
(3*(5*e**x + log(e**x*x**2) + 5*log(x) + x - 5))/(x*(e**x + log(x) - 1))