Integrand size = 100, antiderivative size = 33 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\log \left (3-\frac {e^{4-i \pi +\frac {3 x}{2}-\frac {x}{\log (x)}}}{-1+2 e}\right ) \] Output:
ln(3-exp(4+3/2*x-x/ln(x)-ln(-2*exp(1)+1)))
Time = 0.80 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\frac {1}{2} \left (2 \log \left (e^{4+\frac {3 x}{2}}+6 e^{1+\frac {x}{\log (x)}}-3 e^{\frac {x}{\log (x)}}\right )-\frac {2 x}{\log (x)}\right ) \] Input:
Integrate[(E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[ x]))*(2 - 2*Log[x] + 3*Log[x]^2))/(-6*Log[x]^2 + 2*E^((-2*x + (8 + 3*x - 2 *(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*Log[x]^2),x]
Output:
(2*Log[E^(4 + (3*x)/2) + 6*E^(1 + x/Log[x]) - 3*E^(x/Log[x])] - (2*x)/Log[ 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.
\(\displaystyle \int \frac {\left (3 \log ^2(x)-2 \log (x)+2\right ) \exp \left (\frac {-2 x+(3 x+8-2 (\log (2 e-1)+i \pi )) \log (x)}{2 \log (x)}\right )}{-6 \log ^2(x)+2 \log ^2(x) \exp \left (\frac {-2 x+(3 x+8-2 (\log (2 e-1)+i \pi )) \log (x)}{2 \log (x)}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {3 x}{2}+4} \left (3 \log ^2(x)-2 \log (x)+2\right )}{2 \left (e^{\frac {3 x}{2}+4}+3 (2 e-1) e^{\frac {x}{\log (x)}}\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{\frac {3 x}{2}+4} \left (3 \log ^2(x)-2 \log (x)+2\right )}{\left (e^{\frac {3 x}{2}+4}-3 (1-2 e) e^{\frac {x}{\log (x)}}\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {2 e^{\frac {3 x}{2}+4}}{\left (-e^{\frac {3 x}{2}+4}-6 \left (1-\frac {1}{2 e}\right ) e^{\frac {x}{\log (x)}+1}\right ) \log (x)}+\frac {3 e^{\frac {3 x}{2}+4}}{e^{\frac {3 x}{2}+4}+6 \left (1-\frac {1}{2 e}\right ) e^{\frac {x}{\log (x)}+1}}+\frac {2 e^{\frac {3 x}{2}+4}}{\log ^2(x) \left (e^{\frac {3 x}{2}+4}+6 \left (1-\frac {1}{2 e}\right ) e^{\frac {x}{\log (x)}+1}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {e^{\frac {3 x}{2}+4}}{\left (e^{\frac {3 x}{2}+4}+6 \left (1-\frac {1}{2 e}\right ) e^{\frac {x}{\log (x)}+1}\right ) \log ^2(x)}dx+3 \int \frac {e^{\frac {3 x}{2}+4}}{e^{\frac {3 x}{2}+4}+6 \left (1-\frac {1}{2 e}\right ) e^{\frac {x}{\log (x)}+1}}dx+2 \int \frac {e^{\frac {3 x}{2}+4}}{\left (-e^{\frac {3 x}{2}+4}-6 \left (1-\frac {1}{2 e}\right ) e^{\frac {x}{\log (x)}+1}\right ) \log (x)}dx\right )\) |
Input:
Int[(E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*( 2 - 2*Log[x] + 3*Log[x]^2))/(-6*Log[x]^2 + 2*E^((-2*x + (8 + 3*x - 2*(I*Pi + Log[-1 + 2*E]))*Log[x])/(2*Log[x]))*Log[x]^2),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\ln \left ({\mathrm e}^{\frac {\left (-2 \ln \left (-2 \,{\mathrm e}+1\right )+3 x +8\right ) \ln \left (x \right )-2 x}{2 \ln \left (x \right )}}-3\right )\) | \(32\) |
parallelrisch | \(\ln \left ({\mathrm e}^{\frac {\left (-2 \ln \left (-2 \,{\mathrm e}+1\right )+3 x +8\right ) \ln \left (x \right )-2 x}{2 \ln \left (x \right )}}-3\right )\) | \(32\) |
risch | \(\frac {3 x}{2}-\frac {x}{\ln \left (x \right )}-\frac {\left (-2 \ln \left (-2 \,{\mathrm e}+1\right )+3 x +8\right ) \ln \left (x \right )-2 x}{2 \ln \left (x \right )}+\ln \left (\frac {{\mathrm e}^{\frac {3 x \ln \left (x \right )+8 \ln \left (x \right )-2 x}{2 \ln \left (x \right )}}}{-2 \,{\mathrm e}+1}-3\right )\) | \(71\) |
Input:
int((3*ln(x)^2-2*ln(x)+2)*exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/l n(x))/(2*ln(x)^2*exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))-6*l n(x)^2),x,method=_RETURNVERBOSE)
Output:
ln(exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln(x))-3)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\log \left (e^{\left (\frac {{\left (3 \, x - 2 \, \log \left (-2 \, e + 1\right ) + 8\right )} \log \left (x\right ) - 2 \, x}{2 \, \log \left (x\right )}\right )} - 3\right ) \] Input:
integrate((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log (x)-2*x)/log(x))/(2*log(x)^2*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2 *x)/log(x))-6*log(x)^2),x, algorithm="fricas")
Output:
log(e^(1/2*((3*x - 2*log(-2*e + 1) + 8)*log(x) - 2*x)/log(x)) - 3)
Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=- \frac {x}{\log {\left (x \right )}} + \log {\left (e^{\frac {x}{\log {\left (x \right )}}} + \frac {e^{4}}{- 3 e^{- \frac {3 x}{2}} + 6 e e^{- \frac {3 x}{2}}} \right )} \] Input:
integrate((3*ln(x)**2-2*ln(x)+2)*exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x) -2*x)/ln(x))/(2*ln(x)**2*exp(1/2*((-2*ln(-2*exp(1)+1)+3*x+8)*ln(x)-2*x)/ln (x))-6*ln(x)**2),x)
Output:
-x/log(x) + log(exp(x/log(x)) + exp(4)/(-3*exp(-3*x/2) + 6*E*exp(-3*x/2)))
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=-\frac {x}{\log \left (x\right )} + \log \left (\frac {3 \, {\left (2 \, e - 1\right )} e^{\frac {x}{\log \left (x\right )}} + e^{\left (\frac {3}{2} \, x + 4\right )}}{3 \, {\left (2 \, e - 1\right )}}\right ) \] Input:
integrate((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log (x)-2*x)/log(x))/(2*log(x)^2*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2 *x)/log(x))-6*log(x)^2),x, algorithm="maxima")
Output:
-x/log(x) + log(1/3*(3*(2*e - 1)*e^(x/log(x)) + e^(3/2*x + 4))/(2*e - 1))
Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (27) = 54\).
Time = 0.18 (sec) , antiderivative size = 439, normalized size of antiderivative = 13.30 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx =\text {Too large to display} \] Input:
integrate((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log (x)-2*x)/log(x))/(2*log(x)^2*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2 *x)/log(x))-6*log(x)^2),x, algorithm="giac")
Output:
log(sqrt(-6*cos(-pi^3*sgn(x)^2/(pi^2*sgn(x)^2 - 2*pi^2*sgn(x) + pi^2 + 4*l og(abs(x))^2) + 2*pi^3*sgn(x)/(pi^2*sgn(x)^2 - 2*pi^2*sgn(x) + pi^2 + 4*lo g(abs(x))^2) - pi^3/(pi^2*sgn(x)^2 - 2*pi^2*sgn(x) + pi^2 + 4*log(abs(x))^ 2) - 4*pi*log(abs(x))^2/(pi^2*sgn(x)^2 - 2*pi^2*sgn(x) + pi^2 + 4*log(abs( x))^2) - 2*pi*x*sgn(x)/(pi^2*sgn(x)^2 - 2*pi^2*sgn(x) + pi^2 + 4*log(abs(x ))^2) + 2*pi*x/(pi^2*sgn(x)^2 - 2*pi^2*sgn(x) + pi^2 + 4*log(abs(x))^2))*e ^(1/2*(3*pi^2*x*sgn(x) - 2*pi^2*log(2*e - 1)*sgn(x) - 3*pi^2*x + 2*pi^2*lo g(2*e - 1) - 6*x*log(abs(x))^2 + 4*log(2*e - 1)*log(abs(x))^2 + 8*pi^2*sgn (x) - 8*pi^2 + 4*x*log(abs(x)) - 16*log(abs(x))^2)/(pi^2*sgn(x) - pi^2 - 2 *log(abs(x))^2)) + e^((3*pi^2*x*sgn(x) - 2*pi^2*log(2*e - 1)*sgn(x) - 3*pi ^2*x + 2*pi^2*log(2*e - 1) - 6*x*log(abs(x))^2 + 4*log(2*e - 1)*log(abs(x) )^2 + 8*pi^2*sgn(x) - 8*pi^2 + 4*x*log(abs(x)) - 16*log(abs(x))^2)/(pi^2*s gn(x) - pi^2 - 2*log(abs(x))^2)) + 9))
Time = 7.41 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\ln \left (-\frac {{\mathrm {e}}^4\,{\mathrm {e}}^{-\frac {x}{\ln \left (x\right )}}\,{\left ({\mathrm {e}}^x\right )}^{3/2}}{2\,\mathrm {e}-1}-3\right ) \] Input:
int(-(exp(-(x - (log(x)*(3*x - 2*log(1 - 2*exp(1)) + 8))/2)/log(x))*(3*log (x)^2 - 2*log(x) + 2))/(6*log(x)^2 - 2*exp(-(x - (log(x)*(3*x - 2*log(1 - 2*exp(1)) + 8))/2)/log(x))*log(x)^2),x)
Output:
log(- (exp(4)*exp(-x/log(x))*exp(x)^(3/2))/(2*exp(1) - 1) - 3)
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \left (2-2 \log (x)+3 \log ^2(x)\right )}{-6 \log ^2(x)+2 e^{\frac {-2 x+(8+3 x-2 (i \pi +\log (-1+2 e))) \log (x)}{2 \log (x)}} \log ^2(x)} \, dx=\frac {\mathrm {log}\left (6 e^{\frac {x}{\mathrm {log}\left (x \right )}} e -3 e^{\frac {x}{\mathrm {log}\left (x \right )}}+e^{\frac {3 x}{2}} e^{4}\right ) \mathrm {log}\left (x \right )-x}{\mathrm {log}\left (x \right )} \] Input:
int((3*log(x)^2-2*log(x)+2)*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2* x)/log(x))/(2*log(x)^2*exp(1/2*((-2*log(-2*exp(1)+1)+3*x+8)*log(x)-2*x)/lo g(x))-6*log(x)^2),x)
Output:
(log(6*e**(x/log(x))*e - 3*e**(x/log(x)) + e**((3*x)/2)*e**4)*log(x) - x)/ log(x)