Integrand size = 127, antiderivative size = 29 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=-3+e^{\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}} \] Output:
exp(3*(x+5-ln((-1+ln(x)+x)/(4*x+ln(x))))/x)-3
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{3+\frac {15}{x}} \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )^{-3/x} \] Input:
Integrate[(E^((15 + 3*x - 3*Log[(-1 + x + Log[x])/(4*x + Log[x])])/x)*(-3 + 39*x - 60*x^2 + (15 - 66*x)*Log[x] - 15*Log[x]^2 + (-12*x + 12*x^2 + (-3 + 15*x)*Log[x] + 3*Log[x]^2)*Log[(-1 + x + Log[x])/(4*x + Log[x])]))/(-4* x^3 + 4*x^4 + (-x^2 + 5*x^3)*Log[x] + x^2*Log[x]^2),x]
Output:
E^(3 + 15/x)/((-1 + x + Log[x])/(4*x + Log[x]))^(3/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.
| \(\displaystyle \int \frac {e^{\frac {3 x-3 \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+15}{x}} \left (-60 x^2+\left (12 x^2-12 x+3 \log ^2(x)+(15 x-3) \log (x)\right ) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+39 x-15 \log ^2(x)+(15-66 x) \log (x)-3\right )}{4 x^4-4 x^3+x^2 \log ^2(x)+\left (5 x^3-x^2\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}} \left (60 x^2-\left (12 x^2-12 x+3 \log ^2(x)+(15 x-3) \log (x)\right ) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )-39 x+15 \log ^2(x)-(15-66 x) \log (x)+3\right )}{-4 x^4+4 x^3-x^2 \log ^2(x)-\left (5 x^3-x^2\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {15 e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}} \log ^2(x)}{x^2 (x+\log (x)-1) (4 x+\log (x))}-\frac {3 (22 x-5) e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}} \log (x)}{x^2 (x+\log (x)-1) (4 x+\log (x))}+\frac {3 e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}} \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )}{x^2}-\frac {3 e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}}}{x^2 (x+\log (x)-1) (4 x+\log (x))}-\frac {60 e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}}}{(x+\log (x)-1) (4 x+\log (x))}+\frac {39 e^{\frac {3 \left (x-\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )}{x}}}{x (x+\log (x)-1) (4 x+\log (x))}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {3 e^{\frac {15}{x}+3} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}} \left (-20 x^2+13 x+\log ^2(x) \left (\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )-5\right )+4 (x-1) x \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+\log (x) \left (-22 x+(5 x-1) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )-1\right )}{x^2 (-x-\log (x)+1)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 3 \int -\frac {e^{3+\frac {15}{x}} \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}} \left (20 x^2+4 (1-x) \log \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right ) x-13 x+\log ^2(x) \left (5-\log \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )\right )-\log (x) \left (-22 x-(1-5 x) \log \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )+5\right )+1\right )}{x^2 (-x-\log (x)+1)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -3 \int \frac {e^{3+\frac {15}{x}} \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}} \left (20 x^2+4 (1-x) \log \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right ) x-13 x+\log ^2(x) \left (5-\log \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )\right )-\log (x) \left (-22 x-(1-5 x) \log \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )+5\right )+1\right )}{x^2 (-x-\log (x)+1)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -3 \int \left (-\frac {e^{3+\frac {15}{x}} (4 x+\log (x)) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right ) \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)}+\frac {20 e^{3+\frac {15}{x}} \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{(x+\log (x)-1)^2}+\frac {5 e^{3+\frac {15}{x}} \log ^2(x) \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}+\frac {22 e^{3+\frac {15}{x}} \log (x) \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}-\frac {5 e^{3+\frac {15}{x}} \log (x) \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}-\frac {13 e^{3+\frac {15}{x}} \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}+\frac {e^{3+\frac {15}{x}} \left (-\frac {-x-\log (x)+1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -3 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}} \left (20 x^2-4 (x-1) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right ) x-13 x-\log ^2(x) \left (\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )-5\right )-\log (x) \left (-22 x+(5 x-1) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )+1\right )}{x^2 (-x-\log (x)+1)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -3 \int \left (-\frac {e^{3+\frac {15}{x}} (4 x+\log (x)) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right ) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)}+\frac {20 e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{(x+\log (x)-1)^2}+\frac {5 e^{3+\frac {15}{x}} \log ^2(x) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}+\frac {22 e^{3+\frac {15}{x}} \log (x) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}-\frac {5 e^{3+\frac {15}{x}} \log (x) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}-\frac {13 e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}+\frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -3 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}} \left (20 x^2-4 (x-1) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right ) x-13 x-\log ^2(x) \left (\log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )-5\right )-\log (x) \left (-22 x+(5 x-1) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )+5\right )+1\right )}{x^2 (-x-\log (x)+1)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -3 \int \left (-\frac {e^{3+\frac {15}{x}} (4 x+\log (x)) \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right ) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)}+\frac {20 e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{(x+\log (x)-1)^2}+\frac {5 e^{3+\frac {15}{x}} \log ^2(x) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}+\frac {22 e^{3+\frac {15}{x}} \log (x) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}-\frac {5 e^{3+\frac {15}{x}} \log (x) \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}-\frac {13 e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}+\frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -3 \left (5 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2}dx+\int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)^2}dx+5 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x^2 (x+\log (x)-1)}dx-\int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{-3/x} \log \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )}{x^2}dx+3 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{(x+\log (x)-1)^2}dx+4 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)^2}dx+12 \int \frac {e^{3+\frac {15}{x}} \left (\frac {x+\log (x)-1}{4 x+\log (x)}\right )^{1-\frac {3}{x}}}{x (x+\log (x)-1)}dx\right )\) |
Input:
Int[(E^((15 + 3*x - 3*Log[(-1 + x + Log[x])/(4*x + Log[x])])/x)*(-3 + 39*x - 60*x^2 + (15 - 66*x)*Log[x] - 15*Log[x]^2 + (-12*x + 12*x^2 + (-3 + 15* x)*Log[x] + 3*Log[x]^2)*Log[(-1 + x + Log[x])/(4*x + Log[x])]))/(-4*x^3 + 4*x^4 + (-x^2 + 5*x^3)*Log[x] + x^2*Log[x]^2),x]
Output:
$Aborted
Time = 8.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
| method | result | size |
| parallelrisch | \({\mathrm e}^{-\frac {3 \left (\ln \left (\frac {-1+\ln \left (x \right )+x}{4 x +\ln \left (x \right )}\right )-5-x \right )}{x}}\) | \(27\) |
| risch | \(64^{\frac {1}{x}} \left (x +\frac {\ln \left (x \right )}{4}\right )^{\frac {3}{x}} \left (-1+\ln \left (x \right )+x \right )^{-\frac {3}{x}} {\mathrm e}^{\frac {\frac {3 i \pi \operatorname {csgn}\left (\frac {i \left (-1+\ln \left (x \right )+x \right )}{x +\frac {\ln \left (x \right )}{4}}\right )^{3}}{2}-\frac {3 i \pi \operatorname {csgn}\left (\frac {i \left (-1+\ln \left (x \right )+x \right )}{x +\frac {\ln \left (x \right )}{4}}\right )^{2} \operatorname {csgn}\left (\frac {i}{x +\frac {\ln \left (x \right )}{4}}\right )}{2}-\frac {3 i \pi \operatorname {csgn}\left (\frac {i \left (-1+\ln \left (x \right )+x \right )}{x +\frac {\ln \left (x \right )}{4}}\right )^{2} \operatorname {csgn}\left (i \left (-1+\ln \left (x \right )+x \right )\right )}{2}+\frac {3 i \pi \,\operatorname {csgn}\left (\frac {i \left (-1+\ln \left (x \right )+x \right )}{x +\frac {\ln \left (x \right )}{4}}\right ) \operatorname {csgn}\left (\frac {i}{x +\frac {\ln \left (x \right )}{4}}\right ) \operatorname {csgn}\left (i \left (-1+\ln \left (x \right )+x \right )\right )}{2}+3 x +15}{x}}\) | \(173\) |
Input:
int(((3*ln(x)^2+(15*x-3)*ln(x)+12*x^2-12*x)*ln((-1+ln(x)+x)/(4*x+ln(x)))-1 5*ln(x)^2+(-66*x+15)*ln(x)-60*x^2+39*x-3)*exp((-3*ln((-1+ln(x)+x)/(4*x+ln( x)))+15+3*x)/x)/(x^2*ln(x)^2+(5*x^3-x^2)*ln(x)+4*x^4-4*x^3),x,method=_RETU RNVERBOSE)
Output:
exp(-3*(ln((-1+ln(x)+x)/(4*x+ln(x)))-5-x)/x)
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {3 \, {\left (x - \log \left (\frac {x + \log \left (x\right ) - 1}{4 \, x + \log \left (x\right )}\right ) + 5\right )}}{x}\right )} \] Input:
integrate(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x +log(x)))-15*log(x)^2+(-66*x+15)*log(x)-60*x^2+39*x-3)*exp((-3*log((-1+log (x)+x)/(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4*x ^3),x, algorithm="fricas")
Output:
e^(3*(x - log((x + log(x) - 1)/(4*x + log(x))) + 5)/x)
Time = 1.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\frac {3 x - 3 \log {\left (\frac {x + \log {\left (x \right )} - 1}{4 x + \log {\left (x \right )}} \right )} + 15}{x}} \] Input:
integrate(((3*ln(x)**2+(15*x-3)*ln(x)+12*x**2-12*x)*ln((-1+ln(x)+x)/(4*x+l n(x)))-15*ln(x)**2+(-66*x+15)*ln(x)-60*x**2+39*x-3)*exp((-3*ln((-1+ln(x)+x )/(4*x+ln(x)))+15+3*x)/x)/(x**2*ln(x)**2+(5*x**3-x**2)*ln(x)+4*x**4-4*x**3 ),x)
Output:
exp((3*x - 3*log((x + log(x) - 1)/(4*x + log(x))) + 15)/x)
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (\frac {3 \, \log \left (4 \, x + \log \left (x\right )\right )}{x} - \frac {3 \, \log \left (x + \log \left (x\right ) - 1\right )}{x} + \frac {15}{x} + 3\right )} \] Input:
integrate(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x +log(x)))-15*log(x)^2+(-66*x+15)*log(x)-60*x^2+39*x-3)*exp((-3*log((-1+log (x)+x)/(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4*x ^3),x, algorithm="maxima")
Output:
e^(3*log(4*x + log(x))/x - 3*log(x + log(x) - 1)/x + 15/x + 3)
Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=e^{\left (-\frac {3 \, \log \left (\frac {x}{4 \, x + \log \left (x\right )} + \frac {\log \left (x\right )}{4 \, x + \log \left (x\right )} - \frac {1}{4 \, x + \log \left (x\right )}\right )}{x} + \frac {15}{x} + 3\right )} \] Input:
integrate(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x +log(x)))-15*log(x)^2+(-66*x+15)*log(x)-60*x^2+39*x-3)*exp((-3*log((-1+log (x)+x)/(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4*x ^3),x, algorithm="giac")
Output:
e^(-3*log(x/(4*x + log(x)) + log(x)/(4*x + log(x)) - 1/(4*x + log(x)))/x + 15/x + 3)
Time = 4.51 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^3\,{\mathrm {e}}^{15/x}}{{\left (\frac {x+\ln \left (x\right )-1}{4\,x+\ln \left (x\right )}\right )}^{3/x}} \] Input:
int(-(exp((3*x - 3*log((x + log(x) - 1)/(4*x + log(x))) + 15)/x)*(15*log(x )^2 - 39*x + log(x)*(66*x - 15) + 60*x^2 - log((x + log(x) - 1)/(4*x + log (x)))*(3*log(x)^2 - 12*x + log(x)*(15*x - 3) + 12*x^2) + 3))/(x^2*log(x)^2 - log(x)*(x^2 - 5*x^3) - 4*x^3 + 4*x^4),x)
Output:
(exp(3)*exp(15/x))/((x + log(x) - 1)/(4*x + log(x)))^(3/x)
Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}} \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx=\frac {e^{\frac {15}{x}} e^{3}}{e^{\frac {3 \,\mathrm {log}\left (\frac {\mathrm {log}\left (x \right )+x -1}{\mathrm {log}\left (x \right )+4 x}\right )}{x}}} \] Input:
int(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x+log(x )))-15*log(x)^2+(-66*x+15)*log(x)-60*x^2+39*x-3)*exp((-3*log((-1+log(x)+x) /(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4*x^3),x)
Output:
(e**(15/x)*e**3)/e**((3*log((log(x) + x - 1)/(log(x) + 4*x)))/x)