Integrand size = 299, antiderivative size = 36 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=-3+\frac {1}{2} \left (\frac {2}{x}-\frac {2}{3+\frac {3}{4 \log ^2\left (\log \left (\left (5+e^x-x\right ) x\right )\right )}}\right ) \] Output:
1/x-1/(3/4/ln(ln(x*(exp(x)+5-x)))^2+3)-3
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {3+x+12 \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )}{3 \left (x+4 x \log ^2\left (\log \left (-x \left (-5-e^x+x\right )\right )\right )\right )} \] Input:
Integrate[((-15 - 3*E^x + 3*x)*Log[5*x + E^x*x - x^2] + (-40*x + 16*x^2 + E^x*(-8*x - 8*x^2))*Log[Log[5*x + E^x*x - x^2]] + (-120 - 24*E^x + 24*x)*L og[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (-240 - 48*E^x + 48* x)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4)/((15*x^2 + 3*E^x* x^2 - 3*x^3)*Log[5*x + E^x*x - x^2] + (120*x^2 + 24*E^x*x^2 - 24*x^3)*Log[ 5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (240*x^2 + 48*E^x*x^2 - 48*x^3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4),x]
Output:
(3 + x + 12*Log[Log[-(x*(-5 - E^x + x))]]^2)/(3*(x + 4*x*Log[Log[-(x*(-5 - 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.
| \(\displaystyle \int \frac {\left (48 x-48 e^x-240\right ) \log \left (-x^2+e^x x+5 x\right ) \log ^4\left (\log \left (-x^2+e^x x+5 x\right )\right )+\left (24 x-24 e^x-120\right ) \log \left (-x^2+e^x x+5 x\right ) \log ^2\left (\log \left (-x^2+e^x x+5 x\right )\right )+\left (16 x^2+e^x \left (-8 x^2-8 x\right )-40 x\right ) \log \left (\log \left (-x^2+e^x x+5 x\right )\right )+\left (3 x-3 e^x-15\right ) \log \left (-x^2+e^x x+5 x\right )}{\left (-48 x^3+48 e^x x^2+240 x^2\right ) \log \left (-x^2+e^x x+5 x\right ) \log ^4\left (\log \left (-x^2+e^x x+5 x\right )\right )+\left (-24 x^3+24 e^x x^2+120 x^2\right ) \log \left (-x^2+e^x x+5 x\right ) \log ^2\left (\log \left (-x^2+e^x x+5 x\right )\right )+\left (-3 x^3+3 e^x x^2+15 x^2\right ) \log \left (-x^2+e^x x+5 x\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-3 \left (-x+e^x+5\right ) \log \left (-x \left (x-e^x-5\right )\right ) \left (4 \log ^2\left (\log \left (-x \left (x-e^x-5\right )\right )\right )+1\right )^2-8 x \left (-2 x+e^x (x+1)+5\right ) \log \left (\log \left (-x \left (x-e^x-5\right )\right )\right )}{3 \left (-x+e^x+5\right ) \log \left (-x^2+e^x x+5 x\right ) \left (x+4 x \log ^2\left (\log \left (-x \left (x-e^x-5\right )\right )\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int -\frac {3 \left (-x+e^x+5\right ) \log \left (\left (-x+e^x+5\right ) x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2+8 x \left (-2 x+e^x (x+1)+5\right ) \log \left (\log \left (\left (-x+e^x+5\right ) x\right )\right )}{\left (-x+e^x+5\right ) \log \left (-x^2+e^x x+5 x\right ) \left (4 x \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+x\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{3} \int \frac {3 \left (-x+e^x+5\right ) \log \left (\left (-x+e^x+5\right ) x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2+8 x \left (-2 x+e^x (x+1)+5\right ) \log \left (\log \left (\left (-x+e^x+5\right ) x\right )\right )}{\left (-x+e^x+5\right ) \log \left (-x^2+e^x x+5 x\right ) \left (4 x \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+x\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{3} \int \left (\frac {48 \log \left (\left (-x+e^x+5\right ) x\right ) \log ^4\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+24 \log \left (\left (-x+e^x+5\right ) x\right ) \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+8 x^2 \log \left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+8 x \log \left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+3 \log \left (\left (-x+e^x+5\right ) x\right )}{x^2 \log \left (-x^2+e^x x+5 x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2}+\frac {8 (x-6) \log \left (\log \left (-x^2+e^x x+5 x\right )\right )}{\left (-x+e^x+5\right ) \log \left (-x^2+e^x x+5 x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-8 \int \frac {\log \left (\log \left (-x^2+e^x x+5 x\right )\right )}{\log \left (-x^2+e^x x+5 x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2}dx-8 \int \frac {\log \left (\log \left (-x^2+e^x x+5 x\right )\right )}{x \log \left (-x^2+e^x x+5 x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2}dx-8 \int \frac {x \log \left (\log \left (-x^2+e^x x+5 x\right )\right )}{\left (-x+e^x+5\right ) \log \left (-x^2+e^x x+5 x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2}dx-48 \int \frac {\log \left (\log \left (-x^2+e^x x+5 x\right )\right )}{\left (x-e^x-5\right ) \log \left (-x^2+e^x x+5 x\right ) \left (4 \log ^2\left (\log \left (\left (-x+e^x+5\right ) x\right )\right )+1\right )^2}dx+\frac {3}{x}\right )\) |
Input:
Int[((-15 - 3*E^x + 3*x)*Log[5*x + E^x*x - x^2] + (-40*x + 16*x^2 + E^x*(- 8*x - 8*x^2))*Log[Log[5*x + E^x*x - x^2]] + (-120 - 24*E^x + 24*x)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (-240 - 48*E^x + 48*x)*Log [5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4)/((15*x^2 + 3*E^x*x^2 - 3*x^3)*Log[5*x + E^x*x - x^2] + (120*x^2 + 24*E^x*x^2 - 24*x^3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^2 + (240*x^2 + 48*E^x*x^2 - 48*x^ 3)*Log[5*x + E^x*x - x^2]*Log[Log[5*x + E^x*x - x^2]]^4),x]
Output:
$Aborted
Time = 55.94 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25
| method | result | size |
| parallelrisch | \(\frac {24+96 {\ln \left (\ln \left (x \left ({\mathrm e}^{x}+5-x \right )\right )\right )}^{2}+8 x}{24 x \left (4 {\ln \left (\ln \left (x \left ({\mathrm e}^{x}+5-x \right )\right )\right )}^{2}+1\right )}\) | \(45\) |
| risch | \(\frac {1}{x}+\frac {1}{12 {\ln \left (i \pi +\ln \left (x \right )+\ln \left (-{\mathrm e}^{x}-5+x \right )+\frac {i \pi \,\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right ) \left (\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )-\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5-x \right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )}^{2} \left (-\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+5-x \right )\right )-1\right )\right )}^{2}+3}\) | \(125\) |
Input:
int(((-48*exp(x)+48*x-240)*ln(exp(x)*x-x^2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^4 +(-24*exp(x)+24*x-120)*ln(exp(x)*x-x^2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^2+((- 8*x^2-8*x)*exp(x)+16*x^2-40*x)*ln(ln(exp(x)*x-x^2+5*x))+(-3*exp(x)+3*x-15) *ln(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*ln(exp(x)*x-x^2+5*x )*ln(ln(exp(x)*x-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x^2)*ln(exp(x)*x-x^ 2+5*x)*ln(ln(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3*x^3+15*x^2)*ln(exp(x)*x- x^2+5*x)),x,method=_RETURNVERBOSE)
Output:
1/24*(24+96*ln(ln(x*(exp(x)+5-x)))^2+8*x)/x/(4*ln(ln(x*(exp(x)+5-x)))^2+1)
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {12 \, \log \left (\log \left (-x^{2} + x e^{x} + 5 \, x\right )\right )^{2} + x + 3}{3 \, {\left (4 \, x \log \left (\log \left (-x^{2} + x e^{x} + 5 \, x\right )\right )^{2} + x\right )}} \] Input:
integrate(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^ 2+5*x))^4+(-24*exp(x)+24*x-120)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2 +5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-3* exp(x)+3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log( exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x ^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3*x^3 +15*x^2)*log(exp(x)*x-x^2+5*x)),x, algorithm="fricas")
Output:
1/3*(12*log(log(-x^2 + x*e^x + 5*x))^2 + x + 3)/(4*x*log(log(-x^2 + x*e^x + 5*x))^2 + x)
Time = 0.98 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {1}{12 \log {\left (\log {\left (- x^{2} + x e^{x} + 5 x \right )} \right )}^{2} + 3} + \frac {1}{x} \] Input:
integrate(((-48*exp(x)+48*x-240)*ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2 +5*x))**4+(-24*exp(x)+24*x-120)*ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2+ 5*x))**2+((-8*x**2-8*x)*exp(x)+16*x**2-40*x)*ln(ln(exp(x)*x-x**2+5*x))+(-3 *exp(x)+3*x-15)*ln(exp(x)*x-x**2+5*x))/((48*exp(x)*x**2-48*x**3+240*x**2)* ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2+5*x))**4+(24*exp(x)*x**2-24*x**3 +120*x**2)*ln(exp(x)*x-x**2+5*x)*ln(ln(exp(x)*x-x**2+5*x))**2+(3*exp(x)*x* *2-3*x**3+15*x**2)*ln(exp(x)*x-x**2+5*x)),x)
Output:
1/(12*log(log(-x**2 + x*exp(x) + 5*x))**2 + 3) + 1/x
Time = 0.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {12 \, \log \left (\log \left (x\right ) + \log \left (-x + e^{x} + 5\right )\right )^{2} + x + 3}{3 \, {\left (4 \, x \log \left (\log \left (x\right ) + \log \left (-x + e^{x} + 5\right )\right )^{2} + x\right )}} \] Input:
integrate(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^ 2+5*x))^4+(-24*exp(x)+24*x-120)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2 +5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-3* exp(x)+3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log( exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x ^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3*x^3 +15*x^2)*log(exp(x)*x-x^2+5*x)),x, algorithm="maxima")
Output:
1/3*(12*log(log(x) + log(-x + e^x + 5))^2 + x + 3)/(4*x*log(log(x) + log(- x + e^x + 5))^2 + x)
Time = 78.78 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {12 \, \log \left (\log \left (-x^{2} + x e^{x} + 5 \, x\right )\right )^{2} + x + 3}{3 \, {\left (4 \, x \log \left (\log \left (-x^{2} + x e^{x} + 5 \, x\right )\right )^{2} + x\right )}} \] Input:
integrate(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^ 2+5*x))^4+(-24*exp(x)+24*x-120)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2 +5*x))^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-3* exp(x)+3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log( exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x ^2)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3*x^3 +15*x^2)*log(exp(x)*x-x^2+5*x)),x, algorithm="giac")
Output:
1/3*(12*log(log(-x^2 + x*e^x + 5*x))^2 + x + 3)/(4*x*log(log(-x^2 + x*e^x + 5*x))^2 + x)
Time = 3.68 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {1}{3\,\left (4\,{\ln \left (\ln \left (5\,x+x\,{\mathrm {e}}^x-x^2\right )\right )}^2+1\right )}+\frac {1}{x} \] Input:
int(-(log(5*x + x*exp(x) - x^2)*(3*exp(x) - 3*x + 15) + log(log(5*x + x*ex p(x) - x^2))*(40*x + exp(x)*(8*x + 8*x^2) - 16*x^2) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^2*(24*exp(x) - 24*x + 120) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^4*(48*exp(x) - 48*x + 240 ))/(log(5*x + x*exp(x) - x^2)*(3*x^2*exp(x) + 15*x^2 - 3*x^3) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^2*(24*x^2*exp(x) + 120*x^2 - 24*x^3) + log(5*x + x*exp(x) - x^2)*log(log(5*x + x*exp(x) - x^2))^4*(48 *x^2*exp(x) + 240*x^2 - 48*x^3)),x)
Output:
1/(3*(4*log(log(5*x + x*exp(x) - x^2))^2 + 1)) + 1/x
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.00 \[ \int \frac {\left (-15-3 e^x+3 x\right ) \log \left (5 x+e^x x-x^2\right )+\left (-40 x+16 x^2+e^x \left (-8 x-8 x^2\right )\right ) \log \left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-120-24 e^x+24 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (-240-48 e^x+48 x\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )}{\left (15 x^2+3 e^x x^2-3 x^3\right ) \log \left (5 x+e^x x-x^2\right )+\left (120 x^2+24 e^x x^2-24 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^2\left (\log \left (5 x+e^x x-x^2\right )\right )+\left (240 x^2+48 e^x x^2-48 x^3\right ) \log \left (5 x+e^x x-x^2\right ) \log ^4\left (\log \left (5 x+e^x x-x^2\right )\right )} \, dx=\frac {-4 {\mathrm {log}\left (\mathrm {log}\left (e^{x} x -x^{2}+5 x \right )\right )}^{2} x +12 {\mathrm {log}\left (\mathrm {log}\left (e^{x} x -x^{2}+5 x \right )\right )}^{2}+3}{3 x \left (4 {\mathrm {log}\left (\mathrm {log}\left (e^{x} x -x^{2}+5 x \right )\right )}^{2}+1\right )} \] Input:
int(((-48*exp(x)+48*x-240)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x) )^4+(-24*exp(x)+24*x-120)*log(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x)) ^2+((-8*x^2-8*x)*exp(x)+16*x^2-40*x)*log(log(exp(x)*x-x^2+5*x))+(-3*exp(x) +3*x-15)*log(exp(x)*x-x^2+5*x))/((48*exp(x)*x^2-48*x^3+240*x^2)*log(exp(x) *x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^4+(24*exp(x)*x^2-24*x^3+120*x^2)*lo g(exp(x)*x-x^2+5*x)*log(log(exp(x)*x-x^2+5*x))^2+(3*exp(x)*x^2-3*x^3+15*x^ 2)*log(exp(x)*x-x^2+5*x)),x)
Output:
( - 4*log(log(e**x*x - x**2 + 5*x))**2*x + 12*log(log(e**x*x - x**2 + 5*x) )**2 + 3)/(3*x*(4*log(log(e**x*x - x**2 + 5*x))**2 + 1))