Integrand size = 253, antiderivative size = 32 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {\left (5+e^{5 e^x}\right )^2 \log ^2\left (3+\frac {1}{3} \left (2-e^5\right ) x\right )}{x^2} \] Output:
(exp(5*exp(x))+5)^2/x^2*ln(3+1/3*(2-exp(5))*x)^2
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.25 (sec) , antiderivative size = 447, normalized size of antiderivative = 13.97 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {200 x^2 \log (x)-200 e^5 x^2 \log (x)+50 e^{10} x^2 \log (x)-200 x^2 \log (3) \log (x)+200 e^5 x^2 \log (3) \log (x)-50 e^{10} x^2 \log (3) \log (x)-200 x^2 \log \left (\frac {1}{9} \left (-2+e^5\right ) x\right )+200 e^5 x^2 \log \left (\frac {1}{9} \left (-2+e^5\right ) x\right )-50 e^{10} x^2 \log \left (\frac {1}{9} \left (-2+e^5\right ) x\right )-200 x^2 \log \left (9+2 x-e^5 x\right )+200 e^5 x^2 \log \left (9+2 x-e^5 x\right )-50 e^{10} x^2 \log \left (9+2 x-e^5 x\right )+200 x^2 \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )-200 e^5 x^2 \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+50 e^{10} x^2 \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+200 x^2 \log \left (\frac {1}{9} \left (-2+e^5\right ) x\right ) \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )-200 e^5 x^2 \log \left (\frac {1}{9} \left (-2+e^5\right ) x\right ) \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+50 e^{10} x^2 \log \left (\frac {1}{9} \left (-2+e^5\right ) x\right ) \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+2025 \log ^2\left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+810 e^{5 e^x} \log ^2\left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+81 e^{10 e^x} \log ^2\left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )+50 \left (-2+e^5\right )^2 x^2 \operatorname {PolyLog}\left (2,\frac {1}{9} \left (-2+e^5\right ) x\right )+50 \left (-2+e^5\right )^2 x^2 \operatorname {PolyLog}\left (2,1-\frac {1}{9} \left (-2+e^5\right ) x\right )}{81 x^2} \] Input:
Integrate[((-100*x + 50*E^5*x)*Log[(9 + 2*x - E^5*x)/3] + (450 + 100*x - 5 0*E^5*x)*Log[(9 + 2*x - E^5*x)/3]^2 + E^(10*E^x)*((-4*x + 2*E^5*x)*Log[(9 + 2*x - E^5*x)/3] + (18 + 4*x - 2*E^5*x + E^x*(-90*x - 20*x^2 + 10*E^5*x^2 ))*Log[(9 + 2*x - E^5*x)/3]^2) + E^(5*E^x)*((-40*x + 20*E^5*x)*Log[(9 + 2* x - E^5*x)/3] + (180 + 40*x - 20*E^5*x + E^x*(-450*x - 100*x^2 + 50*E^5*x^ 2))*Log[(9 + 2*x - E^5*x)/3]^2))/(-9*x^3 - 2*x^4 + E^5*x^4),x]
Output:
(200*x^2*Log[x] - 200*E^5*x^2*Log[x] + 50*E^10*x^2*Log[x] - 200*x^2*Log[3] *Log[x] + 200*E^5*x^2*Log[3]*Log[x] - 50*E^10*x^2*Log[3]*Log[x] - 200*x^2* Log[((-2 + E^5)*x)/9] + 200*E^5*x^2*Log[((-2 + E^5)*x)/9] - 50*E^10*x^2*Lo g[((-2 + E^5)*x)/9] - 200*x^2*Log[9 + 2*x - E^5*x] + 200*E^5*x^2*Log[9 + 2 *x - E^5*x] - 50*E^10*x^2*Log[9 + 2*x - E^5*x] + 200*x^2*Log[3 - ((-2 + E^ 5)*x)/3] - 200*E^5*x^2*Log[3 - ((-2 + E^5)*x)/3] + 50*E^10*x^2*Log[3 - ((- 2 + E^5)*x)/3] + 200*x^2*Log[((-2 + E^5)*x)/9]*Log[3 - ((-2 + E^5)*x)/3] - 200*E^5*x^2*Log[((-2 + E^5)*x)/9]*Log[3 - ((-2 + E^5)*x)/3] + 50*E^10*x^2 *Log[((-2 + E^5)*x)/9]*Log[3 - ((-2 + E^5)*x)/3] + 2025*Log[3 - ((-2 + E^5 )*x)/3]^2 + 810*E^(5*E^x)*Log[3 - ((-2 + E^5)*x)/3]^2 + 81*E^(10*E^x)*Log[ 3 - ((-2 + E^5)*x)/3]^2 + 50*(-2 + E^5)^2*x^2*PolyLog[2, ((-2 + E^5)*x)/9] + 50*(-2 + E^5)^2*x^2*PolyLog[2, 1 - ((-2 + E^5)*x)/9])/(81*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.
| \(\displaystyle \int \frac {e^{10 e^x} \left (\left (e^x \left (10 e^5 x^2-20 x^2-90 x\right )-2 e^5 x+4 x+18\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (2 e^5 x-4 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )\right )+e^{5 e^x} \left (\left (e^x \left (50 e^5 x^2-100 x^2-450 x\right )-20 e^5 x+40 x+180\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (20 e^5 x-40 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )\right )+\left (-50 e^5 x+100 x+450\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (50 e^5 x-100 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )}{e^5 x^4-2 x^4-9 x^3} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{10 e^x} \left (\left (e^x \left (10 e^5 x^2-20 x^2-90 x\right )-2 e^5 x+4 x+18\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (2 e^5 x-4 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )\right )+e^{5 e^x} \left (\left (e^x \left (50 e^5 x^2-100 x^2-450 x\right )-20 e^5 x+40 x+180\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (20 e^5 x-40 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )\right )+\left (-50 e^5 x+100 x+450\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (50 e^5 x-100 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )}{\left (e^5-2\right ) x^4-9 x^3}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{10 e^x} \left (\left (e^x \left (10 e^5 x^2-20 x^2-90 x\right )-2 e^5 x+4 x+18\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (2 e^5 x-4 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )\right )+e^{5 e^x} \left (\left (e^x \left (50 e^5 x^2-100 x^2-450 x\right )-20 e^5 x+40 x+180\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (20 e^5 x-40 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )\right )+\left (-50 e^5 x+100 x+450\right ) \log ^2\left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )+\left (50 e^5 x-100 x\right ) \log \left (\frac {1}{3} \left (-e^5 x+2 x+9\right )\right )}{x^3 \left (-\left (\left (2-e^5\right ) x\right )-9\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (e^{5 e^x}+5\right ) \log \left (3-\frac {1}{3} \left (e^5-2\right ) x\right ) \left (-\left (\left (e^5-2\right ) \left (e^{5 e^x}+5\right ) x\right )-\left (5 e^{x+5 e^x} x-e^{5 e^x}-5\right ) \left (\left (e^5-2\right ) x-9\right ) \log \left (3-\frac {1}{3} \left (e^5-2\right ) x\right )\right )}{x^3 \left (9-\left (e^5-2\right ) x\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {\left (5+e^{5 e^x}\right ) \log \left (\frac {1}{3} \left (2-e^5\right ) x+3\right ) \left (\left (2-e^5\right ) \left (5+e^{5 e^x}\right ) x-\left (-5 e^{x+5 e^x} x+e^{5 e^x}+5\right ) \left (\left (2-e^5\right ) x+9\right ) \log \left (\frac {1}{3} \left (2-e^5\right ) x+3\right )\right )}{x^3 \left (\left (2-e^5\right ) x+9\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (\frac {\log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right ) \left (-2 \left (1-\frac {e^5}{2}\right ) \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right ) x+2 \left (1-\frac {e^5}{2}\right ) x-9 \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )\right ) \left (5+e^{5 e^x}\right )^2}{x^3 \left (\left (2-e^5\right ) x+9\right )}+\frac {5 e^{x+5 e^x} \log ^2\left (3-\frac {1}{3} \left (-2+e^5\right ) x\right ) \left (5+e^{5 e^x}\right )}{x^2}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle 2 \int \left (\frac {\log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right ) \left (-2 \left (1-\frac {e^5}{2}\right ) \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right ) x+2 \left (1-\frac {e^5}{2}\right ) x-9 \log \left (3-\frac {1}{3} \left (-2+e^5\right ) x\right )\right ) \left (5+e^{5 e^x}\right )^2}{x^3 \left (\left (2-e^5\right ) x+9\right )}+\frac {5 e^{x+5 e^x} \log ^2\left (3-\frac {1}{3} \left (-2+e^5\right ) x\right ) \left (5+e^{5 e^x}\right )}{x^2}\right )dx\) |
Input:
Int[((-100*x + 50*E^5*x)*Log[(9 + 2*x - E^5*x)/3] + (450 + 100*x - 50*E^5* x)*Log[(9 + 2*x - E^5*x)/3]^2 + E^(10*E^x)*((-4*x + 2*E^5*x)*Log[(9 + 2*x - E^5*x)/3] + (18 + 4*x - 2*E^5*x + E^x*(-90*x - 20*x^2 + 10*E^5*x^2))*Log [(9 + 2*x - E^5*x)/3]^2) + E^(5*E^x)*((-40*x + 20*E^5*x)*Log[(9 + 2*x - E^ 5*x)/3] + (180 + 40*x - 20*E^5*x + E^x*(-450*x - 100*x^2 + 50*E^5*x^2))*Lo g[(9 + 2*x - E^5*x)/3]^2))/(-9*x^3 - 2*x^4 + E^5*x^4),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(66\) vs. \(2(27)=54\).
Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.09
\[\frac {25 \ln \left (-\frac {x \,{\mathrm e}^{5}}{3}+\frac {2 x}{3}+3\right )^{2}}{x^{2}}+\frac {\ln \left (-\frac {x \,{\mathrm e}^{5}}{3}+\frac {2 x}{3}+3\right )^{2} {\mathrm e}^{10 \,{\mathrm e}^{x}}}{x^{2}}+\frac {10 \ln \left (-\frac {x \,{\mathrm e}^{5}}{3}+\frac {2 x}{3}+3\right )^{2} {\mathrm e}^{5 \,{\mathrm e}^{x}}}{x^{2}}\]
Input:
int(((((10*x^2*exp(5)-20*x^2-90*x)*exp(x)-2*x*exp(5)+4*x+18)*ln(-1/3*x*exp (5)+2/3*x+3)^2+(2*x*exp(5)-4*x)*ln(-1/3*x*exp(5)+2/3*x+3))*exp(5*exp(x))^2 +(((50*x^2*exp(5)-100*x^2-450*x)*exp(x)-20*x*exp(5)+40*x+180)*ln(-1/3*x*ex p(5)+2/3*x+3)^2+(20*x*exp(5)-40*x)*ln(-1/3*x*exp(5)+2/3*x+3))*exp(5*exp(x) )+(-50*x*exp(5)+100*x+450)*ln(-1/3*x*exp(5)+2/3*x+3)^2+(50*x*exp(5)-100*x) *ln(-1/3*x*exp(5)+2/3*x+3))/(x^4*exp(5)-2*x^4-9*x^3),x)
Output:
25/x^2*ln(-1/3*x*exp(5)+2/3*x+3)^2+1/x^2*ln(-1/3*x*exp(5)+2/3*x+3)^2*exp(5 *exp(x))^2+10/x^2*ln(-1/3*x*exp(5)+2/3*x+3)^2*exp(5*exp(x))
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {e^{\left (10 \, e^{x}\right )} \log \left (-\frac {1}{3} \, x e^{5} + \frac {2}{3} \, x + 3\right )^{2} + 10 \, e^{\left (5 \, e^{x}\right )} \log \left (-\frac {1}{3} \, x e^{5} + \frac {2}{3} \, x + 3\right )^{2} + 25 \, \log \left (-\frac {1}{3} \, x e^{5} + \frac {2}{3} \, x + 3\right )^{2}}{x^{2}} \] Input:
integrate(((((10*x^2*exp(5)-20*x^2-90*x)*exp(x)-2*x*exp(5)+4*x+18)*log(-1/ 3*x*exp(5)+2/3*x+3)^2+(2*x*exp(5)-4*x)*log(-1/3*x*exp(5)+2/3*x+3))*exp(5*e xp(x))^2+(((50*x^2*exp(5)-100*x^2-450*x)*exp(x)-20*x*exp(5)+40*x+180)*log( -1/3*x*exp(5)+2/3*x+3)^2+(20*x*exp(5)-40*x)*log(-1/3*x*exp(5)+2/3*x+3))*ex p(5*exp(x))+(-50*x*exp(5)+100*x+450)*log(-1/3*x*exp(5)+2/3*x+3)^2+(50*x*ex p(5)-100*x)*log(-1/3*x*exp(5)+2/3*x+3))/(x^4*exp(5)-2*x^4-9*x^3),x, algori thm="fricas")
Output:
(e^(10*e^x)*log(-1/3*x*e^5 + 2/3*x + 3)^2 + 10*e^(5*e^x)*log(-1/3*x*e^5 + 2/3*x + 3)^2 + 25*log(-1/3*x*e^5 + 2/3*x + 3)^2)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (27) = 54\).
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.50 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {25 \log {\left (- \frac {x e^{5}}{3} + \frac {2 x}{3} + 3 \right )}^{2}}{x^{2}} + \frac {x^{2} e^{10 e^{x}} \log {\left (- \frac {x e^{5}}{3} + \frac {2 x}{3} + 3 \right )}^{2} + 10 x^{2} e^{5 e^{x}} \log {\left (- \frac {x e^{5}}{3} + \frac {2 x}{3} + 3 \right )}^{2}}{x^{4}} \] Input:
integrate(((((10*x**2*exp(5)-20*x**2-90*x)*exp(x)-2*x*exp(5)+4*x+18)*ln(-1 /3*x*exp(5)+2/3*x+3)**2+(2*x*exp(5)-4*x)*ln(-1/3*x*exp(5)+2/3*x+3))*exp(5* exp(x))**2+(((50*x**2*exp(5)-100*x**2-450*x)*exp(x)-20*x*exp(5)+40*x+180)* ln(-1/3*x*exp(5)+2/3*x+3)**2+(20*x*exp(5)-40*x)*ln(-1/3*x*exp(5)+2/3*x+3)) *exp(5*exp(x))+(-50*x*exp(5)+100*x+450)*ln(-1/3*x*exp(5)+2/3*x+3)**2+(50*x *exp(5)-100*x)*ln(-1/3*x*exp(5)+2/3*x+3))/(x**4*exp(5)-2*x**4-9*x**3),x)
Output:
25*log(-x*exp(5)/3 + 2*x/3 + 3)**2/x**2 + (x**2*exp(10*exp(x))*log(-x*exp( 5)/3 + 2*x/3 + 3)**2 + 10*x**2*exp(5*exp(x))*log(-x*exp(5)/3 + 2*x/3 + 3)* *2)/x**4
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (25) = 50\).
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.91 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {e^{\left (10 \, e^{x}\right )} \log \left (3\right )^{2} + 10 \, e^{\left (5 \, e^{x}\right )} \log \left (3\right )^{2} + {\left (e^{\left (10 \, e^{x}\right )} + 10 \, e^{\left (5 \, e^{x}\right )} + 25\right )} \log \left (-x {\left (e^{5} - 2\right )} + 9\right )^{2} + 25 \, \log \left (3\right )^{2} - 2 \, {\left (e^{\left (10 \, e^{x}\right )} \log \left (3\right ) + 10 \, e^{\left (5 \, e^{x}\right )} \log \left (3\right ) + 25 \, \log \left (3\right )\right )} \log \left (-x {\left (e^{5} - 2\right )} + 9\right )}{x^{2}} \] Input:
integrate(((((10*x^2*exp(5)-20*x^2-90*x)*exp(x)-2*x*exp(5)+4*x+18)*log(-1/ 3*x*exp(5)+2/3*x+3)^2+(2*x*exp(5)-4*x)*log(-1/3*x*exp(5)+2/3*x+3))*exp(5*e xp(x))^2+(((50*x^2*exp(5)-100*x^2-450*x)*exp(x)-20*x*exp(5)+40*x+180)*log( -1/3*x*exp(5)+2/3*x+3)^2+(20*x*exp(5)-40*x)*log(-1/3*x*exp(5)+2/3*x+3))*ex p(5*exp(x))+(-50*x*exp(5)+100*x+450)*log(-1/3*x*exp(5)+2/3*x+3)^2+(50*x*ex p(5)-100*x)*log(-1/3*x*exp(5)+2/3*x+3))/(x^4*exp(5)-2*x^4-9*x^3),x, algori thm="maxima")
Output:
(e^(10*e^x)*log(3)^2 + 10*e^(5*e^x)*log(3)^2 + (e^(10*e^x) + 10*e^(5*e^x) + 25)*log(-x*(e^5 - 2) + 9)^2 + 25*log(3)^2 - 2*(e^(10*e^x)*log(3) + 10*e^ (5*e^x)*log(3) + 25*log(3))*log(-x*(e^5 - 2) + 9))/x^2
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (25) = 50\).
Time = 0.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 4.41 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {e^{\left (10 \, e^{x}\right )} \log \left (3\right )^{2} + 10 \, e^{\left (5 \, e^{x}\right )} \log \left (3\right )^{2} - 2 \, e^{\left (10 \, e^{x}\right )} \log \left (3\right ) \log \left (-x e^{5} + 2 \, x + 9\right ) - 20 \, e^{\left (5 \, e^{x}\right )} \log \left (3\right ) \log \left (-x e^{5} + 2 \, x + 9\right ) + e^{\left (10 \, e^{x}\right )} \log \left (-x e^{5} + 2 \, x + 9\right )^{2} + 10 \, e^{\left (5 \, e^{x}\right )} \log \left (-x e^{5} + 2 \, x + 9\right )^{2} + 25 \, \log \left (3\right )^{2} - 50 \, \log \left (3\right ) \log \left (-x e^{5} + 2 \, x + 9\right ) + 25 \, \log \left (-x e^{5} + 2 \, x + 9\right )^{2}}{x^{2}} \] Input:
integrate(((((10*x^2*exp(5)-20*x^2-90*x)*exp(x)-2*x*exp(5)+4*x+18)*log(-1/ 3*x*exp(5)+2/3*x+3)^2+(2*x*exp(5)-4*x)*log(-1/3*x*exp(5)+2/3*x+3))*exp(5*e xp(x))^2+(((50*x^2*exp(5)-100*x^2-450*x)*exp(x)-20*x*exp(5)+40*x+180)*log( -1/3*x*exp(5)+2/3*x+3)^2+(20*x*exp(5)-40*x)*log(-1/3*x*exp(5)+2/3*x+3))*ex p(5*exp(x))+(-50*x*exp(5)+100*x+450)*log(-1/3*x*exp(5)+2/3*x+3)^2+(50*x*ex p(5)-100*x)*log(-1/3*x*exp(5)+2/3*x+3))/(x^4*exp(5)-2*x^4-9*x^3),x, algori thm="giac")
Output:
(e^(10*e^x)*log(3)^2 + 10*e^(5*e^x)*log(3)^2 - 2*e^(10*e^x)*log(3)*log(-x* e^5 + 2*x + 9) - 20*e^(5*e^x)*log(3)*log(-x*e^5 + 2*x + 9) + e^(10*e^x)*lo g(-x*e^5 + 2*x + 9)^2 + 10*e^(5*e^x)*log(-x*e^5 + 2*x + 9)^2 + 25*log(3)^2 - 50*log(3)*log(-x*e^5 + 2*x + 9) + 25*log(-x*e^5 + 2*x + 9)^2)/x^2
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx={\ln \left (\frac {2\,x}{3}-\frac {x\,{\mathrm {e}}^5}{3}+3\right )}^2\,\left (\frac {25}{x^2}+\frac {10\,{\mathrm {e}}^{5\,{\mathrm {e}}^x}}{x^2}+\frac {{\mathrm {e}}^{10\,{\mathrm {e}}^x}}{x^2}\right ) \] Input:
int(-(log((2*x)/3 - (x*exp(5))/3 + 3)^2*(100*x - 50*x*exp(5) + 450) + exp( 10*exp(x))*(log((2*x)/3 - (x*exp(5))/3 + 3)^2*(4*x - 2*x*exp(5) - exp(x)*( 90*x - 10*x^2*exp(5) + 20*x^2) + 18) - log((2*x)/3 - (x*exp(5))/3 + 3)*(4* x - 2*x*exp(5))) + exp(5*exp(x))*(log((2*x)/3 - (x*exp(5))/3 + 3)^2*(40*x - 20*x*exp(5) - exp(x)*(450*x - 50*x^2*exp(5) + 100*x^2) + 180) - log((2*x )/3 - (x*exp(5))/3 + 3)*(40*x - 20*x*exp(5))) - log((2*x)/3 - (x*exp(5))/3 + 3)*(100*x - 50*x*exp(5)))/(9*x^3 - x^4*exp(5) + 2*x^4),x)
Output:
log((2*x)/3 - (x*exp(5))/3 + 3)^2*(25/x^2 + (10*exp(5*exp(x)))/x^2 + exp(1 0*exp(x))/x^2)
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {\left (-100 x+50 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (450+100 x-50 e^5 x\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+e^{10 e^x} \left (\left (-4 x+2 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (18+4 x-2 e^5 x+e^x \left (-90 x-20 x^2+10 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )+e^{5 e^x} \left (\left (-40 x+20 e^5 x\right ) \log \left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )+\left (180+40 x-20 e^5 x+e^x \left (-450 x-100 x^2+50 e^5 x^2\right )\right ) \log ^2\left (\frac {1}{3} \left (9+2 x-e^5 x\right )\right )\right )}{-9 x^3-2 x^4+e^5 x^4} \, dx=\frac {\mathrm {log}\left (-\frac {1}{3} e^{5} x +\frac {2}{3} x +3\right )^{2} \left (e^{10 e^{x}}+10 e^{5 e^{x}}+25\right )}{x^{2}} \] Input:
int(((((10*x^2*exp(5)-20*x^2-90*x)*exp(x)-2*x*exp(5)+4*x+18)*log(-1/3*x*ex p(5)+2/3*x+3)^2+(2*x*exp(5)-4*x)*log(-1/3*x*exp(5)+2/3*x+3))*exp(5*exp(x)) ^2+(((50*x^2*exp(5)-100*x^2-450*x)*exp(x)-20*x*exp(5)+40*x+180)*log(-1/3*x *exp(5)+2/3*x+3)^2+(20*x*exp(5)-40*x)*log(-1/3*x*exp(5)+2/3*x+3))*exp(5*ex p(x))+(-50*x*exp(5)+100*x+450)*log(-1/3*x*exp(5)+2/3*x+3)^2+(50*x*exp(5)-1 00*x)*log(-1/3*x*exp(5)+2/3*x+3))/(x^4*exp(5)-2*x^4-9*x^3),x)
Output:
(log(( - e**5*x + 2*x + 9)/3)**2*(e**(10*e**x) + 10*e**(5*e**x) + 25))/x** 2