Integrand size = 226, antiderivative size = 30 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {25}{\left (-2+\frac {1}{3} \left (1+\log \left (\left (1-e^x-\log (20-x)\right )^2\right )\right )\right )^2} \] Output:
25/(-5/3+1/3*ln((1-exp(x)-ln(-x+20))^2))^2
Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\left (-5+\log \left (\left (-1+e^x+\log (20-x)\right )^2\right )\right )^2} \] Input:
Integrate[(-900 + E^x*(18000 - 900*x))/(-2500 + E^x*(2500 - 125*x) + 125*x + (2500 - 125*x)*Log[20 - x] + (1500 - 75*x + E^x*(-1500 + 75*x) + (-1500 + 75*x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2] + (-300 + E^x*(300 - 15*x) + 15*x + (300 - 15*x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2]^2 + (20 + E^x*(-20 + x) - x + (-20 + x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2]^3),x]
Output:
225/(-5 + Log[(-1 + E^x + Log[20 - x])^2])^2
Time = 0.83 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7239, 27, 7237}
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.
| \(\displaystyle \int \frac {e^x (18000-900 x)-900}{e^x (2500-125 x)+125 x+\left (e^x (300-15 x)+15 x+(300-15 x) \log (20-x)-300\right ) \log ^2\left (-2 e^x+e^{2 x}+\log ^2(20-x)+\left (2 e^x-2\right ) \log (20-x)+1\right )+\left (-75 x+e^x (75 x-1500)+(75 x-1500) \log (20-x)+1500\right ) \log \left (-2 e^x+e^{2 x}+\log ^2(20-x)+\left (2 e^x-2\right ) \log (20-x)+1\right )+\left (e^x (x-20)-x+(x-20) \log (20-x)+20\right ) \log ^3\left (-2 e^x+e^{2 x}+\log ^2(20-x)+\left (2 e^x-2\right ) \log (20-x)+1\right )+(2500-125 x) \log (20-x)-2500} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {900 \left (e^x (x-20)+1\right )}{(20-x) \left (-e^x-\log (20-x)+1\right ) \left (5-\log \left (\left (e^x+\log (20-x)-1\right )^2\right )\right )^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 900 \int \frac {1-e^x (20-x)}{(20-x) \left (-\log (20-x)-e^x+1\right ) \left (5-\log \left (\left (\log (20-x)+e^x-1\right )^2\right )\right )^3}dx\) |
| \(\Big \downarrow \) 7237 |
| \(\displaystyle \frac {225}{\left (5-\log \left (\left (e^x+\log (20-x)-1\right )^2\right )\right )^2}\) |
Input:
Int[(-900 + E^x*(18000 - 900*x))/(-2500 + E^x*(2500 - 125*x) + 125*x + (25 00 - 125*x)*Log[20 - x] + (1500 - 75*x + E^x*(-1500 + 75*x) + (-1500 + 75* x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[2 0 - x]^2] + (-300 + E^x*(300 - 15*x) + 15*x + (300 - 15*x)*Log[20 - x])*Lo g[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2]^2 + (20 + E^x*(-20 + x) - x + (-20 + x)*Log[20 - x])*Log[1 - 2*E^x + E^(2*x) + (-2 + 2*E^x)*Log[20 - x] + Log[20 - x]^2]^3),x]
Output:
225/(5 - Log[(-1 + E^x + Log[20 - x])^2])^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 9.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.70
| method | result | size |
| risch | \(-\frac {900}{{\left (\pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )\right )}^{2} \operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )\right ) {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )^{2}\right )}^{3}+4 i \ln \left (-1+{\mathrm e}^{x}+\ln \left (-x +20\right )\right )-10 i\right )}^{2}}\) | \(111\) |
| parallelrisch | \(\frac {45000 \ln \left (x -20\right )-45000 \ln \left (-x +20\right )+36000 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )-1800 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )^{2} \ln \left (-x +20\right )+18000 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right ) \ln \left (-x +20\right )-3600 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )^{2}+1800 \ln \left (x -20\right ) \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )^{2}-18000 \ln \left (x -20\right ) \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )}{400 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )^{2}-4000 \ln \left (\ln \left (-x +20\right )^{2}+\left (2 \,{\mathrm e}^{x}-2\right ) \ln \left (-x +20\right )+{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right )+10000}\) | \(320\) |
Input:
int(((-900*x+18000)*exp(x)-900)/(((x-20)*ln(-x+20)+(x-20)*exp(x)-x+20)*ln( ln(-x+20)^2+(2*exp(x)-2)*ln(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15*x+300)*ln( -x+20)+(-15*x+300)*exp(x)+15*x-300)*ln(ln(-x+20)^2+(2*exp(x)-2)*ln(-x+20)+ exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*ln(-x+20)+(75*x-1500)*exp(x)-75*x+1500 )*ln(ln(-x+20)^2+(2*exp(x)-2)*ln(-x+20)+exp(x)^2-2*exp(x)+1)+(-125*x+2500) *ln(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x,method=_RETURNVERBOSE)
Output:
-900/(Pi*csgn(I*(-1+exp(x)+ln(-x+20)))^2*csgn(I*(-1+exp(x)+ln(-x+20))^2)-2 *Pi*csgn(I*(-1+exp(x)+ln(-x+20)))*csgn(I*(-1+exp(x)+ln(-x+20))^2)^2+Pi*csg n(I*(-1+exp(x)+ln(-x+20))^2)^3+4*I*ln(-1+exp(x)+ln(-x+20))-10*I)^2
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (19) = 38\).
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\log \left (2 \, {\left (e^{x} - 1\right )} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )^{2} - 10 \, \log \left (2 \, {\left (e^{x} - 1\right )} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right ) + 25} \] Input:
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+ 20)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15* x+300)*log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x)- 2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*e xp(x)-75*x+1500)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x )+1)+(-125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x, algorith m="fricas")
Output:
225/(log(2*(e^x - 1)*log(-x + 20) + log(-x + 20)^2 + e^(2*x) - 2*e^x + 1)^ 2 - 10*log(2*(e^x - 1)*log(-x + 20) + log(-x + 20)^2 + e^(2*x) - 2*e^x + 1 ) + 25)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (20) = 40\).
Time = 0.60 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\log {\left (\left (2 e^{x} - 2\right ) \log {\left (20 - x \right )} + e^{2 x} - 2 e^{x} + \log {\left (20 - x \right )}^{2} + 1 \right )}^{2} - 10 \log {\left (\left (2 e^{x} - 2\right ) \log {\left (20 - x \right )} + e^{2 x} - 2 e^{x} + \log {\left (20 - x \right )}^{2} + 1 \right )} + 25} \] Input:
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*ln(-x+20)+(x-20)*exp(x)-x+2 0)*ln(ln(-x+20)**2+(2*exp(x)-2)*ln(-x+20)+exp(x)**2-2*exp(x)+1)**3+((-15*x +300)*ln(-x+20)+(-15*x+300)*exp(x)+15*x-300)*ln(ln(-x+20)**2+(2*exp(x)-2)* ln(-x+20)+exp(x)**2-2*exp(x)+1)**2+((75*x-1500)*ln(-x+20)+(75*x-1500)*exp( x)-75*x+1500)*ln(ln(-x+20)**2+(2*exp(x)-2)*ln(-x+20)+exp(x)**2-2*exp(x)+1) +(-125*x+2500)*ln(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x)
Output:
225/(log((2*exp(x) - 2)*log(20 - x) + exp(2*x) - 2*exp(x) + log(20 - x)**2 + 1)**2 - 10*log((2*exp(x) - 2)*log(20 - x) + exp(2*x) - 2*exp(x) + log(2 0 - x)**2 + 1) + 25)
Time = 1.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{4 \, \log \left (e^{x} + \log \left (-x + 20\right ) - 1\right )^{2} - 20 \, \log \left (e^{x} + \log \left (-x + 20\right ) - 1\right ) + 25} \] Input:
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+ 20)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15* x+300)*log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x)- 2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*e xp(x)-75*x+1500)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x )+1)+(-125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x, algorith m="maxima")
Output:
225/(4*log(e^x + log(-x + 20) - 1)^2 - 20*log(e^x + log(-x + 20) - 1) + 25 )
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (19) = 38\).
Time = 16.54 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.80 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\log \left (2 \, e^{x} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} - 2 \, \log \left (-x + 20\right ) + 1\right )^{2} - 10 \, \log \left (2 \, e^{x} \log \left (-x + 20\right ) + \log \left (-x + 20\right )^{2} + e^{\left (2 \, x\right )} - 2 \, e^{x} - 2 \, \log \left (-x + 20\right ) + 1\right ) + 25} \] Input:
integrate(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+ 20)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15* x+300)*log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x)- 2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*e xp(x)-75*x+1500)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x )+1)+(-125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x, algorith m="giac")
Output:
225/(log(2*e^x*log(-x + 20) + log(-x + 20)^2 + e^(2*x) - 2*e^x - 2*log(-x + 20) + 1)^2 - 10*log(2*e^x*log(-x + 20) + log(-x + 20)^2 + e^(2*x) - 2*e^ x - 2*log(-x + 20) + 1) + 25)
Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{{\left (\ln \left ({\ln \left (20-x\right )}^2+\left (2\,{\mathrm {e}}^x-2\right )\,\ln \left (20-x\right )+{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )-5\right )}^2} \] Input:
int((exp(x)*(900*x - 18000) + 900)/(log(20 - x)*(125*x - 2500) - log(exp(2 *x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*exp(x) - 2) + 1)*(log(20 - x)*(75*x - 1500) - 75*x + exp(x)*(75*x - 1500) + 1500) - 125*x + exp(x)*( 125*x - 2500) + log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*e xp(x) - 2) + 1)^2*(log(20 - x)*(15*x - 300) - 15*x + exp(x)*(15*x - 300) + 300) - log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*exp(x) - 2) + 1)^3*(exp(x)*(x - 20) - x + log(20 - x)*(x - 20) + 20) + 2500),x)
Output:
225/(log(exp(2*x) - 2*exp(x) + log(20 - x)^2 + log(20 - x)*(2*exp(x) - 2) + 1) - 5)^2
Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.00 \[ \int \frac {-900+e^x (18000-900 x)}{-2500+e^x (2500-125 x)+125 x+(2500-125 x) \log (20-x)+\left (1500-75 x+e^x (-1500+75 x)+(-1500+75 x) \log (20-x)\right ) \log \left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (-300+e^x (300-15 x)+15 x+(300-15 x) \log (20-x)\right ) \log ^2\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )+\left (20+e^x (-20+x)-x+(-20+x) \log (20-x)\right ) \log ^3\left (1-2 e^x+e^{2 x}+\left (-2+2 e^x\right ) \log (20-x)+\log ^2(20-x)\right )} \, dx=\frac {225}{\mathrm {log}\left (e^{2 x}+2 e^{x} \mathrm {log}\left (-x +20\right )-2 e^{x}+\mathrm {log}\left (-x +20\right )^{2}-2 \,\mathrm {log}\left (-x +20\right )+1\right )^{2}-10 \,\mathrm {log}\left (e^{2 x}+2 e^{x} \mathrm {log}\left (-x +20\right )-2 e^{x}+\mathrm {log}\left (-x +20\right )^{2}-2 \,\mathrm {log}\left (-x +20\right )+1\right )+25} \] Input:
int(((-900*x+18000)*exp(x)-900)/(((x-20)*log(-x+20)+(x-20)*exp(x)-x+20)*lo g(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)^3+((-15*x+300) *log(-x+20)+(-15*x+300)*exp(x)+15*x-300)*log(log(-x+20)^2+(2*exp(x)-2)*log (-x+20)+exp(x)^2-2*exp(x)+1)^2+((75*x-1500)*log(-x+20)+(75*x-1500)*exp(x)- 75*x+1500)*log(log(-x+20)^2+(2*exp(x)-2)*log(-x+20)+exp(x)^2-2*exp(x)+1)+( -125*x+2500)*log(-x+20)+(-125*x+2500)*exp(x)+125*x-2500),x)
Output:
225/(log(e**(2*x) + 2*e**x*log( - x + 20) - 2*e**x + log( - x + 20)**2 - 2 *log( - x + 20) + 1)**2 - 10*log(e**(2*x) + 2*e**x*log( - x + 20) - 2*e**x + log( - x + 20)**2 - 2*log( - x + 20) + 1) + 25)