Integrand size = 196, antiderivative size = 28 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\log \left (x+\left (-\frac {e^2}{5}+x+\frac {2}{3} \left (25-\log \left (e^x+x\right )\right )\right )^2\right ) \] Output:
ln(x+(x-exp(2-ln(5))+50/3-2/3*ln(exp(x)+x))^2)
Leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(28)=56\).
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.32 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\log \left (62500-1500 e^2+9 e^4+7725 x-90 e^2 x+225 x^2-5000 \log \left (e^x+x\right )+60 e^2 \log \left (e^x+x\right )-300 x \log \left (e^x+x\right )+100 \log ^2\left (e^x+x\right )\right ) \] Input:
Integrate[((5*(12 - 6*E^x - 18*x))/E^2 + (25*(-200 + 297*x + 18*x^2 + E^x* (109 + 6*x)))/E^4 + (25*(8 - 4*E^x - 12*x)*Log[E^x + x])/E^4)/(9*E^x + 9*x + (5*(E^x*(-300 - 18*x) - 300*x - 18*x^2))/E^2 + (25*(2500*x + 309*x^2 + 9*x^3 + E^x*(2500 + 309*x + 9*x^2)))/E^4 + ((5*(12*E^x + 12*x))/E^2 + (25* (E^x*(-200 - 12*x) - 200*x - 12*x^2))/E^4)*Log[E^x + x] + (25*(4*E^x + 4*x )*Log[E^x + x]^2)/E^4),x]
Output:
Log[62500 - 1500*E^2 + 9*E^4 + 7725*x - 90*E^2*x + 225*x^2 - 5000*Log[E^x + x] + 60*E^2*Log[E^x + x] - 300*x*Log[E^x + x] + 100*Log[E^x + x]^2]
Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(28)=56\).
Time = 1.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {7292, 27, 27, 7235}
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.
| \(\displaystyle \int \frac {\frac {25 \left (18 x^2+297 x+e^x (6 x+109)-200\right )}{e^4}+\frac {5 \left (-18 x-6 e^x+12\right )}{e^2}+\frac {25 \left (-12 x-4 e^x+8\right ) \log \left (x+e^x\right )}{e^4}}{\frac {5 \left (-18 x^2-300 x+e^x (-18 x-300)\right )}{e^2}+\left (\frac {25 \left (-12 x^2-200 x+e^x (-12 x-200)\right )}{e^4}+\frac {5 \left (12 x+12 e^x\right )}{e^2}\right ) \log \left (x+e^x\right )+\frac {25 \left (9 x^3+309 x^2+e^x \left (9 x^2+309 x+2500\right )+2500 x\right )}{e^4}+9 e^x+9 x+\frac {25 \left (4 x+4 e^x\right ) \log ^2\left (x+e^x\right )}{e^4}} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^4 \left (\frac {25 \left (18 x^2+297 x+e^x (6 x+109)-200\right )}{e^4}+\frac {5 \left (-18 x-6 e^x+12\right )}{e^2}+\frac {25 \left (-12 x-4 e^x+8\right ) \log \left (x+e^x\right )}{e^4}\right )}{\left (x+e^x\right ) \left (225 x^2+7725 \left (1-\frac {6 e^2}{515}\right ) x+100 \log ^2\left (x+e^x\right )-300 x \log \left (x+e^x\right )-5000 \left (1-\frac {3 e^2}{250}\right ) \log \left (x+e^x\right )+62500 \left (1+\frac {3 e^2 \left (3 e^2-500\right )}{62500}\right )\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^4 \int \frac {5 \left (\frac {20 \log \left (x+e^x\right ) \left (-3 x-e^x+2\right )}{e^4}+\frac {6 \left (-3 x-e^x+2\right )}{e^2}-\frac {5 \left (-18 x^2-297 x-e^x (6 x+109)+200\right )}{e^4}\right )}{\left (x+e^x\right ) \left (225 x^2-300 \log \left (x+e^x\right ) x+15 \left (515-6 e^2\right ) x+100 \log ^2\left (x+e^x\right )-20 \left (250-3 e^2\right ) \log \left (x+e^x\right )+\left (250-3 e^2\right )^2\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 5 e^4 \int \frac {\frac {20 \log \left (x+e^x\right ) \left (-3 x-e^x+2\right )}{e^4}+\frac {6 \left (-3 x-e^x+2\right )}{e^2}-\frac {5 \left (-18 x^2-297 x-e^x (6 x+109)+200\right )}{e^4}}{\left (x+e^x\right ) \left (225 x^2-300 \log \left (x+e^x\right ) x+15 \left (515-6 e^2\right ) x+100 \log ^2\left (x+e^x\right )-20 \left (250-3 e^2\right ) \log \left (x+e^x\right )+\left (250-3 e^2\right )^2\right )}dx\) |
| \(\Big \downarrow \) 7235 |
| \(\displaystyle \log \left (225 x^2+15 \left (515-6 e^2\right ) x+100 \log ^2\left (x+e^x\right )-300 x \log \left (x+e^x\right )-20 \left (250-3 e^2\right ) \log \left (x+e^x\right )+\left (250-3 e^2\right )^2\right )\) |
Input:
Int[((5*(12 - 6*E^x - 18*x))/E^2 + (25*(-200 + 297*x + 18*x^2 + E^x*(109 + 6*x)))/E^4 + (25*(8 - 4*E^x - 12*x)*Log[E^x + x])/E^4)/(9*E^x + 9*x + (5* (E^x*(-300 - 18*x) - 300*x - 18*x^2))/E^2 + (25*(2500*x + 309*x^2 + 9*x^3 + E^x*(2500 + 309*x + 9*x^2)))/E^4 + ((5*(12*E^x + 12*x))/E^2 + (25*(E^x*( -200 - 12*x) - 200*x - 12*x^2))/E^4)*Log[E^x + x] + (25*(4*E^x + 4*x)*Log[ E^x + x]^2)/E^4),x]
Output:
Log[(250 - 3*E^2)^2 + 15*(515 - 6*E^2)*x + 225*x^2 - 20*(250 - 3*E^2)*Log[ E^x + x] - 300*x*Log[E^x + x] + 100*Log[E^x + x]^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_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(24)=48\).
Time = 1.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00
| method | result | size |
| risch | \(\ln \left (\ln \left ({\mathrm e}^{x}+x \right )^{2}-3 x \ln \left ({\mathrm e}^{x}+x \right )-50 \ln \left ({\mathrm e}^{x}+x \right )+\frac {3 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{x}+x \right )}{5}+\frac {9 x^{2}}{4}+\frac {309 x}{4}-\frac {9 \,{\mathrm e}^{2} x}{10}+625-15 \,{\mathrm e}^{2}+\frac {9 \,{\mathrm e}^{4}}{100}\right )\) | \(56\) |
| parallelrisch | \(\ln \left (\frac {\left (100 \,{\mathrm e}^{-4} \ln \left ({\mathrm e}^{x}+x \right )^{2}-300 \,{\mathrm e}^{-4} \ln \left ({\mathrm e}^{x}+x \right ) x +225 \,{\mathrm e}^{-4} x^{2}-5000 \,{\mathrm e}^{-4} \ln \left ({\mathrm e}^{x}+x \right )+7725 \,{\mathrm e}^{-4} x +62500 \,{\mathrm e}^{-4}+12 \,{\mathrm e}^{\ln \left (5\right )-2} \ln \left ({\mathrm e}^{x}+x \right )-18 \,{\mathrm e}^{\ln \left (5\right )-2} x -300 \,{\mathrm e}^{\ln \left (5\right )-2}+9\right ) {\mathrm e}^{4}}{225}\right )\) | \(116\) |
Input:
int(((-4*exp(x)-12*x+8)*exp(ln(5)-2)^2*ln(exp(x)+x)+((6*x+109)*exp(x)+18*x ^2+297*x-200)*exp(ln(5)-2)^2+(-6*exp(x)-18*x+12)*exp(ln(5)-2))/((4*exp(x)+ 4*x)*exp(ln(5)-2)^2*ln(exp(x)+x)^2+(((-12*x-200)*exp(x)-12*x^2-200*x)*exp( ln(5)-2)^2+(12*exp(x)+12*x)*exp(ln(5)-2))*ln(exp(x)+x)+((9*x^2+309*x+2500) *exp(x)+9*x^3+309*x^2+2500*x)*exp(ln(5)-2)^2+((-18*x-300)*exp(x)-18*x^2-30 0*x)*exp(ln(5)-2)+9*exp(x)+9*x),x,method=_RETURNVERBOSE)
Output:
ln(ln(exp(x)+x)^2-3*x*ln(exp(x)+x)-50*ln(exp(x)+x)+3/5*exp(2)*ln(exp(x)+x) +9/4*x^2+309/4*x-9/10*exp(2)*x+625-15*exp(2)+9/100*exp(4))
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (28) = 56\).
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.75 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\log \left (4 \, e^{\left (2 \, \log \left (5\right ) - 4\right )} \log \left (x + e^{x}\right )^{2} + {\left (9 \, x^{2} + 309 \, x + 2500\right )} e^{\left (2 \, \log \left (5\right ) - 4\right )} - 6 \, {\left (3 \, x + 50\right )} e^{\left (\log \left (5\right ) - 2\right )} - 4 \, {\left ({\left (3 \, x + 50\right )} e^{\left (2 \, \log \left (5\right ) - 4\right )} - 3 \, e^{\left (\log \left (5\right ) - 2\right )}\right )} \log \left (x + e^{x}\right ) + 9\right ) \] Input:
integrate(((-4*exp(x)-12*x+8)*exp(log(5)-2)^2*log(exp(x)+x)+((6*x+109)*exp (x)+18*x^2+297*x-200)*exp(log(5)-2)^2+(-6*exp(x)-18*x+12)*exp(log(5)-2))/( (4*exp(x)+4*x)*exp(log(5)-2)^2*log(exp(x)+x)^2+(((-12*x-200)*exp(x)-12*x^2 -200*x)*exp(log(5)-2)^2+(12*exp(x)+12*x)*exp(log(5)-2))*log(exp(x)+x)+((9* x^2+309*x+2500)*exp(x)+9*x^3+309*x^2+2500*x)*exp(log(5)-2)^2+((-18*x-300)* exp(x)-18*x^2-300*x)*exp(log(5)-2)+9*exp(x)+9*x),x, algorithm="fricas")
Output:
log(4*e^(2*log(5) - 4)*log(x + e^x)^2 + (9*x^2 + 309*x + 2500)*e^(2*log(5) - 4) - 6*(3*x + 50)*e^(log(5) - 2) - 4*((3*x + 50)*e^(2*log(5) - 4) - 3*e ^(log(5) - 2))*log(x + e^x) + 9)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).
Time = 0.45 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.18 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\log {\left (\frac {9 x^{2}}{4} - \frac {9 x e^{2}}{10} + \frac {309 x}{4} + \left (- 3 x - 50 + \frac {3 e^{2}}{5}\right ) \log {\left (x + e^{x} \right )} + \log {\left (x + e^{x} \right )}^{2} - 15 e^{2} + \frac {9 e^{4}}{100} + 625 \right )} \] Input:
integrate(((-4*exp(x)-12*x+8)*exp(ln(5)-2)**2*ln(exp(x)+x)+((6*x+109)*exp( x)+18*x**2+297*x-200)*exp(ln(5)-2)**2+(-6*exp(x)-18*x+12)*exp(ln(5)-2))/(( 4*exp(x)+4*x)*exp(ln(5)-2)**2*ln(exp(x)+x)**2+(((-12*x-200)*exp(x)-12*x**2 -200*x)*exp(ln(5)-2)**2+(12*exp(x)+12*x)*exp(ln(5)-2))*ln(exp(x)+x)+((9*x* *2+309*x+2500)*exp(x)+9*x**3+309*x**2+2500*x)*exp(ln(5)-2)**2+((-18*x-300) *exp(x)-18*x**2-300*x)*exp(ln(5)-2)+9*exp(x)+9*x),x)
Output:
log(9*x**2/4 - 9*x*exp(2)/10 + 309*x/4 + (-3*x - 50 + 3*exp(2)/5)*log(x + exp(x)) + log(x + exp(x))**2 - 15*exp(2) + 9*exp(4)/100 + 625)
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\log \left (\frac {9}{4} \, x^{2} - \frac {3}{20} \, x {\left (6 \, e^{2} - 515\right )} - \frac {1}{5} \, {\left (15 \, x - 3 \, e^{2} + 250\right )} \log \left (x + e^{x}\right ) + \log \left (x + e^{x}\right )^{2} + \frac {9}{100} \, e^{4} - 15 \, e^{2} + 625\right ) \] Input:
integrate(((-4*exp(x)-12*x+8)*exp(log(5)-2)^2*log(exp(x)+x)+((6*x+109)*exp (x)+18*x^2+297*x-200)*exp(log(5)-2)^2+(-6*exp(x)-18*x+12)*exp(log(5)-2))/( (4*exp(x)+4*x)*exp(log(5)-2)^2*log(exp(x)+x)^2+(((-12*x-200)*exp(x)-12*x^2 -200*x)*exp(log(5)-2)^2+(12*exp(x)+12*x)*exp(log(5)-2))*log(exp(x)+x)+((9* x^2+309*x+2500)*exp(x)+9*x^3+309*x^2+2500*x)*exp(log(5)-2)^2+((-18*x-300)* exp(x)-18*x^2-300*x)*exp(log(5)-2)+9*exp(x)+9*x),x, algorithm="maxima")
Output:
log(9/4*x^2 - 3/20*x*(6*e^2 - 515) - 1/5*(15*x - 3*e^2 + 250)*log(x + e^x) + log(x + e^x)^2 + 9/100*e^4 - 15*e^2 + 625)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.04 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\log \left (225 \, x^{2} - 90 \, x e^{2} - 300 \, x \log \left (x + e^{x}\right ) + 60 \, e^{2} \log \left (x + e^{x}\right ) + 100 \, \log \left (x + e^{x}\right )^{2} + 7725 \, x + 9 \, e^{4} - 1500 \, e^{2} - 5000 \, \log \left (x + e^{x}\right ) + 62500\right ) \] Input:
integrate(((-4*exp(x)-12*x+8)*exp(log(5)-2)^2*log(exp(x)+x)+((6*x+109)*exp (x)+18*x^2+297*x-200)*exp(log(5)-2)^2+(-6*exp(x)-18*x+12)*exp(log(5)-2))/( (4*exp(x)+4*x)*exp(log(5)-2)^2*log(exp(x)+x)^2+(((-12*x-200)*exp(x)-12*x^2 -200*x)*exp(log(5)-2)^2+(12*exp(x)+12*x)*exp(log(5)-2))*log(exp(x)+x)+((9* x^2+309*x+2500)*exp(x)+9*x^3+309*x^2+2500*x)*exp(log(5)-2)^2+((-18*x-300)* exp(x)-18*x^2-300*x)*exp(log(5)-2)+9*exp(x)+9*x),x, algorithm="giac")
Output:
log(225*x^2 - 90*x*e^2 - 300*x*log(x + e^x) + 60*e^2*log(x + e^x) + 100*lo g(x + e^x)^2 + 7725*x + 9*e^4 - 1500*e^2 - 5000*log(x + e^x) + 62500)
Time = 5.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\ln \left (\frac {309\,x}{4}-50\,\ln \left (x+{\mathrm {e}}^x\right )-15\,{\mathrm {e}}^2+\frac {9\,{\mathrm {e}}^4}{100}-\frac {9\,x\,{\mathrm {e}}^2}{10}+{\ln \left (x+{\mathrm {e}}^x\right )}^2+\frac {9\,x^2}{4}+\frac {3\,\ln \left (x+{\mathrm {e}}^x\right )\,{\mathrm {e}}^2}{5}-3\,x\,\ln \left (x+{\mathrm {e}}^x\right )+625\right ) \] Input:
int(-(exp(log(5) - 2)*(18*x + 6*exp(x) - 12) - exp(2*log(5) - 4)*(297*x + exp(x)*(6*x + 109) + 18*x^2 - 200) + log(x + exp(x))*exp(2*log(5) - 4)*(12 *x + 4*exp(x) - 8))/(9*x + 9*exp(x) + exp(2*log(5) - 4)*(2500*x + exp(x)*( 309*x + 9*x^2 + 2500) + 309*x^2 + 9*x^3) - exp(log(5) - 2)*(300*x + exp(x) *(18*x + 300) + 18*x^2) + log(x + exp(x))*(exp(log(5) - 2)*(12*x + 12*exp( x)) - exp(2*log(5) - 4)*(200*x + exp(x)*(12*x + 200) + 12*x^2)) + log(x + exp(x))^2*exp(2*log(5) - 4)*(4*x + 4*exp(x))),x)
Output:
log((309*x)/4 - 50*log(x + exp(x)) - 15*exp(2) + (9*exp(4))/100 - (9*x*exp (2))/10 + log(x + exp(x))^2 + (9*x^2)/4 + (3*log(x + exp(x))*exp(2))/5 - 3 *x*log(x + exp(x)) + 625)
\[ \int \frac {\frac {5 \left (12-6 e^x-18 x\right )}{e^2}+\frac {25 \left (-200+297 x+18 x^2+e^x (109+6 x)\right )}{e^4}+\frac {25 \left (8-4 e^x-12 x\right ) \log \left (e^x+x\right )}{e^4}}{9 e^x+9 x+\frac {5 \left (e^x (-300-18 x)-300 x-18 x^2\right )}{e^2}+\frac {25 \left (2500 x+309 x^2+9 x^3+e^x \left (2500+309 x+9 x^2\right )\right )}{e^4}+\left (\frac {5 \left (12 e^x+12 x\right )}{e^2}+\frac {25 \left (e^x (-200-12 x)-200 x-12 x^2\right )}{e^4}\right ) \log \left (e^x+x\right )+\frac {25 \left (4 e^x+4 x\right ) \log ^2\left (e^x+x\right )}{e^4}} \, dx=\text {too large to display} \] Input:
int(((-4*exp(x)-12*x+8)*exp(log(5)-2)^2*log(exp(x)+x)+((6*x+109)*exp(x)+18 *x^2+297*x-200)*exp(log(5)-2)^2+(-6*exp(x)-18*x+12)*exp(log(5)-2))/((4*exp (x)+4*x)*exp(log(5)-2)^2*log(exp(x)+x)^2+(((-12*x-200)*exp(x)-12*x^2-200*x )*exp(log(5)-2)^2+(12*exp(x)+12*x)*exp(log(5)-2))*log(exp(x)+x)+((9*x^2+30 9*x+2500)*exp(x)+9*x^3+309*x^2+2500*x)*exp(log(5)-2)^2+((-18*x-300)*exp(x) -18*x^2-300*x)*exp(log(5)-2)+9*exp(x)+9*x),x)
Output:
5*( - 6*int(e**x/(100*e**x*log(e**x + x)**2 + 60*e**x*log(e**x + x)*e**2 - 300*e**x*log(e**x + x)*x - 5000*e**x*log(e**x + x) + 9*e**x*e**4 - 90*e** x*e**2*x - 1500*e**x*e**2 + 225*e**x*x**2 + 7725*e**x*x + 62500*e**x + 100 *log(e**x + x)**2*x + 60*log(e**x + x)*e**2*x - 300*log(e**x + x)*x**2 - 5 000*log(e**x + x)*x + 9*e**4*x - 90*e**2*x**2 - 1500*e**2*x + 225*x**3 + 7 725*x**2 + 62500*x),x)*e**2 + 545*int(e**x/(100*e**x*log(e**x + x)**2 + 60 *e**x*log(e**x + x)*e**2 - 300*e**x*log(e**x + x)*x - 5000*e**x*log(e**x + x) + 9*e**x*e**4 - 90*e**x*e**2*x - 1500*e**x*e**2 + 225*e**x*x**2 + 7725 *e**x*x + 62500*e**x + 100*log(e**x + x)**2*x + 60*log(e**x + x)*e**2*x - 300*log(e**x + x)*x**2 - 5000*log(e**x + x)*x + 9*e**4*x - 90*e**2*x**2 - 1500*e**2*x + 225*x**3 + 7725*x**2 + 62500*x),x) + 90*int(x**2/(100*e**x*l og(e**x + x)**2 + 60*e**x*log(e**x + x)*e**2 - 300*e**x*log(e**x + x)*x - 5000*e**x*log(e**x + x) + 9*e**x*e**4 - 90*e**x*e**2*x - 1500*e**x*e**2 + 225*e**x*x**2 + 7725*e**x*x + 62500*e**x + 100*log(e**x + x)**2*x + 60*log (e**x + x)*e**2*x - 300*log(e**x + x)*x**2 - 5000*log(e**x + x)*x + 9*e**4 *x - 90*e**2*x**2 - 1500*e**2*x + 225*x**3 + 7725*x**2 + 62500*x),x) + 40* int(log(e**x + x)/(100*e**x*log(e**x + x)**2 + 60*e**x*log(e**x + x)*e**2 - 300*e**x*log(e**x + x)*x - 5000*e**x*log(e**x + x) + 9*e**x*e**4 - 90*e* *x*e**2*x - 1500*e**x*e**2 + 225*e**x*x**2 + 7725*e**x*x + 62500*e**x + 10 0*log(e**x + x)**2*x + 60*log(e**x + x)*e**2*x - 300*log(e**x + x)*x**2...