Integrand size = 120, antiderivative size = 21 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x \log \left (2 \left (2+e^x\right )\right ) \left (3+(5+\log (\log (12)))^2\right )^2 \] Output:
x*ln(2*exp(x)+4)*((5+ln(ln(12)))^2+3)^2
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.90 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=\left (28+10 \log (\log (12))+\log ^2(\log (12))\right )^2 \left (\frac {x^2}{2}+x \log (4)+x \log \left (1+2 e^{-x}\right )-\operatorname {PolyLog}\left (2,-2 e^{-x}\right )-\operatorname {PolyLog}\left (2,-\frac {e^x}{2}\right )\right ) \] Input:
Integrate[(784*E^x*x + 560*E^x*x*Log[Log[12]] + 156*E^x*x*Log[Log[12]]^2 + 20*E^x*x*Log[Log[12]]^3 + E^x*x*Log[Log[12]]^4 + Log[4 + 2*E^x]*(1568 + 7 84*E^x + (1120 + 560*E^x)*Log[Log[12]] + (312 + 156*E^x)*Log[Log[12]]^2 + (40 + 20*E^x)*Log[Log[12]]^3 + (2 + E^x)*Log[Log[12]]^4))/(2 + E^x),x]
Output:
(28 + 10*Log[Log[12]] + Log[Log[12]]^2)^2*(x^2/2 + x*Log[4] + x*Log[1 + 2/ E^x] - PolyLog[2, -2/E^x] - PolyLog[2, -1/2*E^x])
Time = 0.61 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6, 6, 6, 6, 7239, 27, 7293, 2009}
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.
| \(\displaystyle \int \frac {784 e^x x+e^x x \log ^4(\log (12))+20 e^x x \log ^3(\log (12))+156 e^x x \log ^2(\log (12))+\log \left (2 e^x+4\right ) \left (784 e^x+\left (e^x+2\right ) \log ^4(\log (12))+\left (20 e^x+40\right ) \log ^3(\log (12))+\left (156 e^x+312\right ) \log ^2(\log (12))+\left (560 e^x+1120\right ) \log (\log (12))+1568\right )+560 e^x x \log (\log (12))}{e^x+2} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^x x \log ^4(\log (12))+20 e^x x \log ^3(\log (12))+156 e^x x \log ^2(\log (12))+\log \left (2 e^x+4\right ) \left (784 e^x+\left (e^x+2\right ) \log ^4(\log (12))+\left (20 e^x+40\right ) \log ^3(\log (12))+\left (156 e^x+312\right ) \log ^2(\log (12))+\left (560 e^x+1120\right ) \log (\log (12))+1568\right )+e^x x (784+560 \log (\log (12)))}{e^x+2}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^x x \log ^4(\log (12))+e^x x \left (20 \log ^3(\log (12))+156 \log ^2(\log (12))\right )+\log \left (2 e^x+4\right ) \left (784 e^x+\left (e^x+2\right ) \log ^4(\log (12))+\left (20 e^x+40\right ) \log ^3(\log (12))+\left (156 e^x+312\right ) \log ^2(\log (12))+\left (560 e^x+1120\right ) \log (\log (12))+1568\right )+e^x x (784+560 \log (\log (12)))}{e^x+2}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^x x \left (784+\log ^4(\log (12))+560 \log (\log (12))\right )+e^x x \left (20 \log ^3(\log (12))+156 \log ^2(\log (12))\right )+\log \left (2 e^x+4\right ) \left (784 e^x+\left (e^x+2\right ) \log ^4(\log (12))+\left (20 e^x+40\right ) \log ^3(\log (12))+\left (156 e^x+312\right ) \log ^2(\log (12))+\left (560 e^x+1120\right ) \log (\log (12))+1568\right )}{e^x+2}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^x x \left (784+\log ^4(\log (12))+20 \log ^3(\log (12))+156 \log ^2(\log (12))+560 \log (\log (12))\right )+\log \left (2 e^x+4\right ) \left (784 e^x+\left (e^x+2\right ) \log ^4(\log (12))+\left (20 e^x+40\right ) \log ^3(\log (12))+\left (156 e^x+312\right ) \log ^2(\log (12))+\left (560 e^x+1120\right ) \log (\log (12))+1568\right )}{e^x+2}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (28+\log ^2(\log (12))+10 \log (\log (12))\right )^2 \left (e^x x+\left (e^x+2\right ) \log \left (2 \left (e^x+2\right )\right )\right )}{e^x+2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \left (28+\log ^2(\log (12))+10 \log (\log (12))\right )^2 \int \frac {e^x x+\left (2+e^x\right ) \log \left (2 \left (2+e^x\right )\right )}{2+e^x}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \left (28+\log ^2(\log (12))+10 \log (\log (12))\right )^2 \int \left (-\frac {2 x}{2+e^x}+x+\log \left (2 \left (2+e^x\right )\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \left (28+\log ^2(\log (12))+10 \log (\log (12))\right )^2 \left (x \log \left (\frac {e^x}{2}+1\right )+x \log (4)\right )\) |
Input:
Int[(784*E^x*x + 560*E^x*x*Log[Log[12]] + 156*E^x*x*Log[Log[12]]^2 + 20*E^ x*x*Log[Log[12]]^3 + E^x*x*Log[Log[12]]^4 + Log[4 + 2*E^x]*(1568 + 784*E^x + (1120 + 560*E^x)*Log[Log[12]] + (312 + 156*E^x)*Log[Log[12]]^2 + (40 + 20*E^x)*Log[Log[12]]^3 + (2 + E^x)*Log[Log[12]]^4))/(2 + E^x),x]
Output:
(x*Log[4] + x*Log[1 + E^x/2])*(28 + 10*Log[Log[12]] + Log[Log[12]]^2)^2
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71
| method | result | size |
| norman | \(\left (\ln \left (\ln \left (12\right )\right )^{4}+20 \ln \left (\ln \left (12\right )\right )^{3}+156 \ln \left (\ln \left (12\right )\right )^{2}+560 \ln \left (\ln \left (12\right )\right )+784\right ) x \ln \left (2 \,{\mathrm e}^{x}+4\right )\) | \(36\) |
| risch | \(x \left (\ln \left (2 \ln \left (2\right )+\ln \left (3\right )\right )^{4}+20 \ln \left (2 \ln \left (2\right )+\ln \left (3\right )\right )^{3}+156 \ln \left (2 \ln \left (2\right )+\ln \left (3\right )\right )^{2}+560 \ln \left (2 \ln \left (2\right )+\ln \left (3\right )\right )+784\right ) \ln \left (2 \,{\mathrm e}^{x}+4\right )\) | \(56\) |
| parallelrisch | \(\ln \left (\ln \left (12\right )\right )^{4} x \ln \left (2 \,{\mathrm e}^{x}+4\right )+20 \ln \left (\ln \left (12\right )\right )^{3} x \ln \left (2 \,{\mathrm e}^{x}+4\right )+156 \ln \left (\ln \left (12\right )\right )^{2} x \ln \left (2 \,{\mathrm e}^{x}+4\right )+560 \ln \left (\ln \left (12\right )\right ) x \ln \left (2 \,{\mathrm e}^{x}+4\right )+784 x \ln \left (2 \,{\mathrm e}^{x}+4\right )\) | \(69\) |
| default | \(784 x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )+\ln \left (\ln \left (12\right )\right )^{4} \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+20 \ln \left (\ln \left (12\right )\right )^{3} \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+156 \ln \left (\ln \left (12\right )\right )^{2} \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+560 \ln \left (\ln \left (12\right )\right ) \left (\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )+x \ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+784 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )+\ln \left (\ln \left (12\right )\right )^{4} \left (\left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-\operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+\ln \left (\ln \left (12\right )\right )^{3} \left (20 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-20 \operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+\ln \left (\ln \left (12\right )\right )^{2} \left (156 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-156 \operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )+\ln \left (\ln \left (12\right )\right ) \left (560 \left (\ln \left (2 \,{\mathrm e}^{x}+4\right )-\ln \left (\frac {{\mathrm e}^{x}}{2}+1\right )\right ) \ln \left (-\frac {{\mathrm e}^{x}}{2}\right )-560 \operatorname {dilog}\left (\frac {{\mathrm e}^{x}}{2}+1\right )\right )\) | \(286\) |
Input:
int((((exp(x)+2)*ln(ln(12))^4+(20*exp(x)+40)*ln(ln(12))^3+(156*exp(x)+312) *ln(ln(12))^2+(560*exp(x)+1120)*ln(ln(12))+784*exp(x)+1568)*ln(2*exp(x)+4) +x*exp(x)*ln(ln(12))^4+20*x*exp(x)*ln(ln(12))^3+156*x*exp(x)*ln(ln(12))^2+ 560*x*exp(x)*ln(ln(12))+784*exp(x)*x)/(exp(x)+2),x,method=_RETURNVERBOSE)
Output:
(ln(ln(12))^4+20*ln(ln(12))^3+156*ln(ln(12))^2+560*ln(ln(12))+784)*x*ln(2* exp(x)+4)
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx={\left (x \log \left (\log \left (12\right )\right )^{4} + 20 \, x \log \left (\log \left (12\right )\right )^{3} + 156 \, x \log \left (\log \left (12\right )\right )^{2} + 560 \, x \log \left (\log \left (12\right )\right ) + 784 \, x\right )} \log \left (2 \, e^{x} + 4\right ) \] Input:
integrate((((2+exp(x))*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*e xp(x)+312)*log(log(12))^2+(560*exp(x)+1120)*log(log(12))+784*exp(x)+1568)* log(2*exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x*e xp(x)*log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(2+exp(x)),x, algorithm="fricas")
Output:
(x*log(log(12))^4 + 20*x*log(log(12))^3 + 156*x*log(log(12))^2 + 560*x*log (log(12)) + 784*x)*log(2*e^x + 4)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=\left (x \log {\left (\log {\left (12 \right )} \right )}^{4} + 20 x \log {\left (\log {\left (12 \right )} \right )}^{3} + 156 x \log {\left (\log {\left (12 \right )} \right )}^{2} + 560 x \log {\left (\log {\left (12 \right )} \right )} + 784 x\right ) \log {\left (2 e^{x} + 4 \right )} \] Input:
integrate((((2+exp(x))*ln(ln(12))**4+(20*exp(x)+40)*ln(ln(12))**3+(156*exp (x)+312)*ln(ln(12))**2+(560*exp(x)+1120)*ln(ln(12))+784*exp(x)+1568)*ln(2* exp(x)+4)+x*exp(x)*ln(ln(12))**4+20*x*exp(x)*ln(ln(12))**3+156*x*exp(x)*ln (ln(12))**2+560*x*exp(x)*ln(ln(12))+784*exp(x)*x)/(2+exp(x)),x)
Output:
(x*log(log(12))**4 + 20*x*log(log(12))**3 + 156*x*log(log(12))**2 + 560*x* log(log(12)) + 784*x)*log(2*exp(x) + 4)
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (20) = 40\).
Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.95 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx={\left (\log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{4} + 20 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{3} + 156 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{2} + 560 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right ) + 784\right )} x \log \left (2\right ) + {\left (\log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{4} + 20 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{3} + 156 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )^{2} + 560 \, \log \left (\log \left (3\right ) + 2 \, \log \left (2\right )\right ) + 784\right )} x \log \left (e^{x} + 2\right ) \] Input:
integrate((((2+exp(x))*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*e xp(x)+312)*log(log(12))^2+(560*exp(x)+1120)*log(log(12))+784*exp(x)+1568)* log(2*exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x*e xp(x)*log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(2+exp(x)),x, algorithm="maxima")
Output:
(log(log(3) + 2*log(2))^4 + 20*log(log(3) + 2*log(2))^3 + 156*log(log(3) + 2*log(2))^2 + 560*log(log(3) + 2*log(2)) + 784)*x*log(2) + (log(log(3) + 2*log(2))^4 + 20*log(log(3) + 2*log(2))^3 + 156*log(log(3) + 2*log(2))^2 + 560*log(log(3) + 2*log(2)) + 784)*x*log(e^x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.76 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x \log \left (2\right ) \log \left (\log \left (12\right )\right )^{4} + x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right )^{4} + 20 \, x \log \left (2\right ) \log \left (\log \left (12\right )\right )^{3} + 20 \, x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right )^{3} + 156 \, x \log \left (2\right ) \log \left (\log \left (12\right )\right )^{2} + 156 \, x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right )^{2} + 560 \, x \log \left (2\right ) \log \left (\log \left (12\right )\right ) + 560 \, x \log \left (e^{x} + 2\right ) \log \left (\log \left (12\right )\right ) + 784 \, x \log \left (2\right ) + 784 \, x \log \left (e^{x} + 2\right ) \] Input:
integrate((((2+exp(x))*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*e xp(x)+312)*log(log(12))^2+(560*exp(x)+1120)*log(log(12))+784*exp(x)+1568)* log(2*exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x*e xp(x)*log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(2+exp(x)),x, algorithm="giac")
Output:
x*log(2)*log(log(12))^4 + x*log(e^x + 2)*log(log(12))^4 + 20*x*log(2)*log( log(12))^3 + 20*x*log(e^x + 2)*log(log(12))^3 + 156*x*log(2)*log(log(12))^ 2 + 156*x*log(e^x + 2)*log(log(12))^2 + 560*x*log(2)*log(log(12)) + 560*x* log(e^x + 2)*log(log(12)) + 784*x*log(2) + 784*x*log(e^x + 2)
Time = 4.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=x\,\left (\ln \left (2\right )+\ln \left ({\mathrm {e}}^x+2\right )\right )\,{\left (10\,\ln \left (\ln \left (12\right )\right )+{\ln \left (\ln \left (12\right )\right )}^2+28\right )}^2 \] Input:
int((log(2*exp(x) + 4)*(784*exp(x) + log(log(12))^3*(20*exp(x) + 40) + log (log(12))^2*(156*exp(x) + 312) + log(log(12))*(560*exp(x) + 1120) + log(lo g(12))^4*(exp(x) + 2) + 1568) + 784*x*exp(x) + 560*x*exp(x)*log(log(12)) + 156*x*exp(x)*log(log(12))^2 + 20*x*exp(x)*log(log(12))^3 + x*exp(x)*log(l og(12))^4)/(exp(x) + 2),x)
Output:
x*(log(2) + log(exp(x) + 2))*(10*log(log(12)) + log(log(12))^2 + 28)^2
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {784 e^x x+560 e^x x \log (\log (12))+156 e^x x \log ^2(\log (12))+20 e^x x \log ^3(\log (12))+e^x x \log ^4(\log (12))+\log \left (4+2 e^x\right ) \left (1568+784 e^x+\left (1120+560 e^x\right ) \log (\log (12))+\left (312+156 e^x\right ) \log ^2(\log (12))+\left (40+20 e^x\right ) \log ^3(\log (12))+\left (2+e^x\right ) \log ^4(\log (12))\right )}{2+e^x} \, dx=\mathrm {log}\left (2 e^{x}+4\right ) x \left (\mathrm {log}\left (\mathrm {log}\left (12\right )\right )^{4}+20 \mathrm {log}\left (\mathrm {log}\left (12\right )\right )^{3}+156 \mathrm {log}\left (\mathrm {log}\left (12\right )\right )^{2}+560 \,\mathrm {log}\left (\mathrm {log}\left (12\right )\right )+784\right ) \] Input:
int((((2+exp(x))*log(log(12))^4+(20*exp(x)+40)*log(log(12))^3+(156*exp(x)+ 312)*log(log(12))^2+(560*exp(x)+1120)*log(log(12))+784*exp(x)+1568)*log(2* exp(x)+4)+x*exp(x)*log(log(12))^4+20*x*exp(x)*log(log(12))^3+156*x*exp(x)* log(log(12))^2+560*x*exp(x)*log(log(12))+784*exp(x)*x)/(2+exp(x)),x)
Output:
log(2*e**x + 4)*x*(log(log(12))**4 + 20*log(log(12))**3 + 156*log(log(12)) **2 + 560*log(log(12)) + 784)