Integrand size = 213, antiderivative size = 31 \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=x+\left (5+e^{e^x}\right ) \left (25-\log \left (1-\frac {x}{2+e^{2+x}}+\log (3)\right )\right ) \] Output:
x+(25-ln(1-x/(exp(2+x)+2)+ln(3)))*(exp(exp(x))+5)
\[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=\int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx \] Input:
Integrate[(14 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(9 - 6*x + 4*Log[3]) + E^E^x*(2 + E^(4 + 3*x)*(25 + 25*Log[3]) + E^x*(100 - 50 *x + 100*Log[3]) + E^(2 + x)*(1 - x + E^x*(100 - 25*x + 100*Log[3]))) + E^ E^x*(E^(2 + 2*x)*(-4 + x - 4*Log[3]) + E^x*(-4 + 2*x - 4*Log[3]) + E^(4 + 3*x)*(-1 - Log[3]))*Log[(2 - x + 2*Log[3] + E^(2 + x)*(1 + Log[3]))/(2 + E ^(2 + x))])/(4 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(4 - x + 4*Log[3])),x]
Output:
Integrate[(14 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(9 - 6*x + 4*Log[3]) + E^E^x*(2 + E^(4 + 3*x)*(25 + 25*Log[3]) + E^x*(100 - 50 *x + 100*Log[3]) + E^(2 + x)*(1 - x + E^x*(100 - 25*x + 100*Log[3]))) + E^ E^x*(E^(2 + 2*x)*(-4 + x - 4*Log[3]) + E^x*(-4 + 2*x - 4*Log[3]) + E^(4 + 3*x)*(-1 - Log[3]))*Log[(2 - x + 2*Log[3] + E^(2 + x)*(1 + Log[3]))/(2 + E ^(2 + x))])/(4 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(4 - x + 4*Log[3])), x]
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.
| \(\displaystyle \int \frac {-2 x+e^{x+2} (-6 x+9+4 \log (3))+e^{e^x} \left (e^x (-50 x+100+100 \log (3))+e^{x+2} \left (-x+e^x (-25 x+100+100 \log (3))+1\right )+e^{3 x+4} (25+25 \log (3))+2\right )+e^{e^x} \left (e^{2 x+2} (x-4-4 \log (3))+e^x (2 x-4-4 \log (3))+e^{3 x+4} (-1-\log (3))\right ) \log \left (\frac {-x+e^{x+2} (1+\log (3))+2+2 \log (3)}{e^{x+2}+2}\right )+e^{2 x+4} (1+\log (3))+14+4 \log (3)}{-2 x+e^{x+2} (-x+4+4 \log (3))+e^{2 x+4} (1+\log (3))+4+4 \log (3)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 x-e^{x+2} (-6 x+9+4 \log (3))-e^{e^x} \left (e^x (-50 x+100+100 \log (3))+e^{x+2} \left (-x+e^x (-25 x+100+100 \log (3))+1\right )+e^{3 x+4} (25+25 \log (3))+2\right )-e^{e^x} \left (e^{2 x+2} (x-4-4 \log (3))+e^x (2 x-4-4 \log (3))+e^{3 x+4} (-1-\log (3))\right ) \log \left (\frac {-x+e^{x+2} (1+\log (3))+2+2 \log (3)}{e^{x+2}+2}\right )-e^{2 x+4} (1+\log (3))-14 \left (1+\frac {2 \log (3)}{7}\right )}{\left (e^{x+2}+2\right ) \left (x-e^{x+2} (1+\log (3))-2 (1+\log (3))\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {2 \left (e^{e^x}+5\right )}{e^{x+2}+2}+\frac {\left (e^{e^x}+5\right ) (x-3-\log (9))}{x-e^{x+2} (1+\log (3))-2 (1+\log (3))}-e^{x+e^x} \left (\log \left (\frac {-x+e^{x+2} (1+\log (3))+2+\log (9)}{e^{x+2}+2}\right )-25\right )+1\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle (3+\log (9)) \int \frac {e^{e^x}}{x-e^{x+2} (1+\log (3))-2 (1+\log (3))}dx+5 \int \frac {x}{x-e^{x+2} (1+\log (3))-2 (1+\log (3))}dx+\int \frac {e^{e^x} x}{x-e^{x+2} (1+\log (3))-2 (1+\log (3))}dx+5 (3+\log (9)) \int \frac {1}{-x+e^{x+2} (1+\log (3))+2 (1+\log (3))}dx+(3+\log (9)) \int \frac {e^{e^x}}{-x+e^{x+2} (1+\log (3))+2 (1+\log (3))}dx+\int \frac {e^{e^x} x}{-x+e^{x+2} (1+\log (3))+2 (1+\log (3))}dx-4 x+25 e^{e^x}+5 \log \left (e^{x+2}+2\right )-e^{e^x} \log \left (\frac {-x+e^{x+2} (1+\log (3))+2+\log (9)}{e^{x+2}+2}\right )\) |
Input:
Int[(14 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(9 - 6*x + 4*Log[3]) + E^E^x*(2 + E^(4 + 3*x)*(25 + 25*Log[3]) + E^x*(100 - 50*x + 1 00*Log[3]) + E^(2 + x)*(1 - x + E^x*(100 - 25*x + 100*Log[3]))) + E^E^x*(E ^(2 + 2*x)*(-4 + x - 4*Log[3]) + E^x*(-4 + 2*x - 4*Log[3]) + E^(4 + 3*x)*( -1 - Log[3]))*Log[(2 - x + 2*Log[3] + E^(2 + x)*(1 + Log[3]))/(2 + E^(2 + x))])/(4 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(4 - x + 4*Log[3])),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(75\) vs. \(2(28)=56\).
Time = 85.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.45
| method | result | size |
| parallelrisch | \(12-{\mathrm e}^{{\mathrm e}^{x}} \ln \left (\frac {\left (\ln \left (3\right )+1\right ) {\mathrm e}^{2+x}+2 \ln \left (3\right )+2-x}{{\mathrm e}^{2+x}+2}\right )+12 \ln \left (3\right )+x +25 \,{\mathrm e}^{{\mathrm e}^{x}}-5 \ln \left (\frac {\left (\ln \left (3\right )+1\right ) {\mathrm e}^{2+x}+2 \ln \left (3\right )+2-x}{{\mathrm e}^{2+x}+2}\right )\) | \(76\) |
| risch | \(-{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )+{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{2+x}+2\right )+x +5 \ln \left ({\mathrm e}^{x}+2 \,{\mathrm e}^{-2}\right )-5 \ln \left ({\mathrm e}^{x}+\frac {\left (2 \ln \left (3\right )-x +2\right ) {\mathrm e}^{-2}}{\ln \left (3\right )+1}\right )+\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2+x}+2}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )}{{\mathrm e}^{2+x}+2}\right )}{2}-\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2+x}+2}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )}{{\mathrm e}^{2+x}+2}\right )}^{2}}{2}-\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )}{{\mathrm e}^{2+x}+2}\right )}^{2}}{2}+\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2+x} \ln \left (3\right )+2 \ln \left (3\right )+{\mathrm e}^{2+x}-x +2\right )}{{\mathrm e}^{2+x}+2}\right )}^{3}}{2}+25 \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(313\) |
Input:
int((((-ln(3)-1)*exp(x)*exp(2+x)^2+(-4*ln(3)+x-4)*exp(x)*exp(2+x)+(-4*ln(3 )+2*x-4)*exp(x))*exp(exp(x))*ln(((ln(3)+1)*exp(2+x)+2*ln(3)+2-x)/(exp(2+x) +2))+((25*ln(3)+25)*exp(x)*exp(2+x)^2+((100*ln(3)-25*x+100)*exp(x)-x+1)*ex p(2+x)+(100*ln(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(ln(3)+1)*exp(2+x)^2+(4* ln(3)-6*x+9)*exp(2+x)+4*ln(3)-2*x+14)/((ln(3)+1)*exp(2+x)^2+(4*ln(3)-x+4)* exp(2+x)+4*ln(3)+4-2*x),x,method=_RETURNVERBOSE)
Output:
12-exp(exp(x))*ln(((ln(3)+1)*exp(2+x)+2*ln(3)+2-x)/(exp(2+x)+2))+12*ln(3)+ x+25*exp(exp(x))-5*ln(((ln(3)+1)*exp(2+x)+2*ln(3)+2-x)/(exp(2+x)+2))
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=-{\left (e^{\left (e^{x}\right )} + 5\right )} \log \left (\frac {{\left (\log \left (3\right ) + 1\right )} e^{\left (x + 2\right )} - x + 2 \, \log \left (3\right ) + 2}{e^{\left (x + 2\right )} + 2}\right ) + x + 25 \, e^{\left (e^{x}\right )} \] Input:
integrate((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+ (-4*log(3)+2*x-4)*exp(x))*exp(exp(x))*log(((log(3)+1)*exp(2+x)+2*log(3)+2- x)/(exp(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100)* exp(x)-x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1 )*exp(2+x)^2+(4*log(3)-6*x+9)*exp(2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+ x)^2+(4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x, algorithm="fricas")
Output:
-(e^(e^x) + 5)*log(((log(3) + 1)*e^(x + 2) - x + 2*log(3) + 2)/(e^(x + 2) + 2)) + x + 25*e^(e^x)
Timed out. \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=\text {Timed out} \] Input:
integrate((((-ln(3)-1)*exp(x)*exp(2+x)**2+(-4*ln(3)+x-4)*exp(x)*exp(2+x)+( -4*ln(3)+2*x-4)*exp(x))*exp(exp(x))*ln(((ln(3)+1)*exp(2+x)+2*ln(3)+2-x)/(e xp(2+x)+2))+((25*ln(3)+25)*exp(x)*exp(2+x)**2+((100*ln(3)-25*x+100)*exp(x) -x+1)*exp(2+x)+(100*ln(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(ln(3)+1)*exp(2+ x)**2+(4*ln(3)-6*x+9)*exp(2+x)+4*ln(3)-2*x+14)/((ln(3)+1)*exp(2+x)**2+(4*l n(3)-x+4)*exp(2+x)+4*ln(3)+4-2*x),x)
Output:
Timed out
Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=-{\left (e^{\left (e^{x}\right )} + 5\right )} \log \left ({\left (e^{2} \log \left (3\right ) + e^{2}\right )} e^{x} - x + 2 \, \log \left (3\right ) + 2\right ) + {\left (e^{\left (e^{x}\right )} + 5\right )} \log \left (e^{\left (x + 2\right )} + 2\right ) + x + 25 \, e^{\left (e^{x}\right )} \] Input:
integrate((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+ (-4*log(3)+2*x-4)*exp(x))*exp(exp(x))*log(((log(3)+1)*exp(2+x)+2*log(3)+2- x)/(exp(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100)* exp(x)-x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1 )*exp(2+x)^2+(4*log(3)-6*x+9)*exp(2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+ x)^2+(4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x, algorithm="maxima")
Output:
-(e^(e^x) + 5)*log((e^2*log(3) + e^2)*e^x - x + 2*log(3) + 2) + (e^(e^x) + 5)*log(e^(x + 2) + 2) + x + 25*e^(e^x)
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (27) = 54\).
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.13 \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx={\left (x e^{x} - e^{\left (x + e^{x}\right )} \log \left (e^{\left (x + 2\right )} \log \left (3\right ) - x + e^{\left (x + 2\right )} + 2 \, \log \left (3\right ) + 2\right ) - 5 \, e^{x} \log \left (-e^{\left (x + 2\right )} \log \left (3\right ) + x - e^{\left (x + 2\right )} - 2 \, \log \left (3\right ) - 2\right ) + e^{\left (x + e^{x}\right )} \log \left (e^{\left (x + 2\right )} + 2\right ) + 5 \, e^{x} \log \left (-e^{\left (x + 2\right )} - 2\right ) + 25 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \] Input:
integrate((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+ (-4*log(3)+2*x-4)*exp(x))*exp(exp(x))*log(((log(3)+1)*exp(2+x)+2*log(3)+2- x)/(exp(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100)* exp(x)-x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1 )*exp(2+x)^2+(4*log(3)-6*x+9)*exp(2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+ x)^2+(4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x, algorithm="giac")
Output:
(x*e^x - e^(x + e^x)*log(e^(x + 2)*log(3) - x + e^(x + 2) + 2*log(3) + 2) - 5*e^x*log(-e^(x + 2)*log(3) + x - e^(x + 2) - 2*log(3) - 2) + e^(x + e^x )*log(e^(x + 2) + 2) + 5*e^x*log(-e^(x + 2) - 2) + 25*e^(x + e^x))*e^(-x)
Timed out. \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=\int \frac {4\,\ln \left (3\right )-2\,x+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{x+2}\,\left ({\mathrm {e}}^x\,\left (100\,\ln \left (3\right )-25\,x+100\right )-x+1\right )+{\mathrm {e}}^{3\,x+4}\,\left (25\,\ln \left (3\right )+25\right )+{\mathrm {e}}^x\,\left (100\,\ln \left (3\right )-50\,x+100\right )+2\right )+{\mathrm {e}}^{x+2}\,\left (4\,\ln \left (3\right )-6\,x+9\right )+{\mathrm {e}}^{2\,x+4}\,\left (\ln \left (3\right )+1\right )-\ln \left (\frac {2\,\ln \left (3\right )-x+{\mathrm {e}}^{x+2}\,\left (\ln \left (3\right )+1\right )+2}{{\mathrm {e}}^{x+2}+2}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (4\,\ln \left (3\right )-2\,x+4\right )+{\mathrm {e}}^{3\,x+4}\,\left (\ln \left (3\right )+1\right )+{\mathrm {e}}^{2\,x+2}\,\left (4\,\ln \left (3\right )-x+4\right )\right )+14}{4\,\ln \left (3\right )-2\,x+{\mathrm {e}}^{x+2}\,\left (4\,\ln \left (3\right )-x+4\right )+{\mathrm {e}}^{2\,x+4}\,\left (\ln \left (3\right )+1\right )+4} \,d x \] Input:
int((4*log(3) - 2*x + exp(exp(x))*(exp(x + 2)*(exp(x)*(100*log(3) - 25*x + 100) - x + 1) + exp(x)*(100*log(3) - 50*x + 100) + exp(2*x + 4)*exp(x)*(2 5*log(3) + 25) + 2) + exp(x + 2)*(4*log(3) - 6*x + 9) + exp(2*x + 4)*(log( 3) + 1) - log((2*log(3) - x + exp(x + 2)*(log(3) + 1) + 2)/(exp(x + 2) + 2 ))*exp(exp(x))*(exp(x)*(4*log(3) - 2*x + 4) + exp(x + 2)*exp(x)*(4*log(3) - x + 4) + exp(2*x + 4)*exp(x)*(log(3) + 1)) + 14)/(4*log(3) - 2*x + exp(x + 2)*(4*log(3) - x + 4) + exp(2*x + 4)*(log(3) + 1) + 4),x)
Output:
int((4*log(3) - 2*x + exp(exp(x))*(exp(x + 2)*(exp(x)*(100*log(3) - 25*x + 100) - x + 1) + exp(3*x + 4)*(25*log(3) + 25) + exp(x)*(100*log(3) - 50*x + 100) + 2) + exp(x + 2)*(4*log(3) - 6*x + 9) + exp(2*x + 4)*(log(3) + 1) - log((2*log(3) - x + exp(x + 2)*(log(3) + 1) + 2)/(exp(x + 2) + 2))*exp( exp(x))*(exp(x)*(4*log(3) - 2*x + 4) + exp(3*x + 4)*(log(3) + 1) + exp(2*x + 2)*(4*log(3) - x + 4)) + 14)/(4*log(3) - 2*x + exp(x + 2)*(4*log(3) - x + 4) + exp(2*x + 4)*(log(3) + 1) + 4), x)
Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.03 \[ \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx=-e^{e^{x}} \mathrm {log}\left (\frac {e^{x} \mathrm {log}\left (3\right ) e^{2}+e^{x} e^{2}+2 \,\mathrm {log}\left (3\right )-x +2}{e^{x} e^{2}+2}\right )+25 e^{e^{x}}+5 \,\mathrm {log}\left (e^{x} e^{2}+2\right )-5 \,\mathrm {log}\left (e^{x} \mathrm {log}\left (3\right ) e^{2}+e^{x} e^{2}+2 \,\mathrm {log}\left (3\right )-x +2\right )+x \] Input:
int((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+(-4*lo g(3)+2*x-4)*exp(x))*exp(exp(x))*log(((log(3)+1)*exp(2+x)+2*log(3)+2-x)/(ex p(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100)*exp(x) -x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1)*exp( 2+x)^2+(4*log(3)-6*x+9)*exp(2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+x)^2+( 4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x)
Output:
- e**(e**x)*log((e**x*log(3)*e**2 + e**x*e**2 + 2*log(3) - x + 2)/(e**x*e **2 + 2)) + 25*e**(e**x) + 5*log(e**x*e**2 + 2) - 5*log(e**x*log(3)*e**2 + e**x*e**2 + 2*log(3) - x + 2) + x