Integrand size = 156, antiderivative size = 24 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=e^{17+e^x+\frac {16}{-4+\frac {e^x}{3}+\log (2)}}+x \] Output:
x+exp(17+16/(exp(-ln(3)+x)+ln(2)-4)+exp(x))
Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(24)=48\).
Time = 1.58 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.08 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x+\frac {e^{17+e^x+\frac {48}{-12+e^x+\log (8)}} \left (96-24 e^x+e^{2 x}+9 \log ^2(2)-24 \log (8)+2 e^x \log (8)\right )}{\left (-12+e^x+\log (8)\right )^2 \left (1-\frac {48}{\left (-12+e^x+\log (8)\right )^2}\right )} \] Input:
Integrate[(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3 + E^((-52 + (17*E^x)/3 + 17*Log[2] + E^x*(-4 + E^x/3 + Log[2]))/(-4 + E^x/ 3 + Log[2]))*((-16*E^x)/3 + E^x*(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E ^x*(-8 + 2*Log[2]))/3)))/(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3),x]
Output:
x + (E^(17 + E^x + 48/(-12 + E^x + Log[8]))*(96 - 24*E^x + E^(2*x) + 9*Log [2]^2 - 24*Log[8] + 2*E^x*Log[8]))/((-12 + E^x + Log[8])^2*(1 - 48/(-12 + E^x + Log[8])^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.
| \(\displaystyle \int \frac {\left (e^x \left (\frac {e^{2 x}}{9}+\frac {1}{3} e^x (2 \log (2)-8)+16+\log ^2(2)-8 \log (2)\right )-\frac {16 e^x}{3}\right ) \exp \left (\frac {\frac {17 e^x}{3}+e^x \left (\frac {e^x}{3}-4+\log (2)\right )-52+17 \log (2)}{\frac {e^x}{3}-4+\log (2)}\right )+\frac {e^{2 x}}{9}+\frac {1}{3} e^x (2 \log (2)-8)+16+\log ^2(2)-8 \log (2)}{\frac {e^{2 x}}{9}+\frac {1}{3} e^x (2 \log (2)-8)+16+\log ^2(2)-8 \log (2)} \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \int \frac {e^{-x} \left (2^{-\frac {51}{-e^x+12-\log (8)}} \left (e^{2 x}-6 e^x (4-\log (2))+3 \left (32+3 \log ^2(2)-24 \log (2)\right )\right ) \exp \left (x+\frac {-e^{2 x}-e^x (5+\log (8))+156}{-e^x+12-\log (8)}\right )+e^{2 x}-6 e^x (4-\log (2))+9 (4-\log (2))^2\right )}{\left (-e^x+12-\log (8)\right )^2}de^x\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2^{\frac {51}{e^x-12+\log (8)}} \left (e^{2 x}-6 e^x (4-\log (2))+3 \left (32+3 \log ^2(2)-24 \log (2)\right )\right ) \exp \left (\frac {e^{2 x}+e^x (5+\log (8))-156}{e^x-12+\log (8)}\right )}{\left (-e^x+12-\log (8)\right )^2}+e^{-x}\right )de^x\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\left (48-9 \log ^2(2)-\log ^2(8)+6 \log (2) \log (8)\right ) \int \frac {2^{\frac {51}{-12+e^x+\log (8)}} \exp \left (\frac {-156+e^{2 x}+e^x (5+\log (8))}{-12+e^x+\log (8)}\right )}{\left (-12+e^x+\log (8)\right )^2}de^x+\int 2^{\frac {51}{-12+e^x+\log (8)}} \exp \left (\frac {-156+e^{2 x}+e^x (5+\log (8))}{-12+e^x+\log (8)}\right )de^x+\log \left (e^x\right )\) |
Input:
Int[(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3 + E^(( -52 + (17*E^x)/3 + 17*Log[2] + E^x*(-4 + E^x/3 + Log[2]))/(-4 + E^x/3 + Lo g[2]))*((-16*E^x)/3 + E^x*(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3)))/(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Lo g[2]))/3),x]
Output:
$Aborted
Time = 3.86 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46
| method | result | size |
| risch | \(x +{\mathrm e}^{\frac {3 \,{\mathrm e}^{x} \ln \left (2\right )+5 \,{\mathrm e}^{x}+51 \ln \left (2\right )+{\mathrm e}^{2 x}-156}{{\mathrm e}^{x}+3 \ln \left (2\right )-12}}\) | \(35\) |
| parallelrisch | \(x +{\mathrm e}^{\frac {\left ({\mathrm e}^{-\ln \left (3\right )+x}+\ln \left (2\right )-4\right ) {\mathrm e}^{x}+17 \,{\mathrm e}^{-\ln \left (3\right )+x}+17 \ln \left (2\right )-52}{{\mathrm e}^{-\ln \left (3\right )+x}+\ln \left (2\right )-4}}\) | \(47\) |
| norman | \(\frac {\left (3 \ln \left (2\right )-12\right ) x +\left (3 \ln \left (2\right )-12\right ) {\mathrm e}^{\frac {\left (\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4\right ) {\mathrm e}^{x}+\frac {17 \,{\mathrm e}^{x}}{3}+17 \ln \left (2\right )-52}{\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4}}+{\mathrm e}^{x} x +{\mathrm e}^{x} {\mathrm e}^{\frac {\left (\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4\right ) {\mathrm e}^{x}+\frac {17 \,{\mathrm e}^{x}}{3}+17 \ln \left (2\right )-52}{\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4}}}{{\mathrm e}^{x}+3 \ln \left (2\right )-12}\) | \(101\) |
Input:
int((((exp(-ln(3)+x)^2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)^2-8*ln(2)+16)*exp(x )-16*exp(-ln(3)+x))*exp(((exp(-ln(3)+x)+ln(2)-4)*exp(x)+17*exp(-ln(3)+x)+1 7*ln(2)-52)/(exp(-ln(3)+x)+ln(2)-4))+exp(-ln(3)+x)^2+(2*ln(2)-8)*exp(-ln(3 )+x)+ln(2)^2-8*ln(2)+16)/(exp(-ln(3)+x)^2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)^ 2-8*ln(2)+16),x,method=_RETURNVERBOSE)
Output:
x+exp((3*exp(x)*ln(2)+5*exp(x)+51*ln(2)+exp(2*x)-156)/(exp(x)+3*ln(2)-12))
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x + e^{\left (\frac {{\left (3 \, \log \left (2\right ) + 5\right )} e^{x} + e^{\left (2 \, x\right )} + 51 \, \log \left (2\right ) - 156}{e^{x} + 3 \, \log \left (2\right ) - 12}\right )} \] Input:
integrate((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2 )+16)*exp(x)-16*exp(-log(3)+x))*exp(((exp(-log(3)+x)+log(2)-4)*exp(x)+17*e xp(-log(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+(2 *log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2 )-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16),x, algorithm="fricas")
Output:
x + e^(((3*log(2) + 5)*e^x + e^(2*x) + 51*log(2) - 156)/(e^x + 3*log(2) - 12))
Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x + e^{\frac {\left (\frac {e^{x}}{3} - 4 + \log {\left (2 \right )}\right ) e^{x} + \frac {17 e^{x}}{3} - 52 + 17 \log {\left (2 \right )}}{\frac {e^{x}}{3} - 4 + \log {\left (2 \right )}}} \] Input:
integrate((((exp(-ln(3)+x)**2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)**2-8*ln(2)+1 6)*exp(x)-16*exp(-ln(3)+x))*exp(((exp(-ln(3)+x)+ln(2)-4)*exp(x)+17*exp(-ln (3)+x)+17*ln(2)-52)/(exp(-ln(3)+x)+ln(2)-4))+exp(-ln(3)+x)**2+(2*ln(2)-8)* exp(-ln(3)+x)+ln(2)**2-8*ln(2)+16)/(exp(-ln(3)+x)**2+(2*ln(2)-8)*exp(-ln(3 )+x)+ln(2)**2-8*ln(2)+16),x)
Output:
x + exp(((exp(x)/3 - 4 + log(2))*exp(x) + 17*exp(x)/3 - 52 + 17*log(2))/(e xp(x)/3 - 4 + log(2)))
Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (22) = 44\).
Time = 0.25 (sec) , antiderivative size = 261, normalized size of antiderivative = 10.88 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx={\left (\frac {x}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} - \frac {\log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right )}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {3}{{\left (\log \left (2\right ) - 4\right )} e^{x} + 3 \, \log \left (2\right )^{2} - 24 \, \log \left (2\right ) + 48}\right )} \log \left (2\right )^{2} - 8 \, {\left (\frac {x}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} - \frac {\log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right )}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {3}{{\left (\log \left (2\right ) - 4\right )} e^{x} + 3 \, \log \left (2\right )^{2} - 24 \, \log \left (2\right ) + 48}\right )} \log \left (2\right ) + \frac {16 \, x}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {3 \, {\left (\log \left (2\right ) - 4\right )}}{e^{x} + 3 \, \log \left (2\right ) - 12} - \frac {6 \, \log \left (2\right )}{e^{x} + 3 \, \log \left (2\right ) - 12} - \frac {16 \, \log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right )}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {48}{{\left (\log \left (2\right ) - 4\right )} e^{x} + 3 \, \log \left (2\right )^{2} - 24 \, \log \left (2\right ) + 48} + \frac {24}{e^{x} + 3 \, \log \left (2\right ) - 12} + e^{\left (\frac {48}{e^{x} + 3 \, \log \left (2\right ) - 12} + e^{x} + 17\right )} + \log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right ) \] Input:
integrate((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2 )+16)*exp(x)-16*exp(-log(3)+x))*exp(((exp(-log(3)+x)+log(2)-4)*exp(x)+17*e xp(-log(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+(2 *log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2 )-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16),x, algorithm="maxima")
Output:
(x/(log(2)^2 - 8*log(2) + 16) - log(e^x + 3*log(2) - 12)/(log(2)^2 - 8*log (2) + 16) + 3/((log(2) - 4)*e^x + 3*log(2)^2 - 24*log(2) + 48))*log(2)^2 - 8*(x/(log(2)^2 - 8*log(2) + 16) - log(e^x + 3*log(2) - 12)/(log(2)^2 - 8* log(2) + 16) + 3/((log(2) - 4)*e^x + 3*log(2)^2 - 24*log(2) + 48))*log(2) + 16*x/(log(2)^2 - 8*log(2) + 16) + 3*(log(2) - 4)/(e^x + 3*log(2) - 12) - 6*log(2)/(e^x + 3*log(2) - 12) - 16*log(e^x + 3*log(2) - 12)/(log(2)^2 - 8*log(2) + 16) + 48/((log(2) - 4)*e^x + 3*log(2)^2 - 24*log(2) + 48) + 24/ (e^x + 3*log(2) - 12) + e^(48/(e^x + 3*log(2) - 12) + e^x + 17) + log(e^x + 3*log(2) - 12)
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (22) = 44\).
Time = 0.52 (sec) , antiderivative size = 239, normalized size of antiderivative = 9.96 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=\frac {1}{2} \, {\left (2 \, {\left (x - \log \left (3\right )\right )} e^{\left (x - \log \left (3\right )\right )} + 2 \, e^{\left (x - \log \left (3\right )\right )} \log \left (e^{\left (x - \log \left (3\right )\right )} + \log \left (2\right ) - 4\right ) - 2 \, e^{\left (x - \log \left (3\right )\right )} \log \left (-e^{\left (x - \log \left (3\right )\right )} - \log \left (2\right ) + 4\right ) + e^{\left (\frac {{\left (x - \log \left (3\right )\right )} e^{\left (x - \log \left (3\right )\right )} \log \left (2\right ) + {\left (x - \log \left (3\right )\right )} \log \left (2\right )^{2} + 3 \, e^{\left (x - \log \left (3\right )\right )} \log \left (2\right )^{2} - 4 \, {\left (x - \log \left (3\right )\right )} e^{\left (x - \log \left (3\right )\right )} - 8 \, {\left (x - \log \left (3\right )\right )} \log \left (2\right ) + 3 \, e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} \log \left (2\right ) - 24 \, e^{\left (x - \log \left (3\right )\right )} \log \left (2\right ) + 16 \, x - 12 \, e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + 32 \, e^{\left (x - \log \left (3\right )\right )} - 16 \, \log \left (3\right )}{e^{\left (x - \log \left (3\right )\right )} \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, e^{\left (x - \log \left (3\right )\right )} - 8 \, \log \left (2\right ) + 16} + \frac {\log \left (2\right )^{2} + 13 \, \log \left (2\right ) - 52}{\log \left (2\right ) - 4}\right )}\right )} e^{\left (-x + \log \left (3\right )\right )} \] Input:
integrate((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2 )+16)*exp(x)-16*exp(-log(3)+x))*exp(((exp(-log(3)+x)+log(2)-4)*exp(x)+17*e xp(-log(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+(2 *log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2 )-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16),x, algorithm="giac")
Output:
1/2*(2*(x - log(3))*e^(x - log(3)) + 2*e^(x - log(3))*log(e^(x - log(3)) + log(2) - 4) - 2*e^(x - log(3))*log(-e^(x - log(3)) - log(2) + 4) + e^(((x - log(3))*e^(x - log(3))*log(2) + (x - log(3))*log(2)^2 + 3*e^(x - log(3) )*log(2)^2 - 4*(x - log(3))*e^(x - log(3)) - 8*(x - log(3))*log(2) + 3*e^( 2*x - 2*log(3))*log(2) - 24*e^(x - log(3))*log(2) + 16*x - 12*e^(2*x - 2*l og(3)) + 32*e^(x - log(3)) - 16*log(3))/(e^(x - log(3))*log(2) + log(2)^2 - 4*e^(x - log(3)) - 8*log(2) + 16) + (log(2)^2 + 13*log(2) - 52)/(log(2) - 4)))*e^(-x + log(3))
Time = 1.74 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.12 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x+2^{\frac {{\mathrm {e}}^x}{\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4}}\,2^{\frac {17}{\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{3\,\left (\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4\right )}-\frac {52}{\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4}+\frac {5\,{\mathrm {e}}^x}{3\,\left (\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4\right )}} \] Input:
int((exp(2*x - 2*log(3)) - 8*log(2) + exp(x - log(3))*(2*log(2) - 8) + log (2)^2 - exp((17*exp(x - log(3)) + 17*log(2) + exp(x)*(exp(x - log(3)) + lo g(2) - 4) - 52)/(exp(x - log(3)) + log(2) - 4))*(16*exp(x - log(3)) - exp( x)*(exp(2*x - 2*log(3)) - 8*log(2) + exp(x - log(3))*(2*log(2) - 8) + log( 2)^2 + 16)) + 16)/(exp(2*x - 2*log(3)) - 8*log(2) + exp(x - log(3))*(2*log (2) - 8) + log(2)^2 + 16),x)
Output:
x + 2^(exp(x)/(log(2) + exp(x)/3 - 4))*2^(17/(log(2) + exp(x)/3 - 4))*exp( exp(2*x)/(3*(log(2) + exp(x)/3 - 4)) - 52/(log(2) + exp(x)/3 - 4) + (5*exp (x))/(3*(log(2) + exp(x)/3 - 4)))
Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=e^{\frac {e^{2 x}+3 e^{x} \mathrm {log}\left (2\right )-12 e^{x}+48}{e^{x}+3 \,\mathrm {log}\left (2\right )-12}} e^{17}+x \] Input:
int((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)* exp(x)-16*exp(-log(3)+x))*exp(((exp(-log(3)+x)+log(2)-4)*exp(x)+17*exp(-lo g(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+(2*log(2 )-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2)-8)*e xp(-log(3)+x)+log(2)^2-8*log(2)+16),x)
Output:
e**((e**(2*x) + 3*e**x*log(2) - 12*e**x + 48)/(e**x + 3*log(2) - 12))*e**1 7 + x