Integrand size = 144, antiderivative size = 24 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=20^{\frac {1}{\log \left (\left (2+\frac {x}{4}-\log \left (3+e^{e^x}\right )\right )^2\right )}} \] Output:
exp(ln(20)/ln((1/4*x-ln(exp(exp(x))+3)+2)^2))
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=20^{\frac {1}{\log \left (\frac {1}{16} \left (8+x-4 \log \left (3+e^{e^x}\right )\right )^2\right )}} \] Input:
Integrate[(20^Log[(64 + 16*x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^(-1)*(6*Log[20] + E^E^x*(2*Log[20] - 8*E^x*Log[20])))/((- 24 + E^E^x*(-8 - x) - 3*x + (12 + 4*E^E^x)*Log[3 + E^E^x])*Log[(64 + 16*x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^2),x]
Output:
20^Log[(8 + x - 4*Log[3 + E^E^x])^2/16]^(-1)
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.
| \(\displaystyle \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (x^2+16 x+16 \log ^2\left (e^{e^x}+3\right )+(-8 x-64) \log \left (e^{e^x}+3\right )+64\right )\right )}} \left (e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )+6 \log (20)\right )}{\left (e^{e^x} (-x-8)-3 x+\left (4 e^{e^x}+12\right ) \log \left (e^{e^x}+3\right )-24\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 x+16 \log ^2\left (e^{e^x}+3\right )+(-8 x-64) \log \left (e^{e^x}+3\right )+64\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (-e^{e^x}+4 e^{x+e^x}-3\right ) \log (20) 2^{\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (e^{e^x}+3\right )+8\right )^2\right )}+1} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (e^{e^x}+3\right )+8\right )^2\right )}}}{\left (e^{e^x}+3\right ) \left (x-4 \log \left (e^{e^x}+3\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (e^{e^x}+3\right )+8\right )^2\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \log (20) \int -\frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} \left (3+e^{e^x}-4 e^{x+e^x}\right )}{\left (3+e^{e^x}\right ) \left (x-4 \log \left (3+e^{e^x}\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\log (20) \int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} \left (3+e^{e^x}-4 e^{x+e^x}\right )}{\left (3+e^{e^x}\right ) \left (x-4 \log \left (3+e^{e^x}\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\log (20) \int \left (\frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}}}{\left (x-4 \log \left (3+e^{e^x}\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}-\frac {2^{3+\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} e^{x+e^x}}{\left (3+e^{e^x}\right ) \left (x-4 \log \left (3+e^{e^x}\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\log (20) \left (\int \frac {2^{1+\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}}}{\left (x-4 \log \left (3+e^{e^x}\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}dx-\int \frac {2^{3+\frac {2}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} 5^{\frac {1}{\log \left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}} e^{x+e^x}}{\left (3+e^{e^x}\right ) \left (x-4 \log \left (3+e^{e^x}\right )+8\right ) \log ^2\left (\frac {1}{16} \left (x-4 \log \left (3+e^{e^x}\right )+8\right )^2\right )}dx\right )\) |
Input:
Int[(20^Log[(64 + 16*x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E ^x]^2)/16]^(-1)*(6*Log[20] + E^E^x*(2*Log[20] - 8*E^x*Log[20])))/((-24 + E ^E^x*(-8 - x) - 3*x + (12 + 4*E^E^x)*Log[3 + E^E^x])*Log[(64 + 16*x + x^2 + (-64 - 8*x)*Log[3 + E^E^x] + 16*Log[3 + E^E^x]^2)/16]^2),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.43 (sec) , antiderivative size = 131, normalized size of antiderivative = 5.46
\[{\mathrm e}^{-\frac {2 \left (2 \ln \left (2\right )+\ln \left (5\right )\right )}{i \pi {\operatorname {csgn}\left (i {\left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}^{2}\right )}^{3}-2 i \pi {\operatorname {csgn}\left (i {\left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}^{2}\right )}^{2} \operatorname {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )\right )+i \pi \,\operatorname {csgn}\left (i {\left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}^{2}\right ) {\operatorname {csgn}\left (i \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )\right )}^{2}+8 \ln \left (2\right )-4 \ln \left (-4 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+3\right )+x +8\right )}}\]
Input:
int(((-8*ln(20)*exp(x)+2*ln(20))*exp(exp(x))+6*ln(20))*exp(ln(20)/ln(ln(ex p(exp(x))+3)^2+1/16*(-8*x-64)*ln(exp(exp(x))+3)+1/16*x^2+x+4))/((4*exp(exp (x))+12)*ln(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/ln(ln(exp(exp(x))+3) ^2+1/16*(-8*x-64)*ln(exp(exp(x))+3)+1/16*x^2+x+4)^2,x)
Output:
exp(-2*(2*ln(2)+ln(5))/(I*Pi*csgn(I*(-4*ln(exp(exp(x))+3)+x+8)^2)^3-2*I*Pi *csgn(I*(-4*ln(exp(exp(x))+3)+x+8)^2)^2*csgn(I*(-4*ln(exp(exp(x))+3)+x+8)) +I*Pi*csgn(I*(-4*ln(exp(exp(x))+3)+x+8)^2)*csgn(I*(-4*ln(exp(exp(x))+3)+x+ 8))^2+8*ln(2)-4*ln(-4*ln(exp(exp(x))+3)+x+8)))
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )} \] Input:
integrate(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20 )/log(log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4) )/((4*exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(lo g(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x, al gorithm="fricas")
Output:
20^(1/log(1/16*x^2 - 1/2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4))
Timed out. \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\text {Timed out} \] Input:
integrate(((-8*ln(20)*exp(x)+2*ln(20))*exp(exp(x))+6*ln(20))*exp(ln(20)/ln (ln(exp(exp(x))+3)**2+1/16*(-8*x-64)*ln(exp(exp(x))+3)+1/16*x**2+x+4))/((4 *exp(exp(x))+12)*ln(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/ln(ln(exp(ex p(x))+3)**2+1/16*(-8*x-64)*ln(exp(exp(x))+3)+1/16*x**2+x+4)**2,x)
Output:
Timed out
\[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\int { \frac {2 \, {\left ({\left (4 \, e^{x} \log \left (20\right ) - \log \left (20\right )\right )} e^{\left (e^{x}\right )} - 3 \, \log \left (20\right )\right )} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )}}{{\left ({\left (x + 8\right )} e^{\left (e^{x}\right )} - 4 \, {\left (e^{\left (e^{x}\right )} + 3\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + 3 \, x + 24\right )} \log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )^{2}} \,d x } \] Input:
integrate(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20 )/log(log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4) )/((4*exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(lo g(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x, al gorithm="maxima")
Output:
2*integrate(((4*e^x*log(20) - log(20))*e^(e^x) - 3*log(20))*20^(1/log(1/16 *x^2 - 1/2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4))/(((x + 8)*e^(e^x) - 4*(e^(e^x) + 3)*log(e^(e^x) + 3) + 3*x + 24)*log(1/16*x^2 - 1 /2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4)^2), x)
\[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=\int { \frac {2 \, {\left ({\left (4 \, e^{x} \log \left (20\right ) - \log \left (20\right )\right )} e^{\left (e^{x}\right )} - 3 \, \log \left (20\right )\right )} 20^{\left (\frac {1}{\log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )}\right )}}{{\left ({\left (x + 8\right )} e^{\left (e^{x}\right )} - 4 \, {\left (e^{\left (e^{x}\right )} + 3\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + 3 \, x + 24\right )} \log \left (\frac {1}{16} \, x^{2} - \frac {1}{2} \, {\left (x + 8\right )} \log \left (e^{\left (e^{x}\right )} + 3\right ) + \log \left (e^{\left (e^{x}\right )} + 3\right )^{2} + x + 4\right )^{2}} \,d x } \] Input:
integrate(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20 )/log(log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4) )/((4*exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(lo g(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x, al gorithm="giac")
Output:
integrate(2*((4*e^x*log(20) - log(20))*e^(e^x) - 3*log(20))*20^(1/log(1/16 *x^2 - 1/2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4))/(((x + 8)*e^(e^x) - 4*(e^(e^x) + 3)*log(e^(e^x) + 3) + 3*x + 24)*log(1/16*x^2 - 1 /2*(x + 8)*log(e^(e^x) + 3) + log(e^(e^x) + 3)^2 + x + 4)^2), x)
Time = 4.97 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx={20}^{\frac {1}{\ln \left (\frac {x^2}{16}-\frac {x\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )}{2}+x+{\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )}^2-4\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}+3\right )+4\right )}} \] Input:
int(-(exp(log(20)/log(x + log(exp(exp(x)) + 3)^2 - (log(exp(exp(x)) + 3)*( 8*x + 64))/16 + x^2/16 + 4))*(6*log(20) + exp(exp(x))*(2*log(20) - 8*exp(x )*log(20))))/(log(x + log(exp(exp(x)) + 3)^2 - (log(exp(exp(x)) + 3)*(8*x + 64))/16 + x^2/16 + 4)^2*(3*x - log(exp(exp(x)) + 3)*(4*exp(exp(x)) + 12) + exp(exp(x))*(x + 8) + 24)),x)
Output:
20^(1/log(x - 4*log(exp(exp(x)) + 3) + log(exp(exp(x)) + 3)^2 + x^2/16 - ( x*log(exp(exp(x)) + 3))/2 + 4))
Time = 0.37 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {20^{\frac {1}{\log \left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )}} \left (6 \log (20)+e^{e^x} \left (2 \log (20)-8 e^x \log (20)\right )\right )}{\left (-24+e^{e^x} (-8-x)-3 x+\left (12+4 e^{e^x}\right ) \log \left (3+e^{e^x}\right )\right ) \log ^2\left (\frac {1}{16} \left (64+16 x+x^2+(-64-8 x) \log \left (3+e^{e^x}\right )+16 \log ^2\left (3+e^{e^x}\right )\right )\right )} \, dx=e^{\frac {\mathrm {log}\left (20\right )}{\mathrm {log}\left (\mathrm {log}\left (e^{e^{x}}+3\right )^{2}-\frac {\mathrm {log}\left (e^{e^{x}}+3\right ) x}{2}-4 \,\mathrm {log}\left (e^{e^{x}}+3\right )+\frac {x^{2}}{16}+x +4\right )}} \] Input:
int(((-8*log(20)*exp(x)+2*log(20))*exp(exp(x))+6*log(20))*exp(log(20)/log( log(exp(exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4))/((4* exp(exp(x))+12)*log(exp(exp(x))+3)+(-x-8)*exp(exp(x))-3*x-24)/log(log(exp( exp(x))+3)^2+1/16*(-8*x-64)*log(exp(exp(x))+3)+1/16*x^2+x+4)^2,x)
Output:
e**(log(20)/log((16*log(e**(e**x) + 3)**2 - 8*log(e**(e**x) + 3)*x - 64*lo g(e**(e**x) + 3) + x**2 + 16*x + 64)/16))