Integrand size = 68, antiderivative size = 19 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (5 \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )\right ) \] Output:
ln(5*ln(exp(ln(5+x)+x)*ln(3*exp(x)+3)))
Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (\log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )\right ) \] Input:
Integrate[(E^x*(5 + x) + (6 + x + E^x*(6 + x))*Log[3 + 3*E^x])/((5 + x + E ^x*(5 + x))*Log[3 + 3*E^x]*Log[E^x*(5 + x)*Log[3 + 3*E^x]]),x]
Output:
Log[Log[E^x*(5 + x)*Log[3*(1 + E^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.
\(\displaystyle \int \frac {e^x (x+5)+\left (x+e^x (x+6)+6\right ) \log \left (3 e^x+3\right )}{\left (x+e^x (x+5)+5\right ) \log \left (3 e^x+3\right ) \log \left (e^x (x+5) \log \left (3 e^x+3\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x (x+5)+\left (x+e^x (x+6)+6\right ) \log \left (3 e^x+3\right )}{\left (e^x+1\right ) (x+5) \log \left (3 \left (e^x+1\right )\right ) \log \left (e^x (x+5) \log \left (3 \left (e^x+1\right )\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x+x \log \left (3 \left (e^x+1\right )\right )+6 \log \left (3 \left (e^x+1\right )\right )+5}{(x+5) \log \left (3 \left (e^x+1\right )\right ) \log \left (e^x (x+5) \log \left (3 \left (e^x+1\right )\right )\right )}-\frac {1}{\left (e^x+1\right ) \log \left (3 \left (e^x+1\right )\right ) \log \left (e^x (x+5) \log \left (3 \left (e^x+1\right )\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\log \left (e^x (x+5) \log \left (3 \left (1+e^x\right )\right )\right )}dx+\int \frac {1}{(x+5) \log \left (e^x (x+5) \log \left (3 \left (1+e^x\right )\right )\right )}dx+\int \frac {1}{\log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (x+5) \log \left (3 \left (1+e^x\right )\right )\right )}dx-\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (x+5) \log \left (3 \left (1+e^x\right )\right )\right )}dx\) |
Input:
Int[(E^x*(5 + x) + (6 + x + E^x*(6 + x))*Log[3 + 3*E^x])/((5 + x + E^x*(5 + x))*Log[3 + 3*E^x]*Log[E^x*(5 + x)*Log[3 + 3*E^x]]),x]
Output:
$Aborted
Time = 49.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(\ln \left (\ln \left ({\mathrm e}^{\ln \left (5+x \right )+x} \ln \left (3 \,{\mathrm e}^{x}+3\right )\right )\right )\) | \(18\) |
risch | \(\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (5+x \right )\right ) \operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right )\right ) \operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )-\pi \,\operatorname {csgn}\left (i \left (5+x \right )\right ) {\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right )\right ) {\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )}^{2}+\pi {\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )}^{3}-\pi \,\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )^{3}+2 i \ln \left (5+x \right )+2 i \ln \left (\ln \left (3 \,{\mathrm e}^{x}+3\right )\right )\right )}{2}\right )\) | \(256\) |
Input:
int((((6+x)*exp(x)+x+6)*ln(3*exp(x)+3)+(5+x)*exp(x))/((5+x)*exp(x)+5+x)/ln (3*exp(x)+3)/ln(exp(ln(5+x)+x)*ln(3*exp(x)+3)),x,method=_RETURNVERBOSE)
Output:
ln(ln(exp(ln(5+x)+x)*ln(3*exp(x)+3)))
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (\log \left (e^{\left (x + \log \left (x + 5\right )\right )} \log \left (\frac {3 \, {\left (x + e^{\left (x + \log \left (x + 5\right )\right )} + 5\right )}}{x + 5}\right )\right )\right ) \] Input:
integrate((((6+x)*exp(x)+x+6)*log(3*exp(x)+3)+(5+x)*exp(x))/((5+x)*exp(x)+ 5+x)/log(3*exp(x)+3)/log(exp(log(5+x)+x)*log(3*exp(x)+3)),x, algorithm="fr icas")
Output:
log(log(e^(x + log(x + 5))*log(3*(x + e^(x + log(x + 5)) + 5)/(x + 5))))
Time = 0.84 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log {\left (\log {\left (\left (x + 5\right ) e^{x} \log {\left (3 e^{x} + 3 \right )} \right )} \right )} \] Input:
integrate((((6+x)*exp(x)+x+6)*ln(3*exp(x)+3)+(5+x)*exp(x))/((5+x)*exp(x)+5 +x)/ln(3*exp(x)+3)/ln(exp(ln(5+x)+x)*ln(3*exp(x)+3)),x)
Output:
log(log((x + 5)*exp(x)*log(3*exp(x) + 3)))
Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (x + \log \left (x + 5\right ) + \log \left (\log \left (3\right ) + \log \left (e^{x} + 1\right )\right )\right ) \] Input:
integrate((((6+x)*exp(x)+x+6)*log(3*exp(x)+3)+(5+x)*exp(x))/((5+x)*exp(x)+ 5+x)/log(3*exp(x)+3)/log(exp(log(5+x)+x)*log(3*exp(x)+3)),x, algorithm="ma xima")
Output:
log(x + log(x + 5) + log(log(3) + log(e^x + 1)))
Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (x + \log \left (x + 5\right ) + \log \left (\log \left (3\right ) + \log \left (e^{x} + 1\right )\right )\right ) \] Input:
integrate((((6+x)*exp(x)+x+6)*log(3*exp(x)+3)+(5+x)*exp(x))/((5+x)*exp(x)+ 5+x)/log(3*exp(x)+3)/log(exp(log(5+x)+x)*log(3*exp(x)+3)),x, algorithm="gi ac")
Output:
log(x + log(x + 5) + log(log(3) + log(e^x + 1)))
Time = 3.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\ln \left (x+\ln \left (\ln \left (3\,{\mathrm {e}}^x+3\right )\,\left (x+5\right )\right )\right ) \] Input:
int((exp(x)*(x + 5) + log(3*exp(x) + 3)*(x + exp(x)*(x + 6) + 6))/(log(3*e xp(x) + 3)*log(log(3*exp(x) + 3)*exp(x + log(x + 5)))*(x + exp(x)*(x + 5) + 5)),x)
Output:
log(x + log(log(3*exp(x) + 3)*(x + 5)))
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (e^{x} \mathrm {log}\left (3 e^{x}+3\right ) x +5 e^{x} \mathrm {log}\left (3 e^{x}+3\right )\right )\right ) \] Input:
int((((6+x)*exp(x)+x+6)*log(3*exp(x)+3)+(5+x)*exp(x))/((5+x)*exp(x)+5+x)/l og(3*exp(x)+3)/log(exp(log(5+x)+x)*log(3*exp(x)+3)),x)
Output:
log(log(e**x*log(3*e**x + 3)*x + 5*e**x*log(3*e**x + 3)))