Integrand size = 133, antiderivative size = 23 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\log \left (\log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (x+e^x x\right )\right )\right ) \] Output:
ln(ln(3/5*x*(ln(2*x)+x)^2*ln(exp(x)*x+x)))
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\log \left (\log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (\left (1+e^x\right ) x\right )\right )\right ) \] Input:
Integrate[(x + E^x*(x + x^2) + (1 + E^x*(1 + x))*Log[2*x] + (2 + 3*x + E^x *(2 + 3*x) + (1 + E^x)*Log[2*x])*Log[x + E^x*x])/((x^2 + E^x*x^2 + (x + E^ x*x)*Log[2*x])*Log[x + E^x*x]*Log[((3*x^3 + 6*x^2*Log[2*x] + 3*x*Log[2*x]^ 2)*Log[x + E^x*x])/5]),x]
Output:
Log[Log[(3*x*(x + Log[2*x])^2*Log[(1 + E^x)*x])/5]]
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.
| \(\displaystyle \int \frac {e^x \left (x^2+x\right )+x+\left (e^x (x+1)+1\right ) \log (2 x)+\left (3 x+e^x (3 x+2)+\left (e^x+1\right ) \log (2 x)+2\right ) \log \left (e^x x+x\right )}{\left (e^x x^2+x^2+\left (e^x x+x\right ) \log (2 x)\right ) \log \left (e^x x+x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (e^x x+x\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^x \left (x^2+x\right )+x+\left (e^x (x+1)+1\right ) \log (2 x)+\left (3 x+e^x (3 x+2)+\left (e^x+1\right ) \log (2 x)+2\right ) \log \left (e^x x+x\right )}{\left (e^x+1\right ) x (x+\log (2 x)) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^2+x+x \log (2 x)+3 x \log \left (\left (e^x+1\right ) x\right )+\log (2 x)+\log (2 x) \log \left (\left (e^x+1\right ) x\right )+2 \log \left (\left (e^x+1\right ) x\right )}{x (x+\log (2 x)) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}+\frac {1}{\left (-e^x-1\right ) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \int \frac {1}{(x+\log (2 x)) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+2 \int \frac {1}{x (x+\log (2 x)) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+\int \frac {\log (2 x)}{x (x+\log (2 x)) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+\int \frac {1}{\left (-1-e^x\right ) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+\int \frac {1}{(x+\log (2 x)) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+\int \frac {x}{(x+\log (2 x)) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+\int \frac {\log (2 x)}{(x+\log (2 x)) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx+\int \frac {\log (2 x)}{x (x+\log (2 x)) \log \left (e^x x+x\right ) \log \left (\frac {3}{5} x (x+\log (2 x))^2 \log \left (e^x x+x\right )\right )}dx\) |
Input:
Int[(x + E^x*(x + x^2) + (1 + E^x*(1 + x))*Log[2*x] + (2 + 3*x + E^x*(2 + 3*x) + (1 + E^x)*Log[2*x])*Log[x + E^x*x])/((x^2 + E^x*x^2 + (x + E^x*x)*L og[2*x])*Log[x + E^x*x]*Log[((3*x^3 + 6*x^2*Log[2*x] + 3*x*Log[2*x]^2)*Log [x + E^x*x])/5]),x]
Output:
$Aborted
Time = 101.76 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30
| method | result | size |
| parallelrisch | \(\ln \left (\ln \left (\frac {3 x \left (\ln \left (2 x \right )^{2}+2 x \ln \left (2 x \right )+x^{2}\right ) \ln \left (x \left ({\mathrm e}^{x}+1\right )\right )}{5}\right )\right )\) | \(30\) |
Input:
int((((exp(x)+1)*ln(2*x)+(2+3*x)*exp(x)+3*x+2)*ln(exp(x)*x+x)+((1+x)*exp(x )+1)*ln(2*x)+(x^2+x)*exp(x)+x)/((exp(x)*x+x)*ln(2*x)+exp(x)*x^2+x^2)/ln(ex p(x)*x+x)/ln(1/5*(3*x*ln(2*x)^2+6*x^2*ln(2*x)+3*x^3)*ln(exp(x)*x+x)),x,met hod=_RETURNVERBOSE)
Output:
ln(ln(3/5*x*(ln(2*x)^2+2*x*ln(2*x)+x^2)*ln(x*(exp(x)+1))))
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\log \left (\log \left (\frac {3}{5} \, {\left (x^{3} + 2 \, x^{2} \log \left (2 \, x\right ) + x \log \left (2 \, x\right )^{2}\right )} \log \left (x e^{x} + x\right )\right )\right ) \] Input:
integrate((((1+exp(x))*log(2*x)+(2+3*x)*exp(x)+3*x+2)*log(exp(x)*x+x)+((1+ x)*exp(x)+1)*log(2*x)+(x^2+x)*exp(x)+x)/((exp(x)*x+x)*log(2*x)+exp(x)*x^2+ x^2)/log(exp(x)*x+x)/log(1/5*(3*x*log(2*x)^2+6*x^2*log(2*x)+3*x^3)*log(exp (x)*x+x)),x, algorithm="fricas")
Output:
log(log(3/5*(x^3 + 2*x^2*log(2*x) + x*log(2*x)^2)*log(x*e^x + x)))
Time = 14.99 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\log {\left (\log {\left (\left (\frac {3 x^{3}}{5} + \frac {6 x^{2} \log {\left (2 x \right )}}{5} + \frac {3 x \log {\left (2 x \right )}^{2}}{5}\right ) \log {\left (x e^{x} + x \right )} \right )} \right )} \] Input:
integrate((((1+exp(x))*ln(2*x)+(2+3*x)*exp(x)+3*x+2)*ln(exp(x)*x+x)+((1+x) *exp(x)+1)*ln(2*x)+(x**2+x)*exp(x)+x)/((exp(x)*x+x)*ln(2*x)+exp(x)*x**2+x* *2)/ln(exp(x)*x+x)/ln(1/5*(3*x*ln(2*x)**2+6*x**2*ln(2*x)+3*x**3)*ln(exp(x) *x+x)),x)
Output:
log(log((3*x**3/5 + 6*x**2*log(2*x)/5 + 3*x*log(2*x)**2/5)*log(x*exp(x) + x)))
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\log \left (-\log \left (5\right ) + \log \left (3\right ) + 2 \, \log \left (x + \log \left (2\right ) + \log \left (x\right )\right ) + \log \left (x\right ) + \log \left (\log \left (x\right ) + \log \left (e^{x} + 1\right )\right )\right ) \] Input:
integrate((((1+exp(x))*log(2*x)+(2+3*x)*exp(x)+3*x+2)*log(exp(x)*x+x)+((1+ x)*exp(x)+1)*log(2*x)+(x^2+x)*exp(x)+x)/((exp(x)*x+x)*log(2*x)+exp(x)*x^2+ x^2)/log(exp(x)*x+x)/log(1/5*(3*x*log(2*x)^2+6*x^2*log(2*x)+3*x^3)*log(exp (x)*x+x)),x, algorithm="maxima")
Output:
log(-log(5) + log(3) + 2*log(x + log(2) + log(x)) + log(x) + log(log(x) + log(e^x + 1)))
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (20) = 40\).
Time = 0.97 (sec) , antiderivative size = 116, normalized size of antiderivative = 5.04 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\log \left (-\log \left (5\right ) + \log \left (3 \, x^{2} \log \left (x\right ) + 6 \, x \log \left (2\right ) \log \left (x\right ) + 3 \, \log \left (2\right )^{2} \log \left (x\right ) + 6 \, x \log \left (x\right )^{2} + 6 \, \log \left (2\right ) \log \left (x\right )^{2} + 3 \, \log \left (x\right )^{3} + 3 \, x^{2} \log \left (e^{x} + 1\right ) + 6 \, x \log \left (2\right ) \log \left (e^{x} + 1\right ) + 3 \, \log \left (2\right )^{2} \log \left (e^{x} + 1\right ) + 6 \, x \log \left (x\right ) \log \left (e^{x} + 1\right ) + 6 \, \log \left (2\right ) \log \left (x\right ) \log \left (e^{x} + 1\right ) + 3 \, \log \left (x\right )^{2} \log \left (e^{x} + 1\right )\right ) + \log \left (x\right )\right ) \] Input:
integrate((((1+exp(x))*log(2*x)+(2+3*x)*exp(x)+3*x+2)*log(exp(x)*x+x)+((1+ x)*exp(x)+1)*log(2*x)+(x^2+x)*exp(x)+x)/((exp(x)*x+x)*log(2*x)+exp(x)*x^2+ x^2)/log(exp(x)*x+x)/log(1/5*(3*x*log(2*x)^2+6*x^2*log(2*x)+3*x^3)*log(exp (x)*x+x)),x, algorithm="giac")
Output:
log(-log(5) + log(3*x^2*log(x) + 6*x*log(2)*log(x) + 3*log(2)^2*log(x) + 6 *x*log(x)^2 + 6*log(2)*log(x)^2 + 3*log(x)^3 + 3*x^2*log(e^x + 1) + 6*x*lo g(2)*log(e^x + 1) + 3*log(2)^2*log(e^x + 1) + 6*x*log(x)*log(e^x + 1) + 6* log(2)*log(x)*log(e^x + 1) + 3*log(x)^2*log(e^x + 1)) + log(x))
Time = 3.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\ln \left (\ln \left (\frac {\ln \left (x+x\,{\mathrm {e}}^x\right )\,\left (3\,x^3+6\,x^2\,\ln \left (2\,x\right )+3\,x\,{\ln \left (2\,x\right )}^2\right )}{5}\right )\right ) \] Input:
int((x + log(2*x)*(exp(x)*(x + 1) + 1) + log(x + x*exp(x))*(3*x + exp(x)*( 3*x + 2) + log(2*x)*(exp(x) + 1) + 2) + exp(x)*(x + x^2))/(log((log(x + x* exp(x))*(3*x*log(2*x)^2 + 6*x^2*log(2*x) + 3*x^3))/5)*log(x + x*exp(x))*(x ^2*exp(x) + log(2*x)*(x + x*exp(x)) + x^2)),x)
Output:
log(log((log(x + x*exp(x))*(3*x*log(2*x)^2 + 6*x^2*log(2*x) + 3*x^3))/5))
Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.17 \[ \int \frac {x+e^x \left (x+x^2\right )+\left (1+e^x (1+x)\right ) \log (2 x)+\left (2+3 x+e^x (2+3 x)+\left (1+e^x\right ) \log (2 x)\right ) \log \left (x+e^x x\right )}{\left (x^2+e^x x^2+\left (x+e^x x\right ) \log (2 x)\right ) \log \left (x+e^x x\right ) \log \left (\frac {1}{5} \left (3 x^3+6 x^2 \log (2 x)+3 x \log ^2(2 x)\right ) \log \left (x+e^x x\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {3 \,\mathrm {log}\left (e^{x} x +x \right ) \mathrm {log}\left (2 x \right )^{2} x}{5}+\frac {6 \,\mathrm {log}\left (e^{x} x +x \right ) \mathrm {log}\left (2 x \right ) x^{2}}{5}+\frac {3 \,\mathrm {log}\left (e^{x} x +x \right ) x^{3}}{5}\right )\right ) \] Input:
int((((1+exp(x))*log(2*x)+(2+3*x)*exp(x)+3*x+2)*log(exp(x)*x+x)+((1+x)*exp (x)+1)*log(2*x)+(x^2+x)*exp(x)+x)/((exp(x)*x+x)*log(2*x)+exp(x)*x^2+x^2)/l og(exp(x)*x+x)/log(1/5*(3*x*log(2*x)^2+6*x^2*log(2*x)+3*x^3)*log(exp(x)*x+ x)),x)
Output:
log(log((3*log(e**x*x + x)*log(2*x)**2*x + 6*log(e**x*x + x)*log(2*x)*x**2 + 3*log(e**x*x + x)*x**3)/5))