Integrand size = 126, antiderivative size = 21 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\log \left (\log \left (\frac {x+\left (e^x+x\right )^2}{e^{e^{10}}+x}\right )\right ) \] Output:
ln(ln(((exp(x)+x)^2+x)/(exp(exp(5)^2)+x)))
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\log \left (\log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )\right ) \] Input:
Integrate[(x^2 + 2*E^x*x^2 + E^(2*x)*(-1 + 2*x) + E^E^10*(1 + 2*E^(2*x) + 2*x + E^x*(2 + 2*x)))/((E^(2*x)*x + x^2 + 2*E^x*x^2 + x^3 + E^E^10*(E^(2*x ) + x + 2*E^x*x + x^2))*Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + x)]),x ]
Output:
Log[Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + 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.
| \(\displaystyle \int \frac {2 e^x x^2+x^2+e^{2 x} (2 x-1)+e^{e^{10}} \left (2 x+2 e^{2 x}+e^x (2 x+2)+1\right )}{\left (x^3+2 e^x x^2+x^2+e^{e^{10}} \left (x^2+2 e^x x+x+e^{2 x}\right )+e^{2 x} x\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 e^x x^2+x^2+e^{2 x} (2 x-1)+e^{e^{10}} \left (2 x+2 e^{2 x}+e^x (2 x+2)+1\right )}{\left (x+e^{e^{10}}\right ) \left (x^2+2 e^x x+x+e^{2 x}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x+2 e^{e^{10}}-1}{\left (x+e^{e^{10}}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}-\frac {2 x^2+2 e^x x-2 e^x-1}{\left (x^2+2 e^x x+x+e^{2 x}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int \frac {1}{\log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx-\int \frac {1}{\left (x+e^{e^{10}}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx+\int \frac {1}{\left (x^2+2 e^x x+x+e^{2 x}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx+2 \int \frac {e^x}{\left (x^2+2 e^x x+x+e^{2 x}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx-2 \int \frac {e^x x}{\left (x^2+2 e^x x+x+e^{2 x}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx-2 \int \frac {x^2}{\left (x^2+2 e^x x+x+e^{2 x}\right ) \log \left (\frac {x^2+2 e^x x+x+e^{2 x}}{x+e^{e^{10}}}\right )}dx\) |
Input:
Int[(x^2 + 2*E^x*x^2 + E^(2*x)*(-1 + 2*x) + E^E^10*(1 + 2*E^(2*x) + 2*x + E^x*(2 + 2*x)))/((E^(2*x)*x + x^2 + 2*E^x*x^2 + x^3 + E^E^10*(E^(2*x) + x + 2*E^x*x + x^2))*Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + x)]),x]
Output:
$Aborted
Time = 1.82 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29
| method | result | size |
| parallelrisch | \(\ln \left (\ln \left (\frac {{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}+x}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )\right )\) | \(27\) |
| risch | \(\ln \left (\ln \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right ) \operatorname {csgn}\left (i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )}^{3}-2 i \ln \left ({\mathrm e}^{{\mathrm e}^{10}}+x \right )\right )}{2}\right )\) | \(219\) |
Input:
int(((2*exp(x)^2+(2+2*x)*exp(x)+2*x+1)*exp(exp(5)^2)+(-1+2*x)*exp(x)^2+2*e xp(x)*x^2+x^2)/((exp(x)^2+2*exp(x)*x+x^2+x)*exp(exp(5)^2)+x*exp(x)^2+2*exp (x)*x^2+x^3+x^2)/ln((exp(x)^2+2*exp(x)*x+x^2+x)/(exp(exp(5)^2)+x)),x,metho d=_RETURNVERBOSE)
Output:
ln(ln((exp(x)^2+2*exp(x)*x+x^2+x)/(exp(exp(5)^2)+x)))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\log \left (\log \left (\frac {x^{2} + 2 \, x e^{x} + x + e^{\left (2 \, x\right )}}{x + e^{\left (e^{10}\right )}}\right )\right ) \] Input:
integrate(((2*exp(x)^2+(2+2*x)*exp(x)+2*x+1)*exp(exp(5)^2)+(-1+2*x)*exp(x) ^2+2*exp(x)*x^2+x^2)/((exp(x)^2+2*exp(x)*x+x^2+x)*exp(exp(5)^2)+x*exp(x)^2 +2*exp(x)*x^2+x^3+x^2)/log((exp(x)^2+2*exp(x)*x+x^2+x)/(exp(exp(5)^2)+x)), x, algorithm="fricas")
Output:
log(log((x^2 + 2*x*e^x + x + e^(2*x))/(x + e^(e^10))))
Time = 0.92 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\log {\left (\log {\left (\frac {x^{2} + 2 x e^{x} + x + e^{2 x}}{x + e^{e^{10}}} \right )} \right )} \] Input:
integrate(((2*exp(x)**2+(2+2*x)*exp(x)+2*x+1)*exp(exp(5)**2)+(-1+2*x)*exp( x)**2+2*exp(x)*x**2+x**2)/((exp(x)**2+2*exp(x)*x+x**2+x)*exp(exp(5)**2)+x* exp(x)**2+2*exp(x)*x**2+x**3+x**2)/ln((exp(x)**2+2*exp(x)*x+x**2+x)/(exp(e xp(5)**2)+x)),x)
Output:
log(log((x**2 + 2*x*exp(x) + x + exp(2*x))/(x + exp(exp(10)))))
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\log \left (\log \left (x^{2} + 2 \, x e^{x} + x + e^{\left (2 \, x\right )}\right ) - \log \left (x + e^{\left (e^{10}\right )}\right )\right ) \] Input:
integrate(((2*exp(x)^2+(2+2*x)*exp(x)+2*x+1)*exp(exp(5)^2)+(-1+2*x)*exp(x) ^2+2*exp(x)*x^2+x^2)/((exp(x)^2+2*exp(x)*x+x^2+x)*exp(exp(5)^2)+x*exp(x)^2 +2*exp(x)*x^2+x^3+x^2)/log((exp(x)^2+2*exp(x)*x+x^2+x)/(exp(exp(5)^2)+x)), x, algorithm="maxima")
Output:
log(log(x^2 + 2*x*e^x + x + e^(2*x)) - log(x + e^(e^10)))
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\log \left (\log \left (\frac {x^{2} + 2 \, x e^{x} + x + e^{\left (2 \, x\right )}}{x + e^{\left (e^{10}\right )}}\right )\right ) \] Input:
integrate(((2*exp(x)^2+(2+2*x)*exp(x)+2*x+1)*exp(exp(5)^2)+(-1+2*x)*exp(x) ^2+2*exp(x)*x^2+x^2)/((exp(x)^2+2*exp(x)*x+x^2+x)*exp(exp(5)^2)+x*exp(x)^2 +2*exp(x)*x^2+x^3+x^2)/log((exp(x)^2+2*exp(x)*x+x^2+x)/(exp(exp(5)^2)+x)), x, algorithm="giac")
Output:
log(log((x^2 + 2*x*e^x + x + e^(2*x))/(x + e^(e^10))))
Time = 3.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\ln \left (\ln \left (\frac {x+{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2}{x+{\mathrm {e}}^{{\mathrm {e}}^{10}}}\right )\right ) \] Input:
int((2*x^2*exp(x) + exp(exp(10))*(2*x + 2*exp(2*x) + exp(x)*(2*x + 2) + 1) + exp(2*x)*(2*x - 1) + x^2)/(log((x + exp(2*x) + 2*x*exp(x) + x^2)/(x + e xp(exp(10))))*(x*exp(2*x) + 2*x^2*exp(x) + exp(exp(10))*(x + exp(2*x) + 2* x*exp(x) + x^2) + x^2 + x^3)),x)
Output:
log(log((x + exp(2*x) + 2*x*exp(x) + x^2)/(x + exp(exp(10)))))
\[ \int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx=\text {too large to display} \] Input:
int(((2*exp(x)^2+(2+2*x)*exp(x)+2*x+1)*exp(exp(5)^2)+(-1+2*x)*exp(x)^2+2*e xp(x)*x^2+x^2)/((exp(x)^2+2*exp(x)*x+x^2+x)*exp(exp(5)^2)+x*exp(x)^2+2*exp (x)*x^2+x^3+x^2)/log((exp(x)^2+2*exp(x)*x+x^2+x)/(exp(exp(5)^2)+x)),x)
Output:
2*e**(e**10)*int(e**(2*x)/(e**(e**10 + 2*x)*log((e**(2*x) + 2*e**x*x + x** 2 + x)/(e**(e**10) + x)) + 2*e**(e**10 + x)*log((e**(2*x) + 2*e**x*x + x** 2 + x)/(e**(e**10) + x))*x + e**(e**10)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**2 + e**(e**10)*log((e**(2*x) + 2*e**x*x + x**2 + x )/(e**(e**10) + x))*x + e**(2*x)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e** (e**10) + x))*x + 2*e**x*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**2 + log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**3 + log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**2),x) + 2*e**(e **10)*int(e**x/(e**(e**10 + 2*x)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e** (e**10) + x)) + 2*e**(e**10 + x)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e** (e**10) + x))*x + e**(e**10)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e** 10) + x))*x**2 + e**(e**10)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**1 0) + x))*x + e**(2*x)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x ))*x + 2*e**x*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**2 + log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**3 + log((e**(2 *x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x**2),x) + 2*e**(e**10)*int(( e**x*x)/(e**(e**10 + 2*x)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x)) + 2*e**(e**10 + x)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x))*x + e**(e**10)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) + x ))*x**2 + e**(e**10)*log((e**(2*x) + 2*e**x*x + x**2 + x)/(e**(e**10) +...