Integrand size = 117, antiderivative size = 26 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=1+\frac {x^2}{e^{-3-e^x+x}+\log (\log (2 x \log (4)))} \] Output:
x^2/(exp(-exp(x)+x-3)+ln(ln(4*x*ln(2))))+1
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\frac {e^{3+e^x} x^2}{e^x+e^{3+e^x} \log (\log (x \log (16)))} \] Input:
Integrate[(-x + E^(-3 - E^x + x)*(2*x - x^2 + E^x*x^2)*Log[2*x*Log[4]] + 2 *x*Log[2*x*Log[4]]*Log[Log[2*x*Log[4]]])/(E^(-6 - 2*E^x + 2*x)*Log[2*x*Log [4]] + 2*E^(-3 - E^x + x)*Log[2*x*Log[4]]*Log[Log[2*x*Log[4]]] + Log[2*x*L og[4]]*Log[Log[2*x*Log[4]]]^2),x]
Output:
(E^(3 + E^x)*x^2)/(E^x + E^(3 + E^x)*Log[Log[x*Log[16]]])
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.
| \(\displaystyle \int \frac {e^{x-e^x-3} \left (e^x x^2-x^2+2 x\right ) \log (2 x \log (4))-x+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))+2 e^{x-e^x-3} \log (2 x \log (4)) \log (\log (2 x \log (4)))+e^{2 x-2 e^x-6} \log (2 x \log (4))} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{2 \left (e^x+3\right )} \left (e^{x-e^x-3} \left (e^x x^2-x^2+2 x\right ) \log (2 x \log (4))-x+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))\right )}{\log (x \log (16)) \left (e^x+e^{e^x+3} \log (\log (x \log (16)))\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (e^{2 \left (e^x+3\right )-e^x-3} x^2+\frac {e^{2 \left (e^x+3\right )} x \left (e^{e^x+3} x \log (x \log (16)) \log ^2(\log (x \log (16)))+x \log (x \log (16)) \log (\log (x \log (16)))-1\right )}{\log (x \log (16)) \left (e^x+e^{e^x+3} \log (\log (x \log (16)))\right )^2}-\frac {e^{2 \left (e^x+3\right )-e^x-3} x \left (x+2 e^{e^x+3} x \log (\log (x \log (16)))-2\right )}{e^x+e^{e^x+3} \log (\log (x \log (16)))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int e^{3+e^x} x^2dx+\int \frac {e^{3 \left (3+e^x\right )} x^2 \log ^2(\log (x \log (16)))}{\left (e^{3+e^x} \log (\log (x \log (16)))+e^x\right )^2}dx+\int \frac {e^{2 \left (3+e^x\right )} x^2 \log (\log (x \log (16)))}{\left (e^{3+e^x} \log (\log (x \log (16)))+e^x\right )^2}dx-\int \frac {e^{3+e^x} x^2}{e^{3+e^x} \log (\log (x \log (16)))+e^x}dx-2 \int \frac {e^{2 \left (3+e^x\right )} x^2 \log (\log (x \log (16)))}{e^{3+e^x} \log (\log (x \log (16)))+e^x}dx-\int \frac {e^{2 \left (3+e^x\right )} x}{\log (x \log (16)) \left (e^{3+e^x} \log (\log (x \log (16)))+e^x\right )^2}dx+2 \int \frac {e^{3+e^x} x}{e^{3+e^x} \log (\log (x \log (16)))+e^x}dx\) |
Input:
Int[(-x + E^(-3 - E^x + x)*(2*x - x^2 + E^x*x^2)*Log[2*x*Log[4]] + 2*x*Log [2*x*Log[4]]*Log[Log[2*x*Log[4]]])/(E^(-6 - 2*E^x + 2*x)*Log[2*x*Log[4]] + 2*E^(-3 - E^x + x)*Log[2*x*Log[4]]*Log[Log[2*x*Log[4]]] + Log[2*x*Log[4]] *Log[Log[2*x*Log[4]]]^2),x]
Output:
$Aborted
Time = 9.49 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
| method | result | size |
| risch | \(\frac {x^{2}}{{\mathrm e}^{-{\mathrm e}^{x}+x -3}+\ln \left (\ln \left (4 x \ln \left (2\right )\right )\right )}\) | \(23\) |
| parallelrisch | \(\frac {x^{2}}{{\mathrm e}^{-{\mathrm e}^{x}+x -3}+\ln \left (\ln \left (4 x \ln \left (2\right )\right )\right )}\) | \(23\) |
Input:
int((2*x*ln(4*x*ln(2))*ln(ln(4*x*ln(2)))+(exp(x)*x^2-x^2+2*x)*ln(4*x*ln(2) )*exp(-exp(x)+x-3)-x)/(ln(4*x*ln(2))*ln(ln(4*x*ln(2)))^2+2*ln(4*x*ln(2))*e xp(-exp(x)+x-3)*ln(ln(4*x*ln(2)))+ln(4*x*ln(2))*exp(-exp(x)+x-3)^2),x,meth od=_RETURNVERBOSE)
Output:
x^2/(exp(-exp(x)+x-3)+ln(ln(4*x*ln(2))))
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\frac {x^{2}}{e^{\left (x - e^{x} - 3\right )} + \log \left (\log \left (4 \, x \log \left (2\right )\right )\right )} \] Input:
integrate((2*x*log(4*x*log(2))*log(log(4*x*log(2)))+(exp(x)*x^2-x^2+2*x)*l og(4*x*log(2))*exp(-exp(x)+x-3)-x)/(log(4*x*log(2))*log(log(4*x*log(2)))^2 +2*log(4*x*log(2))*exp(-exp(x)+x-3)*log(log(4*x*log(2)))+log(4*x*log(2))*e xp(-exp(x)+x-3)^2),x, algorithm="fricas")
Output:
x^2/(e^(x - e^x - 3) + log(log(4*x*log(2))))
Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\frac {x^{2}}{e^{x - e^{x} - 3} + \log {\left (\log {\left (4 x \log {\left (2 \right )} \right )} \right )}} \] Input:
integrate((2*x*ln(4*x*ln(2))*ln(ln(4*x*ln(2)))+(exp(x)*x**2-x**2+2*x)*ln(4 *x*ln(2))*exp(-exp(x)+x-3)-x)/(ln(4*x*ln(2))*ln(ln(4*x*ln(2)))**2+2*ln(4*x *ln(2))*exp(-exp(x)+x-3)*ln(ln(4*x*ln(2)))+ln(4*x*ln(2))*exp(-exp(x)+x-3)* *2),x)
Output:
x**2/(exp(x - exp(x) - 3) + log(log(4*x*log(2))))
Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\frac {x^{2} e^{\left (e^{x} + 3\right )}}{e^{\left (e^{x} + 3\right )} \log \left (2 \, \log \left (2\right ) + \log \left (x\right ) + \log \left (\log \left (2\right )\right )\right ) + e^{x}} \] Input:
integrate((2*x*log(4*x*log(2))*log(log(4*x*log(2)))+(exp(x)*x^2-x^2+2*x)*l og(4*x*log(2))*exp(-exp(x)+x-3)-x)/(log(4*x*log(2))*log(log(4*x*log(2)))^2 +2*log(4*x*log(2))*exp(-exp(x)+x-3)*log(log(4*x*log(2)))+log(4*x*log(2))*e xp(-exp(x)+x-3)^2),x, algorithm="maxima")
Output:
x^2*e^(e^x + 3)/(e^(e^x + 3)*log(2*log(2) + log(x) + log(log(2))) + e^x)
Leaf count of result is larger than twice the leaf count of optimal. 3138 vs. \(2 (24) = 48\).
Time = 0.57 (sec) , antiderivative size = 3138, normalized size of antiderivative = 120.69 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\text {Too large to display} \] Input:
integrate((2*x*log(4*x*log(2))*log(log(4*x*log(2)))+(exp(x)*x^2-x^2+2*x)*l og(4*x*log(2))*exp(-exp(x)+x-3)-x)/(log(4*x*log(2))*log(log(4*x*log(2)))^2 +2*log(4*x*log(2))*exp(-exp(x)+x-3)*log(log(4*x*log(2)))+log(4*x*log(2))*e xp(-exp(x)+x-3)^2),x, algorithm="giac")
Output:
(4*x^4*e^(6*x - 2*e^x + 6)*log(2)^2*log(2*log(2) + log(x) + log(log(2))) - 8*x^4*e^(5*x - 2*e^x + 6)*log(2)^2*log(2*log(2) + log(x) + log(log(2))) + 4*x^4*e^(4*x - 2*e^x + 6)*log(2)^2*log(2*log(2) + log(x) + log(log(2))) + 4*x^4*e^(6*x - 2*e^x + 6)*log(2)*log(x)*log(2*log(2) + log(x) + log(log(2 ))) - 8*x^4*e^(5*x - 2*e^x + 6)*log(2)*log(x)*log(2*log(2) + log(x) + log( log(2))) + 4*x^4*e^(4*x - 2*e^x + 6)*log(2)*log(x)*log(2*log(2) + log(x) + log(log(2))) + x^4*e^(6*x - 2*e^x + 6)*log(x)^2*log(2*log(2) + log(x) + l og(log(2))) - 2*x^4*e^(5*x - 2*e^x + 6)*log(x)^2*log(2*log(2) + log(x) + l og(log(2))) + x^4*e^(4*x - 2*e^x + 6)*log(x)^2*log(2*log(2) + log(x) + log (log(2))) + 4*x^4*e^(6*x - 2*e^x + 6)*log(2)*log(2*log(2) + log(x) + log(l og(2)))*log(log(2)) - 8*x^4*e^(5*x - 2*e^x + 6)*log(2)*log(2*log(2) + log( x) + log(log(2)))*log(log(2)) + 4*x^4*e^(4*x - 2*e^x + 6)*log(2)*log(2*log (2) + log(x) + log(log(2)))*log(log(2)) + 2*x^4*e^(6*x - 2*e^x + 6)*log(x) *log(2*log(2) + log(x) + log(log(2)))*log(log(2)) - 4*x^4*e^(5*x - 2*e^x + 6)*log(x)*log(2*log(2) + log(x) + log(log(2)))*log(log(2)) + 2*x^4*e^(4*x - 2*e^x + 6)*log(x)*log(2*log(2) + log(x) + log(log(2)))*log(log(2)) + x^ 4*e^(6*x - 2*e^x + 6)*log(2*log(2) + log(x) + log(log(2)))*log(log(2))^2 - 2*x^4*e^(5*x - 2*e^x + 6)*log(2*log(2) + log(x) + log(log(2)))*log(log(2) )^2 + x^4*e^(4*x - 2*e^x + 6)*log(2*log(2) + log(x) + log(log(2)))*log(log (2))^2 + 4*x^4*e^(7*x - 3*e^x + 3)*log(2)^2 - 8*x^4*e^(6*x - 3*e^x + 3)...
Timed out. \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\int \frac {{\mathrm {e}}^{x-{\mathrm {e}}^x-3}\,\ln \left (4\,x\,\ln \left (2\right )\right )\,\left (2\,x+x^2\,{\mathrm {e}}^x-x^2\right )-x+2\,x\,\ln \left (\ln \left (4\,x\,\ln \left (2\right )\right )\right )\,\ln \left (4\,x\,\ln \left (2\right )\right )}{\ln \left (4\,x\,\ln \left (2\right )\right )\,{\ln \left (\ln \left (4\,x\,\ln \left (2\right )\right )\right )}^2+2\,{\mathrm {e}}^{x-{\mathrm {e}}^x-3}\,\ln \left (4\,x\,\ln \left (2\right )\right )\,\ln \left (\ln \left (4\,x\,\ln \left (2\right )\right )\right )+{\mathrm {e}}^{2\,x-2\,{\mathrm {e}}^x-6}\,\ln \left (4\,x\,\ln \left (2\right )\right )} \,d x \] Input:
int((exp(x - exp(x) - 3)*log(4*x*log(2))*(2*x + x^2*exp(x) - x^2) - x + 2* x*log(log(4*x*log(2)))*log(4*x*log(2)))/(log(log(4*x*log(2)))^2*log(4*x*lo g(2)) + exp(2*x - 2*exp(x) - 6)*log(4*x*log(2)) + 2*log(log(4*x*log(2)))*e xp(x - exp(x) - 3)*log(4*x*log(2))),x)
Output:
int((exp(x - exp(x) - 3)*log(4*x*log(2))*(2*x + x^2*exp(x) - x^2) - x + 2* x*log(log(4*x*log(2)))*log(4*x*log(2)))/(log(log(4*x*log(2)))^2*log(4*x*lo g(2)) + exp(2*x - 2*exp(x) - 6)*log(4*x*log(2)) + 2*log(log(4*x*log(2)))*e xp(x - exp(x) - 3)*log(4*x*log(2))), x)
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-x+e^{-3-e^x+x} \left (2 x-x^2+e^x x^2\right ) \log (2 x \log (4))+2 x \log (2 x \log (4)) \log (\log (2 x \log (4)))}{e^{-6-2 e^x+2 x} \log (2 x \log (4))+2 e^{-3-e^x+x} \log (2 x \log (4)) \log (\log (2 x \log (4)))+\log (2 x \log (4)) \log ^2(\log (2 x \log (4)))} \, dx=\frac {e^{e^{x}} e^{3} x^{2}}{e^{e^{x}} \mathrm {log}\left (\mathrm {log}\left (4 \,\mathrm {log}\left (2\right ) x \right )\right ) e^{3}+e^{x}} \] Input:
int((2*x*log(4*x*log(2))*log(log(4*x*log(2)))+(exp(x)*x^2-x^2+2*x)*log(4*x *log(2))*exp(-exp(x)+x-3)-x)/(log(4*x*log(2))*log(log(4*x*log(2)))^2+2*log (4*x*log(2))*exp(-exp(x)+x-3)*log(log(4*x*log(2)))+log(4*x*log(2))*exp(-ex p(x)+x-3)^2),x)
Output:
(e**(e**x)*e**3*x**2)/(e**(e**x)*log(log(4*log(2)*x))*e**3 + e**x)