Integrand size = 138, antiderivative size = 24 \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=5+e^{2 x} \left (4+e^{e^x}\right ) \left (x+\frac {4}{\log (\log (x))}\right ) \]
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=\frac {e^{2 x} \left (4+e^{e^x}\right ) (4+x \log (\log (x)))}{\log (\log (x))} \]
Integrate[((-16*E^(2*x))/x - (4*E^(E^x + 2*x))/x + (32*E^(2*x)*Log[x] + (E ^(E^x + 2*x)*(8*x^2 + 4*E^x*x^2)*Log[x])/x^2)*Log[Log[x]] + ((E^(2*x)*(4*x ^2 + 8*x^3)*Log[x])/x^2 + (E^(E^x + 2*x)*(x^2 + 2*x^3 + E^x*x^3)*Log[x])/x ^2)*Log[Log[x]]^2)/(Log[x]*Log[Log[x]]^2),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.
| \(\displaystyle \int \frac {\left (\frac {e^{2 x+e^x} \left (4 e^x x^2+8 x^2\right ) \log (x)}{x^2}+32 e^{2 x} \log (x)\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (8 x^3+4 x^2\right ) \log (x)}{x^2}+\frac {e^{2 x+e^x} \left (e^x x^3+2 x^3+x^2\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))-\frac {16 e^{2 x}}{x}-\frac {4 e^{2 x+e^x}}{x}}{\log (x) \log ^2(\log (x))} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{2 x} \left (x \log (x) \log (\log (x)) \left (4 \left (2 e^{e^x}+e^{x+e^x}+8\right )+\left (e^{x+e^x} x+8 x+e^{e^x} (2 x+1)+4\right ) \log (\log (x))\right )-4 \left (e^{e^x}+4\right )\right )}{x \log (x) \log ^2(\log (x))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^{2 x} \left (e^{e^x}+4\right ) \left (2 x^2 \log (x) \log ^2(\log (x))+x \log (x) \log ^2(\log (x))+8 x \log (x) \log (\log (x))-4\right )}{x \log (x) \log ^2(\log (x))}+\frac {e^{3 x+e^x} (x \log (\log (x))+4)}{\log (\log (x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int e^{2 x+e^x} xdx+\int e^{3 x+e^x} xdx-4 \int \frac {e^{2 x+e^x}}{x \log (x) \log ^2(\log (x))}dx+8 \int \frac {e^{2 x+e^x}}{\log (\log (x))}dx+4 \int \frac {e^{3 x+e^x}}{\log (\log (x))}dx+\frac {4 e^{2 x} \left (x^2 \log (x) \log ^2(\log (x))+4 x \log (x) \log (\log (x))\right )}{x \log (x) \log ^2(\log (x))}-e^{e^x}+e^{x+e^x}\) |
Int[((-16*E^(2*x))/x - (4*E^(E^x + 2*x))/x + (32*E^(2*x)*Log[x] + (E^(E^x + 2*x)*(8*x^2 + 4*E^x*x^2)*Log[x])/x^2)*Log[Log[x]] + ((E^(2*x)*(4*x^2 + 8 *x^3)*Log[x])/x^2 + (E^(E^x + 2*x)*(x^2 + 2*x^3 + E^x*x^3)*Log[x])/x^2)*Lo g[Log[x]]^2)/(Log[x]*Log[Log[x]]^2),x]
3.12.19.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 100.48 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42
| method | result | size |
| risch | \({\mathrm e}^{2 x} {\mathrm e}^{{\mathrm e}^{x}} x +4 x \,{\mathrm e}^{2 x}+\frac {4 \,{\mathrm e}^{2 x} \left ({\mathrm e}^{{\mathrm e}^{x}}+4\right )}{\ln \left (\ln \left (x \right )\right )}\) | \(34\) |
| parallelrisch | \(\frac {{\mathrm e}^{2 x} x \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (\ln \left (x \right )\right )+4 \,{\mathrm e}^{2 x} x \ln \left (\ln \left (x \right )\right )+4 \,{\mathrm e}^{2 x} {\mathrm e}^{{\mathrm e}^{x}}+16 \,{\mathrm e}^{2 x}}{\ln \left (\ln \left (x \right )\right )}\) | \(75\) |
int((((exp(x)*x^3+2*x^3+x^2)*ln(x)*exp(x-ln(x))^2*exp(exp(x))+(8*x^3+4*x^2 )*ln(x)*exp(x-ln(x))^2)*ln(ln(x))^2+((4*exp(x)*x^2+8*x^2)*ln(x)*exp(x-ln(x ))^2*exp(exp(x))+32*x^2*ln(x)*exp(x-ln(x))^2)*ln(ln(x))-4*x*exp(x-ln(x))^2 *exp(exp(x))-16*x*exp(x-ln(x))^2)/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOS E)
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=\frac {4 \, x^{2} e^{\left (2 \, x + e^{x} - 2 \, \log \left (x\right )\right )} + {\left (x^{3} e^{\left (2 \, x + e^{x} - 2 \, \log \left (x\right )\right )} + 4 \, x e^{\left (2 \, x\right )}\right )} \log \left (\log \left (x\right )\right ) + 16 \, e^{\left (2 \, x\right )}}{\log \left (\log \left (x\right )\right )} \]
integrate((((exp(x)*x^3+2*x^3+x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+(8*x ^3+4*x^2)*log(x)*exp(x-log(x))^2)*log(log(x))^2+((4*exp(x)*x^2+8*x^2)*log( x)*exp(x-log(x))^2*exp(exp(x))+32*x^2*log(x)*exp(x-log(x))^2)*log(log(x))- 4*x*exp(x-log(x))^2*exp(exp(x))-16*x*exp(x-log(x))^2)/log(x)/log(log(x))^2 ,x, algorithm=\
(4*x^2*e^(2*x + e^x - 2*log(x)) + (x^3*e^(2*x + e^x - 2*log(x)) + 4*x*e^(2 *x))*log(log(x)) + 16*e^(2*x))/log(log(x))
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=\frac {\left (4 x \log {\left (\log {\left (x \right )} \right )} + 16\right ) e^{2 x}}{\log {\left (\log {\left (x \right )} \right )}} + \frac {\left (x e^{2 x} \log {\left (\log {\left (x \right )} \right )} + 4 e^{2 x}\right ) e^{e^{x}}}{\log {\left (\log {\left (x \right )} \right )}} \]
integrate((((exp(x)*x**3+2*x**3+x**2)*ln(x)*exp(x-ln(x))**2*exp(exp(x))+(8 *x**3+4*x**2)*ln(x)*exp(x-ln(x))**2)*ln(ln(x))**2+((4*exp(x)*x**2+8*x**2)* ln(x)*exp(x-ln(x))**2*exp(exp(x))+32*x**2*ln(x)*exp(x-ln(x))**2)*ln(ln(x)) -4*x*exp(x-ln(x))**2*exp(exp(x))-16*x*exp(x-ln(x))**2)/ln(x)/ln(ln(x))**2, x)
(4*x*log(log(x)) + 16)*exp(2*x)/log(log(x)) + (x*exp(2*x)*log(log(x)) + 4* exp(2*x))*exp(exp(x))/log(log(x))
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.75 \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + {\left (e^{x} - 1\right )} e^{\left (e^{x}\right )} + \frac {{\left ({\left (x e^{\left (2 \, x\right )} - e^{x} + 1\right )} \log \left (\log \left (x\right )\right ) + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (e^{x}\right )} + 16 \, e^{\left (2 \, x\right )}}{\log \left (\log \left (x\right )\right )} + 2 \, e^{\left (2 \, x\right )} \]
integrate((((exp(x)*x^3+2*x^3+x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+(8*x ^3+4*x^2)*log(x)*exp(x-log(x))^2)*log(log(x))^2+((4*exp(x)*x^2+8*x^2)*log( x)*exp(x-log(x))^2*exp(exp(x))+32*x^2*log(x)*exp(x-log(x))^2)*log(log(x))- 4*x*exp(x-log(x))^2*exp(exp(x))-16*x*exp(x-log(x))^2)/log(x)/log(log(x))^2 ,x, algorithm=\
2*(2*x - 1)*e^(2*x) + (e^x - 1)*e^(e^x) + (((x*e^(2*x) - e^x + 1)*log(log( x)) + 4*e^(2*x))*e^(e^x) + 16*e^(2*x))/log(log(x)) + 2*e^(2*x)
Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=\frac {4 \, x e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right ) + x e^{\left (2 \, x + e^{x}\right )} \log \left (\log \left (x\right )\right ) + 16 \, e^{\left (2 \, x\right )} + 4 \, e^{\left (2 \, x + e^{x}\right )}}{\log \left (\log \left (x\right )\right )} \]
integrate((((exp(x)*x^3+2*x^3+x^2)*log(x)*exp(x-log(x))^2*exp(exp(x))+(8*x ^3+4*x^2)*log(x)*exp(x-log(x))^2)*log(log(x))^2+((4*exp(x)*x^2+8*x^2)*log( x)*exp(x-log(x))^2*exp(exp(x))+32*x^2*log(x)*exp(x-log(x))^2)*log(log(x))- 4*x*exp(x-log(x))^2*exp(exp(x))-16*x*exp(x-log(x))^2)/log(x)/log(log(x))^2 ,x, algorithm=\
(4*x*e^(2*x)*log(log(x)) + x*e^(2*x + e^x)*log(log(x)) + 16*e^(2*x) + 4*e^ (2*x + e^x))/log(log(x))
Timed out. \[ \int \frac {-\frac {16 e^{2 x}}{x}-\frac {4 e^{e^x+2 x}}{x}+\left (32 e^{2 x} \log (x)+\frac {e^{e^x+2 x} \left (8 x^2+4 e^x x^2\right ) \log (x)}{x^2}\right ) \log (\log (x))+\left (\frac {e^{2 x} \left (4 x^2+8 x^3\right ) \log (x)}{x^2}+\frac {e^{e^x+2 x} \left (x^2+2 x^3+e^x x^3\right ) \log (x)}{x^2}\right ) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx=\int \frac {\left ({\mathrm {e}}^{2\,x-2\,\ln \left (x\right )}\,\ln \left (x\right )\,\left (8\,x^3+4\,x^2\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x-2\,\ln \left (x\right )}\,\ln \left (x\right )\,\left (x^3\,{\mathrm {e}}^x+x^2+2\,x^3\right )\right )\,{\ln \left (\ln \left (x\right )\right )}^2+\left (32\,x^2\,{\mathrm {e}}^{2\,x-2\,\ln \left (x\right )}\,\ln \left (x\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x-2\,\ln \left (x\right )}\,\ln \left (x\right )\,\left (4\,x^2\,{\mathrm {e}}^x+8\,x^2\right )\right )\,\ln \left (\ln \left (x\right )\right )-16\,x\,{\mathrm {e}}^{2\,x-2\,\ln \left (x\right )}-4\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x-2\,\ln \left (x\right )}}{{\ln \left (\ln \left (x\right )\right )}^2\,\ln \left (x\right )} \,d x \]
int((log(log(x))*(32*x^2*exp(2*x - 2*log(x))*log(x) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(4*x^2*exp(x) + 8*x^2)) + log(log(x))^2*(exp(2*x - 2*lo g(x))*log(x)*(4*x^2 + 8*x^3) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(x^3 *exp(x) + x^2 + 2*x^3)) - 16*x*exp(2*x - 2*log(x)) - 4*x*exp(exp(x))*exp(2 *x - 2*log(x)))/(log(log(x))^2*log(x)),x)
int((log(log(x))*(32*x^2*exp(2*x - 2*log(x))*log(x) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(4*x^2*exp(x) + 8*x^2)) + log(log(x))^2*(exp(2*x - 2*lo g(x))*log(x)*(4*x^2 + 8*x^3) + exp(exp(x))*exp(2*x - 2*log(x))*log(x)*(x^3 *exp(x) + x^2 + 2*x^3)) - 16*x*exp(2*x - 2*log(x)) - 4*x*exp(exp(x))*exp(2 *x - 2*log(x)))/(log(log(x))^2*log(x)), x)