Integrand size = 131, antiderivative size = 21 \[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=\frac {\log (2)}{-1+e^{(x+\log (\log (4)))^2}-2 x+\log (x)} \]
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=\frac {\log (2)}{-1-2 x+e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)+\log (x)} \]
Integrate[((-1 + 2*x)*Log[2] + E^(x^2 + 2*x*Log[Log[4]] + Log[Log[4]]^2)*( -2*x^2*Log[2] - 2*x*Log[2]*Log[Log[4]]))/(x + E^(2*x^2 + 4*x*Log[Log[4]] + 2*Log[Log[4]]^2)*x + 4*x^2 + 4*x^3 + (-2*x - 4*x^2)*Log[x] + x*Log[x]^2 + E^(x^2 + 2*x*Log[Log[4]] + Log[Log[4]]^2)*(-2*x - 4*x^2 + 2*x*Log[x])),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.
| \(\displaystyle \int \frac {e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )+(2 x-1) \log (2)}{4 x^3+4 x^2+x e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))}+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-4 x^2-2 x+2 x \log (x)\right )+\left (-4 x^2-2 x\right ) \log (x)+x+x \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )+(2 x-1) \log (2)}{x \left (-e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)+2 x-\log (x)+1\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {2 x \log (2) e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)}{\left (e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)-2 x+\log (x)-1\right )^2}+\frac {\log (4)}{\left (e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)-2 x+\log (x)-1\right )^2}-\frac {2 \log (2) \log (\log (4)) e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)}{\left (e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)-2 x+\log (x)-1\right )^2}-\frac {\log (2)}{x \left (-e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)+2 x-\log (x)+1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\log (2) \int \frac {1}{x \left (2 x-e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)-\log (x)+1\right )^2}dx+\log (4) \int \frac {1}{\left (-2 x+e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)+\log (x)-1\right )^2}dx-2 \log (2) \log (\log (4)) \int \frac {e^{x^2+2 \log (\log (4)) x+\log ^2(\log (4))}}{\left (-2 x+e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)+\log (x)-1\right )^2}dx-2 \log (2) \int \frac {e^{x^2+2 \log (\log (4)) x+\log ^2(\log (4))} x}{\left (-2 x+e^{x^2+\log ^2(\log (4))} \log ^{2 x}(4)+\log (x)-1\right )^2}dx\) |
Int[((-1 + 2*x)*Log[2] + E^(x^2 + 2*x*Log[Log[4]] + Log[Log[4]]^2)*(-2*x^2 *Log[2] - 2*x*Log[2]*Log[Log[4]]))/(x + E^(2*x^2 + 4*x*Log[Log[4]] + 2*Log [Log[4]]^2)*x + 4*x^2 + 4*x^3 + (-2*x - 4*x^2)*Log[x] + x*Log[x]^2 + E^(x^ 2 + 2*x*Log[Log[4]] + Log[Log[4]]^2)*(-2*x - 4*x^2 + 2*x*Log[x])),x]
3.25.70.3.1 Defintions of rubi rules used
Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33
| method | result | size |
| risch | \(-\frac {\ln \left (2\right )}{2 x -{\mathrm e}^{\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+x \right )^{2}}-\ln \left (x \right )+1}\) | \(28\) |
| parallelrisch | \(-\frac {\ln \left (2\right )}{2 x -{\mathrm e}^{\ln \left (2 \ln \left (2\right )\right )^{2}+2 x \ln \left (2 \ln \left (2\right )\right )+x^{2}}-\ln \left (x \right )+1}\) | \(38\) |
int(((-2*x*ln(2)*ln(2*ln(2))-2*x^2*ln(2))*exp(ln(2*ln(2))^2+2*x*ln(2*ln(2) )+x^2)+(-1+2*x)*ln(2))/(x*exp(ln(2*ln(2))^2+2*x*ln(2*ln(2))+x^2)^2+(2*x*ln (x)-4*x^2-2*x)*exp(ln(2*ln(2))^2+2*x*ln(2*ln(2))+x^2)+x*ln(x)^2+(-4*x^2-2* x)*ln(x)+4*x^3+4*x^2+x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=-\frac {\log \left (2\right )}{2 \, x - e^{\left (x^{2} + 2 \, x \log \left (2 \, \log \left (2\right )\right ) + \log \left (2 \, \log \left (2\right )\right )^{2}\right )} - \log \left (x\right ) + 1} \]
integrate(((-2*x*log(2)*log(2*log(2))-2*x^2*log(2))*exp(log(2*log(2))^2+2* x*log(2*log(2))+x^2)+(-1+2*x)*log(2))/(x*exp(log(2*log(2))^2+2*x*log(2*log (2))+x^2)^2+(2*x*log(x)-4*x^2-2*x)*exp(log(2*log(2))^2+2*x*log(2*log(2))+x ^2)+x*log(x)^2+(-4*x^2-2*x)*log(x)+4*x^3+4*x^2+x),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=\frac {\log {\left (2 \right )}}{- 2 x + e^{x^{2} + 2 x \log {\left (2 \log {\left (2 \right )} \right )} + \log {\left (2 \log {\left (2 \right )} \right )}^{2}} + \log {\left (x \right )} - 1} \]
integrate(((-2*x*ln(2)*ln(2*ln(2))-2*x**2*ln(2))*exp(ln(2*ln(2))**2+2*x*ln (2*ln(2))+x**2)+(-1+2*x)*ln(2))/(x*exp(ln(2*ln(2))**2+2*x*ln(2*ln(2))+x**2 )**2+(2*x*ln(x)-4*x**2-2*x)*exp(ln(2*ln(2))**2+2*x*ln(2*ln(2))+x**2)+x*ln( x)**2+(-4*x**2-2*x)*ln(x)+4*x**3+4*x**2+x),x)
Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=\frac {\log \left (2\right )}{2^{2 \, \log \left (\log \left (2\right )\right )} e^{\left (x^{2} + 2 \, x \log \left (2\right ) + \log \left (2\right )^{2} + 2 \, x \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )} - 2 \, x + \log \left (x\right ) - 1} \]
integrate(((-2*x*log(2)*log(2*log(2))-2*x^2*log(2))*exp(log(2*log(2))^2+2* x*log(2*log(2))+x^2)+(-1+2*x)*log(2))/(x*exp(log(2*log(2))^2+2*x*log(2*log (2))+x^2)^2+(2*x*log(x)-4*x^2-2*x)*exp(log(2*log(2))^2+2*x*log(2*log(2))+x ^2)+x*log(x)^2+(-4*x^2-2*x)*log(x)+4*x^3+4*x^2+x),x, algorithm=\
log(2)/(2^(2*log(log(2)))*e^(x^2 + 2*x*log(2) + log(2)^2 + 2*x*log(log(2)) + log(log(2))^2) - 2*x + log(x) - 1)
\[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=\int { -\frac {2 \, {\left (x^{2} \log \left (2\right ) + x \log \left (2\right ) \log \left (2 \, \log \left (2\right )\right )\right )} e^{\left (x^{2} + 2 \, x \log \left (2 \, \log \left (2\right )\right ) + \log \left (2 \, \log \left (2\right )\right )^{2}\right )} - {\left (2 \, x - 1\right )} \log \left (2\right )}{4 \, x^{3} + x \log \left (x\right )^{2} + 4 \, x^{2} + x e^{\left (2 \, x^{2} + 4 \, x \log \left (2 \, \log \left (2\right )\right ) + 2 \, \log \left (2 \, \log \left (2\right )\right )^{2}\right )} - 2 \, {\left (2 \, x^{2} - x \log \left (x\right ) + x\right )} e^{\left (x^{2} + 2 \, x \log \left (2 \, \log \left (2\right )\right ) + \log \left (2 \, \log \left (2\right )\right )^{2}\right )} - 2 \, {\left (2 \, x^{2} + x\right )} \log \left (x\right ) + x} \,d x } \]
integrate(((-2*x*log(2)*log(2*log(2))-2*x^2*log(2))*exp(log(2*log(2))^2+2* x*log(2*log(2))+x^2)+(-1+2*x)*log(2))/(x*exp(log(2*log(2))^2+2*x*log(2*log (2))+x^2)^2+(2*x*log(x)-4*x^2-2*x)*exp(log(2*log(2))^2+2*x*log(2*log(2))+x ^2)+x*log(x)^2+(-4*x^2-2*x)*log(x)+4*x^3+4*x^2+x),x, algorithm=\
Time = 12.59 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.10 \[ \int \frac {(-1+2 x) \log (2)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x^2 \log (2)-2 x \log (2) \log (\log (4))\right )}{x+e^{2 x^2+4 x \log (\log (4))+2 \log ^2(\log (4))} x+4 x^2+4 x^3+\left (-2 x-4 x^2\right ) \log (x)+x \log ^2(x)+e^{x^2+2 x \log (\log (4))+\log ^2(\log (4))} \left (-2 x-4 x^2+2 x \log (x)\right )} \, dx=-\frac {2\,\ln \left (2\right )\,\left (x+\ln \left (\ln \left (4\right )\right )\right )}{\left (2\,x+\ln \left ({\ln \left (4\right )}^2\right )\right )\,\left (2\,x-\ln \left (x\right )-2^{2\,x}\,2^{2\,\ln \left (\ln \left (2\right )\right )}\,{\mathrm {e}}^{x^2+{\ln \left (\ln \left (2\right )\right )}^2+{\ln \left (2\right )}^2}\,{\ln \left (2\right )}^{2\,x}+1\right )} \]
int((log(2)*(2*x - 1) - exp(2*x*log(2*log(2)) + log(2*log(2))^2 + x^2)*(2* x^2*log(2) + 2*x*log(2*log(2))*log(2)))/(x + x*log(x)^2 - log(x)*(2*x + 4* x^2) - exp(2*x*log(2*log(2)) + log(2*log(2))^2 + x^2)*(2*x - 2*x*log(x) + 4*x^2) + 4*x^2 + 4*x^3 + x*exp(4*x*log(2*log(2)) + 2*log(2*log(2))^2 + 2*x ^2)),x)