Integrand size = 77, antiderivative size = 21 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log ^4\left (\frac {5}{4} \left (5-5 e^{e^{x^2}} x\right ) \log (x)\right ) \]
Time = 0.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log ^4\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right ) \]
Integrate[((-4 + E^E^x^2*(4*x + (4*x + 8*E^x^2*x^3)*Log[x]))*Log[(25*Log[x ] - 25*E^E^x^2*x*Log[x])/4]^3)/(-(x*Log[x]) + E^E^x^2*x^2*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.
\(\displaystyle \int \frac {\left (e^{e^{x^2}} \left (\left (8 e^{x^2} x^3+4 x\right ) \log (x)+4 x\right )-4\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{e^{e^{x^2}} x^2 \log (x)-x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (4-e^{e^{x^2}} \left (\left (8 e^{x^2} x^3+4 x\right ) \log (x)+4 x\right )\right ) \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \left (1-e^{e^{x^2}} x\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 \left (e^{e^{x^2}} x+e^{e^{x^2}} x \log (x)-1\right ) \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \left (e^{e^{x^2}} x-1\right ) \log (x)}+\frac {8 e^{x^2+e^{x^2}} x^2 \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{e^{e^{x^2}} x-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {\log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x}dx+4 \int \frac {\log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \left (e^{e^{x^2}} x-1\right )}dx+8 \int \frac {e^{x^2+e^{x^2}} x^2 \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{e^{e^{x^2}} x-1}dx+4 \int \frac {\log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \log (x)}dx\) |
Int[((-4 + E^E^x^2*(4*x + (4*x + 8*E^x^2*x^3)*Log[x]))*Log[(25*Log[x] - 25 *E^E^x^2*x*Log[x])/4]^3)/(-(x*Log[x]) + E^E^x^2*x^2*Log[x]),x]
3.28.76.3.1 Defintions of rubi rules used
Time = 200.90 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81
method | result | size |
parallelrisch | \({\ln \left (-\frac {25 \ln \left (x \right ) \left ({\mathrm e}^{{\mathrm e}^{x^{2}}} x -1\right )}{4}\right )}^{4}\) | \(17\) |
int((((8*x^3*exp(x^2)+4*x)*ln(x)+4*x)*exp(exp(x^2))-4)*ln(-25/4*x*ln(x)*ex p(exp(x^2))+25/4*ln(x))^3/(x^2*ln(x)*exp(exp(x^2))-x*ln(x)),x,method=_RETU RNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) + \frac {25}{4} \, \log \left (x\right )\right )^{4} \]
integrate((((8*x^3*exp(x^2)+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x* log(x)*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x, algorithm=\
Time = 1.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log {\left (- \frac {25 x e^{e^{x^{2}}} \log {\left (x \right )}}{4} + \frac {25 \log {\left (x \right )}}{4} \right )}^{4} \]
integrate((((8*x**3*exp(x**2)+4*x)*ln(x)+4*x)*exp(exp(x**2))-4)*ln(-25/4*x *ln(x)*exp(exp(x**2))+25/4*ln(x))**3/(x**2*ln(x)*exp(exp(x**2))-x*ln(x)),x )
\[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\int { \frac {4 \, {\left ({\left ({\left (2 \, x^{3} e^{\left (x^{2}\right )} + x\right )} \log \left (x\right ) + x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - 1\right )} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) + \frac {25}{4} \, \log \left (x\right )\right )^{3}}{x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - x \log \left (x\right )} \,d x } \]
integrate((((8*x^3*exp(x^2)+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x* log(x)*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x, algorithm=\
4*integrate((((2*x^3*e^(x^2) + x)*log(x) + x)*e^(e^(x^2)) - 1)*log(-25/4*x *e^(e^(x^2))*log(x) + 25/4*log(x))^3/(x^2*e^(e^(x^2))*log(x) - x*log(x)), x)
\[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\int { \frac {4 \, {\left ({\left ({\left (2 \, x^{3} e^{\left (x^{2}\right )} + x\right )} \log \left (x\right ) + x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - 1\right )} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) + \frac {25}{4} \, \log \left (x\right )\right )^{3}}{x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - x \log \left (x\right )} \,d x } \]
integrate((((8*x^3*exp(x^2)+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x* log(x)*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x, algorithm=\
integrate(4*(((2*x^3*e^(x^2) + x)*log(x) + x)*e^(e^(x^2)) - 1)*log(-25/4*x *e^(e^(x^2))*log(x) + 25/4*log(x))^3/(x^2*e^(e^(x^2))*log(x) - x*log(x)), x)
Time = 9.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx={\left (\ln \left (\ln \left (x\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \left (x\right )\right )+\ln \left (\frac {25}{4}\right )\right )}^4 \]