Integrand size = 184, antiderivative size = 28 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log ^2\left (4+\log \left (1+e^{\frac {x}{\log (5)}+\frac {3 (x+\log (x))}{x}}\right )\right )} \]
Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log ^2\left (4+\log \left (1+e^{3+\frac {x}{\log (5)}} x^{3/x}\right )\right )} \]
Integrate[(E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*(-72*x^2 - 216*Log[5] + 216*Log[5]*Log[x]))/((4*x^2*Log[5] + 4*E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5] + (x^2*Log[5] + E^((x^2 + 3*x*Log [5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5])*Log[1 + E^((x^2 + 3*x*Log[5 ] + 3*Log[5]*Log[x])/(x*Log[5]))])*Log[4 + Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))]]^3),x]
Time = 1.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {7237}
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.
\(\displaystyle \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2+216 \log (5) \log (x)-216 \log (5)\right )}{\left (4 x^2 \log (5) e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+4 x^2 \log (5)+\left (x^2 \log (5) e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+x^2 \log (5)\right ) \log \left (e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+1\right )\right ) \log ^3\left (\log \left (e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+1\right )+4\right )} \, dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {36}{\log ^2\left (\log \left (x^{3/x} 5^{\frac {3}{\log (5)}} e^{\frac {x}{\log (5)}}+1\right )+4\right )}\) |
Int[(E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*(-72*x^2 - 216*Lo g[5] + 216*Log[5]*Log[x]))/((4*x^2*Log[5] + 4*E^((x^2 + 3*x*Log[5] + 3*Log [5]*Log[x])/(x*Log[5]))*x^2*Log[5] + (x^2*Log[5] + E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5])*Log[1 + E^((x^2 + 3*x*Log[5] + 3* Log[5]*Log[x])/(x*Log[5]))])*Log[4 + Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[ 5]*Log[x])/(x*Log[5]))]]^3),x]
3.13.14.3.1 Defintions of rubi rules used
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
\[\frac {36}{{\ln \left (\ln \left (x^{\frac {3}{x}} {\mathrm e}^{\frac {3 \ln \left (5\right )+x}{\ln \left (5\right )}}+1\right )+4\right )}^{2}}\]
int((216*ln(5)*ln(x)-216*ln(5)-72*x^2)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x /ln(5))/((x^2*ln(5)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+x^2*ln(5))* ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+1)+4*x^2*ln(5)*exp((3*ln(5)* ln(x)+3*x*ln(5)+x^2)/x/ln(5))+4*x^2*ln(5))/ln(ln(exp((3*ln(5)*ln(x)+3*x*ln (5)+x^2)/x/ln(5))+1)+4)^3,x)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log \left (\log \left (e^{\left (\frac {x^{2} + 3 \, x \log \left (5\right ) + 3 \, \log \left (5\right ) \log \left (x\right )}{x \log \left (5\right )}\right )} + 1\right ) + 4\right )^{2}} \]
integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*l og(5)+x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/l og(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4 *x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/l og(log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm =\
Timed out. \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\text {Timed out} \]
integrate((216*ln(5)*ln(x)-216*ln(5)-72*x**2)*exp((3*ln(5)*ln(x)+3*x*ln(5) +x**2)/x/ln(5))/((x**2*ln(5)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+x **2*ln(5))*ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+1)+4*x**2*ln(5)* exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+4*x**2*ln(5))/ln(ln(exp((3*ln( 5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+1)+4)**3,x)
Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log \left (\log \left (e^{\left (\frac {x}{\log \left (5\right )} + \frac {3 \, \log \left (x\right )}{x} + 3\right )} + 1\right ) + 4\right )^{2}} \]
integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*l og(5)+x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/l og(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4 *x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/l og(log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm =\
Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log \left (\log \left (e^{\left (\frac {x^{2} + 3 \, x \log \left (5\right ) + 3 \, \log \left (5\right ) \log \left (x\right )}{x \log \left (5\right )}\right )} + 1\right ) + 4\right )^{2}} \]
integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*l og(5)+x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/l og(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4 *x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/l og(log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm =\
Time = 13.41 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{{\ln \left (\ln \left (x^{3/x}\,{\mathrm {e}}^{\frac {x}{\ln \left (5\right )}}\,{\mathrm {e}}^3+1\right )+4\right )}^2} \]
int(-(exp((3*x*log(5) + 3*log(5)*log(x) + x^2)/(x*log(5)))*(216*log(5) - 2 16*log(5)*log(x) + 72*x^2))/(log(log(exp((3*x*log(5) + 3*log(5)*log(x) + x ^2)/(x*log(5))) + 1) + 4)^3*(4*x^2*log(5) + log(exp((3*x*log(5) + 3*log(5) *log(x) + x^2)/(x*log(5))) + 1)*(x^2*log(5) + x^2*exp((3*x*log(5) + 3*log( 5)*log(x) + x^2)/(x*log(5)))*log(5)) + 4*x^2*exp((3*x*log(5) + 3*log(5)*lo g(x) + x^2)/(x*log(5)))*log(5))),x)