Integrand size = 109, antiderivative size = 22 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \]
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=\frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2)}{\log (32)} \]
Integrate[(2^((10*x)/(6*Log[x] + 2*E^(x^2/E^5)*Log[x]))*(-15*Log[2] + 15*L og[2]*Log[x] + E^(x^2/E^5)*(-5*Log[2] + 5*Log[2]*Log[x] - (10*x^2*Log[2]*L og[x])/E^5)))/(9*Log[x]^2 + 6*E^(x^2/E^5)*Log[x]^2 + E^((2*x^2)/E^5)*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.
\(\displaystyle \int \frac {2^{\frac {10 x}{2 e^{\frac {x^2}{e^5}} \log (x)+6 \log (x)}} \left (e^{\frac {x^2}{e^5}} \left (-\frac {10 x^2 \log (2) \log (x)}{e^5}+5 \log (2) \log (x)-5 \log (2)\right )+15 \log (2) \log (x)-15 \log (2)\right )}{6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)+9 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2^{\frac {5 x}{\left (e^{\frac {x^2}{e^5}}+3\right ) \log (x)}} \left (e^{\frac {x^2}{e^5}} \left (-\frac {10 x^2 \log (2) \log (x)}{e^5}+5 \log (2) \log (x)-5 \log (2)\right )+15 \log (2) \log (x)-15 \log (2)\right )}{\left (e^{\frac {x^2}{e^5}}+3\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 \log (2) 2^{\frac {5 x}{\left (e^{\frac {x^2}{e^5}}+3\right ) \log (x)}} \left (-2 x^2 \log (x)+e^5 \log (x)-e^5\right )}{e^5 \left (e^{\frac {x^2}{e^5}}+3\right ) \log ^2(x)}+\frac {15 x^2 \log (2) 2^{\frac {5 x}{\left (e^{\frac {x^2}{e^5}}+3\right ) \log (x)}+1}}{e^5 \left (e^{\frac {x^2}{e^5}}+3\right )^2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \log (2) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}dx+5 \log (2) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}dx+\frac {15 \log (2) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}+1} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)}dx}{e^5}-\frac {5 \log (2) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}+1} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}dx}{e^5}\) |
Int[(2^((10*x)/(6*Log[x] + 2*E^(x^2/E^5)*Log[x]))*(-15*Log[2] + 15*Log[2]* Log[x] + E^(x^2/E^5)*(-5*Log[2] + 5*Log[2]*Log[x] - (10*x^2*Log[2]*Log[x]) /E^5)))/(9*Log[x]^2 + 6*E^(x^2/E^5)*Log[x]^2 + E^((2*x^2)/E^5)*Log[x]^2),x ]
3.8.86.3.1 Defintions of rubi rules used
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
\[2^{\frac {5 x}{\ln \left (x \right ) \left ({\mathrm e}^{x^{2} {\mathrm e}^{-5}}+3\right )}}\]
int(((-10*ln(2)*ln(x)*exp(ln(x^2)-5)+5*ln(2)*ln(x)-5*ln(2))*exp(exp(ln(x^2 )-5))+15*ln(2)*ln(x)-15*ln(2))*exp(5*x*ln(2)/(2*ln(x)*exp(exp(ln(x^2)-5))+ 6*ln(x)))^2/(ln(x)^2*exp(exp(ln(x^2)-5))^2+6*ln(x)^2*exp(exp(ln(x^2)-5))+9 *ln(x)^2),x)
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \left (x\right )}} \]
integrate(((-10*log(2)*log(x)*exp(log(x^2)-5)+5*log(2)*log(x)-5*log(2))*ex p(exp(log(x^2)-5))+15*log(2)*log(x)-15*log(2))*exp(5*x*log(2)/(2*log(x)*ex p(exp(log(x^2)-5))+6*log(x)))^2/(log(x)^2*exp(exp(log(x^2)-5))^2+6*log(x)^ 2*exp(exp(log(x^2)-5))+9*log(x)^2),x, algorithm=\
Time = 0.59 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=e^{\frac {10 x \log {\left (2 \right )}}{2 e^{\frac {x^{2}}{e^{5}}} \log {\left (x \right )} + 6 \log {\left (x \right )}}} \]
integrate(((-10*ln(2)*ln(x)*exp(ln(x**2)-5)+5*ln(2)*ln(x)-5*ln(2))*exp(exp (ln(x**2)-5))+15*ln(2)*ln(x)-15*ln(2))*exp(5*x*ln(2)/(2*ln(x)*exp(exp(ln(x **2)-5))+6*ln(x)))**2/(ln(x)**2*exp(exp(ln(x**2)-5))**2+6*ln(x)**2*exp(exp (ln(x**2)-5))+9*ln(x)**2),x)
Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \left (x\right )}} \]
integrate(((-10*log(2)*log(x)*exp(log(x^2)-5)+5*log(2)*log(x)-5*log(2))*ex p(exp(log(x^2)-5))+15*log(2)*log(x)-15*log(2))*exp(5*x*log(2)/(2*log(x)*ex p(exp(log(x^2)-5))+6*log(x)))^2/(log(x)^2*exp(exp(log(x^2)-5))^2+6*log(x)^ 2*exp(exp(log(x^2)-5))+9*log(x)^2),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 \, x}{e^{\left (x^{2} e^{\left (-5\right )}\right )} \log \left (x\right ) + 3 \, \log \left (x\right )}} \]
integrate(((-10*log(2)*log(x)*exp(log(x^2)-5)+5*log(2)*log(x)-5*log(2))*ex p(exp(log(x^2)-5))+15*log(2)*log(x)-15*log(2))*exp(5*x*log(2)/(2*log(x)*ex p(exp(log(x^2)-5))+6*log(x)))^2/(log(x)^2*exp(exp(log(x^2)-5))^2+6*log(x)^ 2*exp(exp(log(x^2)-5))+9*log(x)^2),x, algorithm=\
Time = 11.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx={\mathrm {e}}^{\frac {5\,x\,\ln \left (2\right )}{3\,\ln \left (x\right )+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-5}}\,\ln \left (x\right )}} \]