Integrand size = 70, antiderivative size = 26 \[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx=-2+e^{4 \left (e^x+x\right )} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \]
Time = 1.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx=e^{4 \left (e^x+x\right )} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \]
Integrate[(-4*E^(4*E^x + 4*x)*Log[3/(Log[4]*Log[x])]^3 + E^(4*E^x + 4*x)*( 1 + 4*x + 4*E^x*x)*Log[x]*Log[3/(Log[4]*Log[x])]^4)/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.
\(\displaystyle \int \frac {e^{4 x+4 e^x} \left (4 e^x x+4 x+1\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )-4 e^{4 x+4 e^x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (4 e^{5 x+4 e^x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right )+\frac {e^{4 \left (x+e^x\right )} \left (4 x \log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )+\log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )-4\right ) \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{4 \left (x+e^x\right )} \log ^4\left (\frac {3}{\log (4) \log (x)}\right )dx+4 \int e^{4 \left (x+e^x\right )} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right )dx+4 \int e^{5 x+4 e^x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right )dx-4 \int \frac {e^{4 \left (x+e^x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)}dx\) |
Int[(-4*E^(4*E^x + 4*x)*Log[3/(Log[4]*Log[x])]^3 + E^(4*E^x + 4*x)*(1 + 4* x + 4*E^x*x)*Log[x]*Log[3/(Log[4]*Log[x])]^4)/Log[x],x]
3.13.57.3.1 Defintions of rubi rules used
Time = 11.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \({\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} x \ln \left (\frac {3}{2 \ln \left (2\right ) \ln \left (x \right )}\right )^{4}\) | \(25\) |
risch | \({\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} x \ln \left (\ln \left (x \right )\right )^{4}+2 \left (2 x \ln \left (\ln \left (2\right )\right )-2 x \ln \left (3\right )+2 x \ln \left (2\right )\right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} \ln \left (\ln \left (x \right )\right )^{3}+\frac {3 \left (4 x \ln \left (2\right )^{2}+4 x \ln \left (3\right )^{2}+4 x \ln \left (\ln \left (2\right )\right )^{2}-8 x \ln \left (2\right ) \ln \left (3\right )+8 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right ) x -8 \ln \left (\ln \left (2\right )\right ) \ln \left (3\right ) x \right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} \ln \left (\ln \left (x \right )\right )^{2}}{2}+\frac {\left (24 x \ln \left (2\right ) \ln \left (3\right )^{2}+8 x \ln \left (2\right )^{3}-8 x \ln \left (3\right )^{3}-48 \ln \left (\ln \left (2\right )\right ) \ln \left (3\right ) \ln \left (2\right ) x +8 \ln \left (\ln \left (2\right )\right )^{3} x -24 \ln \left (\ln \left (2\right )\right )^{2} \ln \left (3\right ) x +24 \ln \left (\ln \left (2\right )\right )^{2} \ln \left (2\right ) x +24 \ln \left (\ln \left (2\right )\right ) \ln \left (3\right )^{2} x +24 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )^{2} x -24 \ln \left (3\right ) \ln \left (2\right )^{2} x \right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} \ln \left (\ln \left (x \right )\right )}{2}+\frac {\left (16 x \ln \left (2\right )^{4}+16 x \ln \left (3\right )^{4}-64 x \ln \left (3\right ) \ln \left (2\right )^{3}-192 \ln \left (\ln \left (2\right )\right )^{2} \ln \left (3\right ) \ln \left (2\right ) x +192 \ln \left (\ln \left (2\right )\right ) \ln \left (3\right )^{2} \ln \left (2\right ) x -192 \ln \left (\ln \left (2\right )\right ) \ln \left (3\right ) \ln \left (2\right )^{2} x +16 \ln \left (\ln \left (2\right )\right )^{4} x -64 \ln \left (3\right )^{3} \ln \left (2\right ) x +96 \ln \left (3\right )^{2} \ln \left (2\right )^{2} x -64 \ln \left (\ln \left (2\right )\right )^{3} \ln \left (3\right ) x +64 \ln \left (\ln \left (2\right )\right )^{3} \ln \left (2\right ) x +96 \ln \left (\ln \left (2\right )\right )^{2} \ln \left (3\right )^{2} x +96 \ln \left (\ln \left (2\right )\right )^{2} \ln \left (2\right )^{2} x -64 \ln \left (\ln \left (2\right )\right ) \ln \left (3\right )^{3} x +64 \ln \left (\ln \left (2\right )\right ) \ln \left (2\right )^{3} x \right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x}}{16}\) | \(381\) |
int(((4*exp(x)*x+4*x+1)*ln(x)*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))^4-4*ex p(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))^3)/ln(x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx=x e^{\left (4 \, x + 4 \, e^{x}\right )} \log \left (\frac {3}{2 \, \log \left (2\right ) \log \left (x\right )}\right )^{4} \]
integrate(((4*exp(x)*x+4*x+1)*log(x)*exp(4*exp(x)+4*x)*log(3/2/log(2)/log( x))^4-4*exp(4*exp(x)+4*x)*log(3/2/log(2)/log(x))^3)/log(x),x, algorithm=\
Timed out. \[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx=\text {Timed out} \]
integrate(((4*exp(x)*x+4*x+1)*ln(x)*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))* *4-4*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))**3)/ln(x),x)
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (26) = 52\).
Time = 0.39 (sec) , antiderivative size = 291, normalized size of antiderivative = 11.19 \[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx =\text {Too large to display} \]
integrate(((4*exp(x)*x+4*x+1)*log(x)*exp(4*exp(x)+4*x)*log(3/2/log(2)/log( x))^4-4*exp(4*exp(x)+4*x)*log(3/2/log(2)/log(x))^3)/log(x),x, algorithm=\
-(4*x*(log(3) - log(2) - log(log(2)))*e^(4*x)*log(log(x))^3 - x*e^(4*x)*lo g(log(x))^4 - 6*(log(3)^2 - 2*(log(3) - log(log(2)))*log(2) + log(2)^2 - 2 *log(3)*log(log(2)) + log(log(2))^2)*x*e^(4*x)*log(log(x))^2 + 4*(log(3)^3 + 3*(log(3) - log(log(2)))*log(2)^2 - log(2)^3 - 3*log(3)^2*log(log(2)) + 3*log(3)*log(log(2))^2 - log(log(2))^3 - 3*(log(3)^2 - 2*log(3)*log(log(2 )) + log(log(2))^2)*log(2))*x*e^(4*x)*log(log(x)) - (log(3)^4 - 4*(log(3) - log(log(2)))*log(2)^3 + log(2)^4 - 4*log(3)^3*log(log(2)) + 6*log(3)^2*l og(log(2))^2 - 4*log(3)*log(log(2))^3 + log(log(2))^4 + 6*(log(3)^2 - 2*lo g(3)*log(log(2)) + log(log(2))^2)*log(2)^2 - 4*(log(3)^3 - 3*log(3)^2*log( log(2)) + 3*log(3)*log(log(2))^2 - log(log(2))^3)*log(2))*x*e^(4*x))*e^(4* e^x)
\[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx=\int { \frac {{\left (4 \, x e^{x} + 4 \, x + 1\right )} e^{\left (4 \, x + 4 \, e^{x}\right )} \log \left (x\right ) \log \left (\frac {3}{2 \, \log \left (2\right ) \log \left (x\right )}\right )^{4} - 4 \, e^{\left (4 \, x + 4 \, e^{x}\right )} \log \left (\frac {3}{2 \, \log \left (2\right ) \log \left (x\right )}\right )^{3}}{\log \left (x\right )} \,d x } \]
integrate(((4*exp(x)*x+4*x+1)*log(x)*exp(4*exp(x)+4*x)*log(3/2/log(2)/log( x))^4-4*exp(4*exp(x)+4*x)*log(3/2/log(2)/log(x))^3)/log(x),x, algorithm=\
integrate(((4*x*e^x + 4*x + 1)*e^(4*x + 4*e^x)*log(x)*log(3/2/(log(2)*log( x)))^4 - 4*e^(4*x + 4*e^x)*log(3/2/(log(2)*log(x)))^3)/log(x), x)
Time = 9.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-4 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx=x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\ln \left (\frac {3}{2\,\ln \left (2\right )\,\ln \left (x\right )}\right )}^4 \]