Integrand size = 78, antiderivative size = 29 \[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx=e^{-16/x} \left (1+e^5-x+(-1+x) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right ) \]
Time = 5.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx=e^{-16/x} \left (1+e^5-x+(-1+x) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right ) \]
Integrate[(x - x^2 + (16 + 16*E^5 - 16*x - x^2)*Log[1/(5*x)] + (-16 + 16*x + x^2)*Log[1/(5*x)]*Log[Log[1/(5*x)]])/(E^(16/x)*x^2*Log[1/(5*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.
\(\displaystyle \int \frac {e^{-16/x} \left (-x^2+\left (-x^2-16 x+16 e^5+16\right ) \log \left (\frac {1}{5 x}\right )+\left (x^2+16 x-16\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )+x\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-16/x} \left (-x^2+x^2 \left (-\log \left (\frac {1}{5 x}\right )\right )+x-16 x \log \left (\frac {1}{5 x}\right )+16 \left (1+e^5\right ) \log \left (\frac {1}{5 x}\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )}+\frac {e^{-16/x} \left (x^2+16 x-16\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\text {Subst}\left (\int \frac {e^{-16 x}}{x \log \left (\frac {x}{5}\right )}dx,x,\frac {1}{x}\right )-16 \text {Subst}\left (\int \frac {e^{-16 x} \log \left (\log \left (\frac {x}{5}\right )\right )}{x}dx,x,\frac {1}{x}\right )+80 \text {Subst}\left (\int e^{-80 x} \log (\log (x))dx,x,\frac {1}{5 x}\right )+\int \frac {e^{-16/x}}{\log (5)-\log \left (\frac {1}{x}\right )}dx+\int e^{-16/x} \log \left (\log \left (\frac {1}{5 x}\right )\right )dx+e^{-16/x} \left (-x+e^5+1\right )\) |
Int[(x - x^2 + (16 + 16*E^5 - 16*x - x^2)*Log[1/(5*x)] + (-16 + 16*x + x^2 )*Log[1/(5*x)]*Log[Log[1/(5*x)]])/(E^(16/x)*x^2*Log[1/(5*x)]),x]
3.4.4.3.1 Defintions of rubi rules used
Time = 16.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24
method | result | size |
risch | \(\left (-1+x \right ) {\mathrm e}^{-\frac {16}{x}} \ln \left (-\ln \left (5\right )-\ln \left (x \right )\right )+\left ({\mathrm e}^{5}-x +1\right ) {\mathrm e}^{-\frac {16}{x}}\) | \(36\) |
parallelrisch | \(\frac {\left (\ln \left (\ln \left (\frac {1}{5 x}\right )\right ) x^{2}+x \,{\mathrm e}^{5}-x^{2}-\ln \left (\ln \left (\frac {1}{5 x}\right )\right ) x +x \right ) {\mathrm e}^{-\frac {16}{x}}}{x}\) | \(45\) |
int(((x^2+16*x-16)*ln(1/5/x)*ln(ln(1/5/x))+(16*exp(5)-x^2-16*x+16)*ln(1/5/ x)-x^2+x)/x^2/exp(4/x)^4/ln(1/5/x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx={\left (x - 1\right )} e^{\left (-\frac {16}{x}\right )} \log \left (\log \left (\frac {1}{5 \, x}\right )\right ) - {\left (x - e^{5} - 1\right )} e^{\left (-\frac {16}{x}\right )} \]
integrate(((x^2+16*x-16)*log(1/5/x)*log(log(1/5/x))+(16*exp(5)-x^2-16*x+16 )*log(1/5/x)-x^2+x)/x^2/exp(4/x)^4/log(1/5/x),x, algorithm=\
Time = 52.82 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx=\left (x \log {\left (\log {\left (\frac {1}{5 x} \right )} \right )} - x - \log {\left (\log {\left (\frac {1}{5 x} \right )} \right )} + 1 + e^{5}\right ) e^{- \frac {16}{x}} \]
integrate(((x**2+16*x-16)*ln(1/5/x)*ln(ln(1/5/x))+(16*exp(5)-x**2-16*x+16) *ln(1/5/x)-x**2+x)/x**2/exp(4/x)**4/ln(1/5/x),x)
\[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx=\int { \frac {{\left ({\left (x^{2} + 16 \, x - 16\right )} \log \left (\frac {1}{5 \, x}\right ) \log \left (\log \left (\frac {1}{5 \, x}\right )\right ) - x^{2} - {\left (x^{2} + 16 \, x - 16 \, e^{5} - 16\right )} \log \left (\frac {1}{5 \, x}\right ) + x\right )} e^{\left (-\frac {16}{x}\right )}}{x^{2} \log \left (\frac {1}{5 \, x}\right )} \,d x } \]
integrate(((x^2+16*x-16)*log(1/5/x)*log(log(1/5/x))+(16*exp(5)-x^2-16*x+16 )*log(1/5/x)-x^2+x)/x^2/exp(4/x)^4/log(1/5/x),x, algorithm=\
e^(-16/x + 5) - 16*gamma(-1, 16/x) + integrate((x^2 - x*(16*log(5) + 1) - 16*(x - 1)*log(x) + (x^2*log(5) + 16*x*log(5) + (x^2 + 16*x - 16)*log(x) - 16*log(5))*log(-log(5) - log(x)) + 16*log(5))*e^(-16/x)/(x^2*log(5) + x^2 *log(x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx=x e^{\left (-\frac {16}{x}\right )} \log \left (-\log \left (5 \, x\right )\right ) - x e^{\left (-\frac {16}{x}\right )} - e^{\left (-\frac {16}{x}\right )} \log \left (-\log \left (5 \, x\right )\right ) + e^{\left (-\frac {16}{x}\right )} + e^{\left (-\frac {16}{x} + 5\right )} \]
integrate(((x^2+16*x-16)*log(1/5/x)*log(log(1/5/x))+(16*exp(5)-x^2-16*x+16 )*log(1/5/x)-x^2+x)/x^2/exp(4/x)^4/log(1/5/x),x, algorithm=\
x*e^(-16/x)*log(-log(5*x)) - x*e^(-16/x) - e^(-16/x)*log(-log(5*x)) + e^(- 16/x) + e^(-16/x + 5)
Timed out. \[ \int \frac {e^{-16/x} \left (x-x^2+\left (16+16 e^5-16 x-x^2\right ) \log \left (\frac {1}{5 x}\right )+\left (-16+16 x+x^2\right ) \log \left (\frac {1}{5 x}\right ) \log \left (\log \left (\frac {1}{5 x}\right )\right )\right )}{x^2 \log \left (\frac {1}{5 x}\right )} \, dx=\int \frac {{\mathrm {e}}^{-\frac {16}{x}}\,\left (x-\ln \left (\frac {1}{5\,x}\right )\,\left (x^2+16\,x-16\,{\mathrm {e}}^5-16\right )-x^2+\ln \left (\ln \left (\frac {1}{5\,x}\right )\right )\,\ln \left (\frac {1}{5\,x}\right )\,\left (x^2+16\,x-16\right )\right )}{x^2\,\ln \left (\frac {1}{5\,x}\right )} \,d x \]
int((exp(-16/x)*(x - log(1/(5*x))*(16*x - 16*exp(5) + x^2 - 16) - x^2 + lo g(log(1/(5*x)))*log(1/(5*x))*(16*x + x^2 - 16)))/(x^2*log(1/(5*x))),x)