Integrand size = 84, antiderivative size = 31 \[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=4^{-5+e^{5 e x}} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}} \]
\[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=\int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx \]
Integrate[(4^(-5 + E^(5*E*x))*(E^(-((-5 + x^2)/x)))^(-5 + E^(5*E*x))*(25 + 5*x^2 + E^(5*E*x)*(-5 - x^2) + 5*E^(1 + 5*E*x)*x^2*Log[4/E^((-5 + x^2)/x) ]))/x^2,x]
Integrate[(4^(-5 + E^(5*E*x))*(E^(-((-5 + x^2)/x)))^(-5 + E^(5*E*x))*(25 + 5*x^2 + E^(5*E*x)*(-5 - x^2) + 5*E^(1 + 5*E*x)*x^2*Log[4/E^((-5 + x^2)/x) ]))/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.
\(\displaystyle \int \frac {4^{e^{5 e x}-5} \left (e^{-\frac {x^2-5}{x}}\right )^{e^{5 e x}-5} \left (5 x^2+e^{5 e x} \left (-x^2-5\right )+5 e^{5 e x+1} x^2 \log \left (4 e^{-\frac {x^2-5}{x}}\right )+25\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4^{e^{5 e x}-5} \left (e^{\frac {5}{x}-x}\right )^{e^{5 e x}-5} \left (5 x^2+e^{5 e x} \left (-x^2-5\right )+5 e^{5 e x+1} x^2 \log \left (4 e^{-\frac {x^2-5}{x}}\right )+25\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5\ 4^{e^{5 e x}-5} \left (x^2+5\right ) \left (e^{\frac {5}{x}-x}\right )^{e^{5 e x}-5}}{x^2}+\frac {4^{e^{5 e x}-5} e^{5 e x} \left (e^{\frac {5}{x}-x}\right )^{e^{5 e x}-5} \left (-x^2+5 e x^2 \log \left (4 e^{\frac {5}{x}-x}\right )-5\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 25 \int \frac {4^{-5+e^{5 e x}} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}}{x^2}dx-5 \int \frac {4^{-5+e^{5 e x}} e^{5 e x} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}}{x^2}dx+25 \int \frac {\int 4^{-5+e^{5 e x}} e^{5 e x+1} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}dx}{x^2}dx+5 \int 4^{-5+e^{5 e x}} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}dx-\int 4^{-5+e^{5 e x}} e^{5 e x} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}dx+5 \int \int 4^{-5+e^{5 e x}} e^{5 e x+1} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}dxdx+5 \log \left (4 e^{\frac {5}{x}-x}\right ) \int 4^{-5+e^{5 e x}} e^{5 e x+1} \left (e^{\frac {5}{x}-x}\right )^{-5+e^{5 e x}}dx\) |
Int[(4^(-5 + E^(5*E*x))*(E^(-((-5 + x^2)/x)))^(-5 + E^(5*E*x))*(25 + 5*x^2 + E^(5*E*x)*(-5 - x^2) + 5*E^(1 + 5*E*x)*x^2*Log[4/E^((-5 + x^2)/x)]))/x^ 2,x]
3.23.52.3.1 Defintions of rubi rules used
Time = 28.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84
method | result | size |
default | \({\mathrm e}^{\left ({\mathrm e}^{5 x \,{\mathrm e}}-5\right ) \ln \left (4 \,{\mathrm e}^{-\frac {x^{2}-5}{x}}\right )}\) | \(26\) |
parallelrisch | \({\mathrm e}^{\left ({\mathrm e}^{5 x \,{\mathrm e}}-5\right ) \ln \left (4 \,{\mathrm e}^{-\frac {x^{2}-5}{x}}\right )}\) | \(26\) |
risch | \({\mathrm e}^{\left ({\mathrm e}^{5 x \,{\mathrm e}}-5\right ) \left (2 \ln \left (2\right )-\ln \left ({\mathrm e}^{\frac {x^{2}-5}{x}}\right )\right )}\) | \(29\) |
int((5*x^2*exp(1)*exp(5*x*exp(1))*ln(4/exp((x^2-5)/x))+(-x^2-5)*exp(5*x*ex p(1))+5*x^2+25)*exp((exp(5*x*exp(1))-5)*ln(4/exp((x^2-5)/x)))/x^2,x,method =_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=e^{\left (-\frac {{\left (10 \, x e \log \left (2\right ) - 5 \, {\left (x^{2} - 5\right )} e + {\left (x^{2} - 2 \, x \log \left (2\right ) - 5\right )} e^{\left (5 \, x e + 1\right )}\right )} e^{\left (-1\right )}}{x}\right )} \]
integrate((5*x^2*exp(1)*exp(5*x*exp(1))*log(4/exp((x^2-5)/x))+(-x^2-5)*exp (5*x*exp(1))+5*x^2+25)*exp((exp(5*x*exp(1))-5)*log(4/exp((x^2-5)/x)))/x^2, x, algorithm=\
Timed out. \[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=\text {Timed out} \]
integrate((5*x**2*exp(1)*exp(5*x*exp(1))*ln(4/exp((x**2-5)/x))+(-x**2-5)*e xp(5*x*exp(1))+5*x**2+25)*exp((exp(5*x*exp(1))-5)*ln(4/exp((x**2-5)/x)))/x **2,x)
Time = 0.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=\frac {1}{1024} \, e^{\left (-x e^{\left (5 \, x e\right )} + 2 \, e^{\left (5 \, x e\right )} \log \left (2\right ) + 5 \, x + \frac {5 \, e^{\left (5 \, x e\right )}}{x} - \frac {25}{x}\right )} \]
integrate((5*x^2*exp(1)*exp(5*x*exp(1))*log(4/exp((x^2-5)/x))+(-x^2-5)*exp (5*x*exp(1))+5*x^2+25)*exp((exp(5*x*exp(1))-5)*log(4/exp((x^2-5)/x)))/x^2, x, algorithm=\
\[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=\int { \frac {{\left (5 \, x^{2} e^{\left (5 \, x e + 1\right )} \log \left (4 \, e^{\left (-\frac {x^{2} - 5}{x}\right )}\right ) + 5 \, x^{2} - {\left (x^{2} + 5\right )} e^{\left (5 \, x e\right )} + 25\right )} \left (4 \, e^{\left (-\frac {x^{2} - 5}{x}\right )}\right )^{e^{\left (5 \, x e\right )} - 5}}{x^{2}} \,d x } \]
integrate((5*x^2*exp(1)*exp(5*x*exp(1))*log(4/exp((x^2-5)/x))+(-x^2-5)*exp (5*x*exp(1))+5*x^2+25)*exp((exp(5*x*exp(1))-5)*log(4/exp((x^2-5)/x)))/x^2, x, algorithm=\
integrate((5*x^2*e^(5*x*e + 1)*log(4*e^(-(x^2 - 5)/x)) + 5*x^2 - (x^2 + 5) *e^(5*x*e) + 25)*(4*e^(-(x^2 - 5)/x))^(e^(5*x*e) - 5)/x^2, x)
Time = 11.62 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {4^{-5+e^{5 e x}} \left (e^{-\frac {-5+x^2}{x}}\right )^{-5+e^{5 e x}} \left (25+5 x^2+e^{5 e x} \left (-5-x^2\right )+5 e^{1+5 e x} x^2 \log \left (4 e^{-\frac {-5+x^2}{x}}\right )\right )}{x^2} \, dx=\frac {2^{2\,{\mathrm {e}}^{5\,x\,\mathrm {e}}}\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^{5\,x\,\mathrm {e}}}{x}}\,{\mathrm {e}}^{-\frac {25}{x}}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{5\,x\,\mathrm {e}}}}{1024} \]
int((exp(log(4*exp(-(x^2 - 5)/x))*(exp(5*x*exp(1)) - 5))*(5*x^2 - exp(5*x* exp(1))*(x^2 + 5) + 5*x^2*exp(1)*exp(5*x*exp(1))*log(4*exp(-(x^2 - 5)/x)) + 25))/x^2,x)