Integrand size = 87, antiderivative size = 34 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=5 \left (x+\frac {e^3 \left (4+e^{x-\frac {1+e^4 x}{x}}\right )}{\log \left (\frac {x}{2}\right )}\right ) \] Output:
5*exp(3)/ln(1/2*x)*(4+exp(x-(x*exp(4)+1)/x))+5*x
Time = 0.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=5 x+\frac {5 e^3 \left (4+e^{-e^4-\frac {1}{x}+x}\right )}{\log \left (\frac {x}{2}\right )} \] Input:
Integrate[(-20*E^3*x - 5*E^(3 + (-1 - E^4*x + x^2)/x)*x + E^(3 + (-1 - E^4 *x + x^2)/x)*(5 + 5*x^2)*Log[x/2] + 5*x^2*Log[x/2]^2)/(x^2*Log[x/2]^2),x]
Output:
5*x + (5*E^3*(4 + E^(-E^4 - x^(-1) + x)))/Log[x/2]
Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(34)=68\).
Time = 1.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {-5 e^{\frac {x^2-e^4 x-1}{x}+3} x+5 x^2 \log ^2\left (\frac {x}{2}\right )+e^{\frac {x^2-e^4 x-1}{x}+3} \left (5 x^2+5\right ) \log \left (\frac {x}{2}\right )-20 e^3 x}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 e^{x-\frac {1}{x}+3 \left (1-\frac {e^4}{3}\right )} \left (x^2 \log \left (\frac {x}{2}\right )-x+\log \left (\frac {x}{2}\right )\right )}{x^2 \log ^2\left (\frac {x}{2}\right )}-\frac {5 \left (4 e^3-x \log ^2\left (\frac {x}{2}\right )\right )}{x \log ^2\left (\frac {x}{2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 e^{x-\frac {1}{x}-e^4+3} \left (x^2 \log \left (\frac {x}{2}\right )+\log \left (\frac {x}{2}\right )\right )}{\left (\frac {1}{x^2}+1\right ) x^2 \log ^2\left (\frac {x}{2}\right )}+5 x+\frac {20 e^3}{\log \left (\frac {x}{2}\right )}\) |
Input:
Int[(-20*E^3*x - 5*E^(3 + (-1 - E^4*x + x^2)/x)*x + E^(3 + (-1 - E^4*x + x ^2)/x)*(5 + 5*x^2)*Log[x/2] + 5*x^2*Log[x/2]^2)/(x^2*Log[x/2]^2),x]
Output:
5*x + (20*E^3)/Log[x/2] + (5*E^(3 - E^4 - x^(-1) + x)*(Log[x/2] + x^2*Log[ x/2]))/((1 + x^(-2))*x^2*Log[x/2]^2)
Time = 0.57 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00
method | result | size |
risch | \(5 x +\frac {5 \,{\mathrm e}^{3} \left ({\mathrm e}^{-\frac {x \,{\mathrm e}^{4}-x^{2}+1}{x}}+4\right )}{\ln \left (\frac {x}{2}\right )}\) | \(34\) |
default | \(5 x +\frac {5 \,{\mathrm e}^{3} {\mathrm e}^{\frac {-x \,{\mathrm e}^{4}+x^{2}-1}{x}}}{\ln \left (\frac {x}{2}\right )}+\frac {20 \,{\mathrm e}^{3}}{\ln \left (\frac {x}{2}\right )}\) | \(40\) |
parts | \(5 x +\frac {5 \,{\mathrm e}^{3} {\mathrm e}^{\frac {-x \,{\mathrm e}^{4}+x^{2}-1}{x}}}{\ln \left (\frac {x}{2}\right )}+\frac {20 \,{\mathrm e}^{3}}{\ln \left (\frac {x}{2}\right )}\) | \(40\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{3} {\mathrm e}^{-\frac {x \,{\mathrm e}^{4}-x^{2}+1}{x}}+5 x \ln \left (\frac {x}{2}\right )+20 \,{\mathrm e}^{3}}{\ln \left (\frac {x}{2}\right )}\) | \(41\) |
norman | \(\frac {20 x \,{\mathrm e}^{3}+5 \ln \left (\frac {x}{2}\right ) x^{2}+5 x \,{\mathrm e}^{3} {\mathrm e}^{\frac {-x \,{\mathrm e}^{4}+x^{2}-1}{x}}}{x \ln \left (\frac {x}{2}\right )}\) | \(46\) |
Input:
int((5*x^2*ln(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*ln(1/2*x) -5*x*exp(3)*exp((-x*exp(4)+x^2-1)/x)-20*x*exp(3))/x^2/ln(1/2*x)^2,x,method =_RETURNVERBOSE)
Output:
5*x+5*exp(3)*(exp(-(x*exp(4)-x^2+1)/x)+4)/ln(1/2*x)
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\frac {5 \, {\left (x \log \left (\frac {1}{2} \, x\right ) + 4 \, e^{3} + e^{\left (\frac {x^{2} - x e^{4} + 3 \, x - 1}{x}\right )}\right )}}{\log \left (\frac {1}{2} \, x\right )} \] Input:
integrate((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*lo g(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2 ,x, algorithm="fricas")
Output:
5*(x*log(1/2*x) + 4*e^3 + e^((x^2 - x*e^4 + 3*x - 1)/x))/log(1/2*x)
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=5 x + \frac {5 e^{3} e^{\frac {x^{2} - x e^{4} - 1}{x}}}{\log {\left (\frac {x}{2} \right )}} + \frac {20 e^{3}}{\log {\left (\frac {x}{2} \right )}} \] Input:
integrate((5*x**2*ln(1/2*x)**2+(5*x**2+5)*exp(3)*exp((-x*exp(4)+x**2-1)/x) *ln(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x**2-1)/x)-20*x*exp(3))/x**2/ln(1/2*x )**2,x)
Output:
5*x + 5*exp(3)*exp((x**2 - x*exp(4) - 1)/x)/log(x/2) + 20*exp(3)/log(x/2)
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=5 \, x - \frac {5 \, e^{\left (x - \frac {1}{x} + 3\right )}}{e^{\left (e^{4}\right )} \log \left (2\right ) - e^{\left (e^{4}\right )} \log \left (x\right )} + \frac {20 \, e^{3}}{\log \left (\frac {1}{2} \, x\right )} \] Input:
integrate((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*lo g(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2 ,x, algorithm="maxima")
Output:
5*x - 5*e^(x - 1/x + 3)/(e^(e^4)*log(2) - e^(e^4)*log(x)) + 20*e^3/log(1/2 *x)
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\frac {5 \, {\left (x \log \left (\frac {1}{2} \, x\right ) + 4 \, e^{3} + e^{\left (\frac {x^{2} - x e^{4} + 3 \, x - 1}{x}\right )}\right )}}{\log \left (\frac {1}{2} \, x\right )} \] Input:
integrate((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*lo g(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2 ,x, algorithm="giac")
Output:
5*(x*log(1/2*x) + 4*e^3 + e^((x^2 - x*e^4 + 3*x - 1)/x))/log(1/2*x)
Time = 3.67 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=5\,x-\frac {20\,{\mathrm {e}}^3}{\ln \left (2\right )-\ln \left (x\right )}-\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-\frac {1}{x}}\,{\mathrm {e}}^x}{\ln \left (2\right )-\ln \left (x\right )} \] Input:
int(-(20*x*exp(3) - 5*x^2*log(x/2)^2 + 5*x*exp(3)*exp(-(x*exp(4) - x^2 + 1 )/x) - log(x/2)*exp(3)*exp(-(x*exp(4) - x^2 + 1)/x)*(5*x^2 + 5))/(x^2*log( x/2)^2),x)
Output:
5*x - (20*exp(3))/(log(2) - log(x)) - (5*exp(-exp(4))*exp(3)*exp(-1/x)*exp (x))/(log(2) - log(x))
Time = 1.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.74 \[ \int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} \left (5+5 x^2\right ) \log \left (\frac {x}{2}\right )+5 x^2 \log ^2\left (\frac {x}{2}\right )}{x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\frac {5 e^{\frac {e^{4} x +1}{x}} \mathrm {log}\left (\frac {x}{2}\right ) \mathrm {log}\left (\frac {2}{x}\right ) x +20 e^{\frac {e^{4} x +1}{x}} \mathrm {log}\left (\frac {2}{x}\right ) e^{3}-5 e^{x} \mathrm {log}\left (\frac {x}{2}\right ) e^{3}}{e^{\frac {e^{4} x +1}{x}} \mathrm {log}\left (\frac {x}{2}\right ) \mathrm {log}\left (\frac {2}{x}\right )} \] Input:
int((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*log(1/2* x)-5*x*exp(3)*exp((-x*exp(4)+x^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2,x)
Output:
(5*(e**((e**4*x + 1)/x)*log(x/2)*log(2/x)*x + 4*e**((e**4*x + 1)/x)*log(2/ x)*e**3 - e**x*log(x/2)*e**3))/(e**((e**4*x + 1)/x)*log(x/2)*log(2/x))