Integrand size = 99, antiderivative size = 26 \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=6-\log \left (x \left (-2+\frac {(5+x)^2}{-e^{e^x}+x}\right )\right ) \]
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=\log \left (e^{e^x}-x\right )-\log (x)-\log \left (25+2 e^{e^x}+8 x+x^2\right ) \]
Integrate[(-2*E^(2*E^x) + 8*x^2 + 2*x^3 + E^E^x*(-25 - 16*x - 3*x^2 + E^x* (25*x + 10*x^2 + x^3)))/(2*E^(2*E^x)*x - 25*x^2 - 8*x^3 - x^4 + E^E^x*(25* x + 6*x^2 + x^3)),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.
\(\displaystyle \int \frac {2 x^3+8 x^2+e^{e^x} \left (-3 x^2+e^x \left (x^3+10 x^2+25 x\right )-16 x-25\right )-2 e^{2 e^x}}{-x^4-8 x^3-25 x^2+e^{e^x} \left (x^3+6 x^2+25 x\right )+2 e^{2 e^x} x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right )}-\frac {3 e^{e^x} x}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right )}+\frac {8 x}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right )}+\frac {e^{x+e^x} (x+5)^2}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right )}-\frac {16 e^{e^x}}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right )}-\frac {25 e^{e^x}}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right ) x}-\frac {2 e^{2 e^x}}{\left (e^{e^x}-x\right ) \left (x^2+8 x+2 e^{e^x}+25\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {1}{x^2+8 x+2 e^{e^x}+25}dx-2 \int \frac {e^{x+e^x}}{x^2+8 x+2 e^{e^x}+25}dx+2 \int \frac {e^{e^x}}{x \left (x^2+8 x+2 e^{e^x}+25\right )}dx+\frac {4}{25} \int \frac {e^{2 e^x}}{x \left (x^2+8 x+2 e^{e^x}+25\right )}dx-20 \int \frac {1}{(x+5)^2 \left (x^2+8 x+2 e^{e^x}+25\right )}dx-8 \int \frac {e^{e^x}}{(x+5)^2 \left (x^2+8 x+2 e^{e^x}+25\right )}dx-\frac {4}{5} \int \frac {e^{2 e^x}}{(x+5)^2 \left (x^2+8 x+2 e^{e^x}+25\right )}dx+24 \int \frac {1}{(x+5) \left (x^2+8 x+2 e^{e^x}+25\right )}dx+4 \int \frac {e^{e^x}}{(x+5) \left (x^2+8 x+2 e^{e^x}+25\right )}dx-\frac {4}{25} \int \frac {e^{2 e^x}}{(x+5) \left (x^2+8 x+2 e^{e^x}+25\right )}dx+2 \int \frac {1}{e^{e^x}-x}dx+\int \frac {e^{x+e^x}}{e^{e^x}-x}dx-\int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) x}dx-\frac {2}{25} \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) x}dx+10 \int \frac {1}{\left (e^{e^x}-x\right ) (x+5)^2}dx+4 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (x+5)^2}dx+\frac {2}{5} \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) (x+5)^2}dx-12 \int \frac {1}{\left (e^{e^x}-x\right ) (x+5)}dx-2 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (x+5)}dx+\frac {2}{25} \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) (x+5)}dx\) |
Int[(-2*E^(2*E^x) + 8*x^2 + 2*x^3 + E^E^x*(-25 - 16*x - 3*x^2 + E^x*(25*x + 10*x^2 + x^3)))/(2*E^(2*E^x)*x - 25*x^2 - 8*x^3 - x^4 + E^E^x*(25*x + 6* x^2 + x^3)),x]
3.18.70.3.1 Defintions of rubi rules used
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15
method | result | size |
norman | \(-\ln \left (x \right )-\ln \left (x^{2}+8 x +2 \,{\mathrm e}^{{\mathrm e}^{x}}+25\right )+\ln \left (x -{\mathrm e}^{{\mathrm e}^{x}}\right )\) | \(30\) |
risch | \(-\ln \left (x \right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-x \right )-\ln \left (\frac {x^{2}}{2}+4 x +{\mathrm e}^{{\mathrm e}^{x}}+\frac {25}{2}\right )\) | \(30\) |
parallelrisch | \(-\ln \left (x \right )-\ln \left (x^{2}+8 x +2 \,{\mathrm e}^{{\mathrm e}^{x}}+25\right )+\ln \left (x -{\mathrm e}^{{\mathrm e}^{x}}\right )\) | \(30\) |
int((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(exp(x)) +2*x^3+8*x^2)/(2*x*exp(exp(x))^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8*x^3-25 *x^2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=-\log \left (x^{2} + 8 \, x + 2 \, e^{\left (e^{x}\right )} + 25\right ) - \log \left (x\right ) + \log \left (-x + e^{\left (e^{x}\right )}\right ) \]
integrate((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(e xp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8* x^3-25*x^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=- \log {\left (x \right )} + \log {\left (- x + e^{e^{x}} \right )} - \log {\left (\frac {x^{2}}{2} + 4 x + e^{e^{x}} + \frac {25}{2} \right )} \]
integrate((-2*exp(exp(x))**2+((x**3+10*x**2+25*x)*exp(x)-3*x**2-16*x-25)*e xp(exp(x))+2*x**3+8*x**2)/(2*x*exp(exp(x))**2+(x**3+6*x**2+25*x)*exp(exp(x ))-x**4-8*x**3-25*x**2),x)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=-\log \left (\frac {1}{2} \, x^{2} + 4 \, x + e^{\left (e^{x}\right )} + \frac {25}{2}\right ) - \log \left (x\right ) + \log \left (-x + e^{\left (e^{x}\right )}\right ) \]
integrate((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(e xp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8* x^3-25*x^2),x, algorithm=\
Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=-\log \left (x^{2} + 8 \, x + 2 \, e^{\left (e^{x}\right )} + 25\right ) + \log \left (x - e^{\left (e^{x}\right )}\right ) - \log \left (x\right ) \]
integrate((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(e xp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8* x^3-25*x^2),x, algorithm=\
Timed out. \[ \int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx=\int \frac {2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (16\,x-{\mathrm {e}}^x\,\left (x^3+10\,x^2+25\,x\right )+3\,x^2+25\right )-8\,x^2-2\,x^3}{25\,x^2-2\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+8\,x^3+x^4-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x^3+6\,x^2+25\,x\right )} \,d x \]
int((2*exp(2*exp(x)) + exp(exp(x))*(16*x - exp(x)*(25*x + 10*x^2 + x^3) + 3*x^2 + 25) - 8*x^2 - 2*x^3)/(25*x^2 - 2*x*exp(2*exp(x)) + 8*x^3 + x^4 - e xp(exp(x))*(25*x + 6*x^2 + x^3)),x)