Integrand size = 138, antiderivative size = 31 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4-x+4 \log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )}{x} \]
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \left (1+x+\log \left (1+e^{-\frac {1}{4}-\frac {5}{x}-x+x^2}\right )\right )}{x} \]
Integrate[(-4*x + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*(20 - 4*x - 4*x^2 + 8*x^3) + (-4*x - 4*E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*x)*Log[1 + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))])/(x^3 + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*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.
\(\displaystyle \int \frac {e^{\frac {4 x^3-4 x^2-x-20}{4 x}} \left (8 x^3-4 x^2-4 x+20\right )+\left (-4 e^{\frac {4 x^3-4 x^2-x-20}{4 x}} x-4 x\right ) \log \left (e^{\frac {4 x^3-4 x^2-x-20}{4 x}}+1\right )-4 x}{x^3+e^{\frac {4 x^3-4 x^2-x-20}{4 x}} x^3} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 \left (2 x^3-x^2-x \log \left (e^{x^2-x-\frac {5}{x}-\frac {1}{4}}+1\right )-x+5\right )}{x^3}-\frac {4 e^{x+\frac {5}{x}+\frac {1}{4}} \left (2 x^3-x^2+5\right )}{\left (e^{x^2}+e^{x+\frac {5}{x}+\frac {1}{4}}\right ) x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {e^{x^2}}{e^{x^2}+e^{x+\frac {1}{4}+\frac {5}{x}}}dx-8 \int \frac {e^{x+\frac {1}{4}+\frac {5}{x}}}{e^{x^2}+e^{x+\frac {1}{4}+\frac {5}{x}}}dx+4 \int \frac {e^{x^2}}{\left (e^{x^2}+e^{x+\frac {1}{4}+\frac {5}{x}}\right ) x}dx+4 \int \frac {e^{x+\frac {1}{4}+\frac {5}{x}}}{\left (e^{x^2}+e^{x+\frac {1}{4}+\frac {5}{x}}\right ) x}dx-20 \int \frac {e^{x^2}}{\left (e^{x^2}+e^{x+\frac {1}{4}+\frac {5}{x}}\right ) x^3}dx-20 \int \frac {e^{x+\frac {1}{4}+\frac {5}{x}}}{\left (e^{x^2}+e^{x+\frac {1}{4}+\frac {5}{x}}\right ) x^3}dx-\frac {10}{x^2}+\frac {4 \log \left (e^{x^2-x-\frac {5}{x}-\frac {1}{4}}+1\right )}{x}+8 x+\frac {4}{x}-4 \log (x)\) |
Int[(-4*x + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*(20 - 4*x - 4*x^2 + 8*x^3) + (-4*x - 4*E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*x)*Log[1 + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))])/(x^3 + E^((-20 - x - 4*x^2 + 4*x^3)/(4*x))*x^3),x]
3.23.87.3.1 Defintions of rubi rules used
Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(\frac {4+4 \ln \left ({\mathrm e}^{\frac {4 x^{3}-4 x^{2}-x -20}{4 x}}+1\right )}{x}\) | \(33\) |
norman | \(\frac {4 x +4 x \ln \left ({\mathrm e}^{\frac {4 x^{3}-4 x^{2}-x -20}{4 x}}+1\right )}{x^{2}}\) | \(36\) |
risch | \(\frac {4 \ln \left ({\mathrm e}^{\frac {4 x^{3}-4 x^{2}-x -20}{4 x}}+1\right )}{x}+\frac {4}{x}\) | \(36\) |
int(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*ln(exp(1/4*(4*x^3-4*x^2-x-20 )/x)+1)+(8*x^3-4*x^2-4*x+20)*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^3*exp(1 /4*(4*x^3-4*x^2-x-20)/x)+x^3),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \, {\left (\log \left (e^{\left (\frac {4 \, x^{3} - 4 \, x^{2} - x - 20}{4 \, x}\right )} + 1\right ) + 1\right )}}{x} \]
integrate(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*log(exp(1/4*(4*x^3-4*x ^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^ 3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \log {\left (e^{\frac {x^{3} - x^{2} - \frac {x}{4} - 5}{x}} + 1 \right )}}{x} + \frac {4}{x} \]
integrate(((-4*x*exp(1/4*(4*x**3-4*x**2-x-20)/x)-4*x)*ln(exp(1/4*(4*x**3-4 *x**2-x-20)/x)+1)+(8*x**3-4*x**2-4*x+20)*exp(1/4*(4*x**3-4*x**2-x-20)/x)-4 *x)/(x**3*exp(1/4*(4*x**3-4*x**2-x-20)/x)+x**3),x)
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \, x \log \left (e^{\left (x^{2}\right )} + e^{\left (x + \frac {5}{x} + \frac {1}{4}\right )}\right ) + 3 \, x - 20}{x^{2}} \]
integrate(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*log(exp(1/4*(4*x^3-4*x ^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^ 3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4 \, {\left (\log \left (e^{\left (\frac {4 \, x^{3} - 4 \, x^{2} - x - 20}{4 \, x}\right )} + 1\right ) + 1\right )}}{x} \]
integrate(((-4*x*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)*log(exp(1/4*(4*x^3-4*x ^2-x-20)/x)+1)+(8*x^3-4*x^2-4*x+20)*exp(1/4*(4*x^3-4*x^2-x-20)/x)-4*x)/(x^ 3*exp(1/4*(4*x^3-4*x^2-x-20)/x)+x^3),x, algorithm=\
Time = 8.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-4 x+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} \left (20-4 x-4 x^2+8 x^3\right )+\left (-4 x-4 e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x\right ) \log \left (1+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}}\right )}{x^3+e^{\frac {-20-x-4 x^2+4 x^3}{4 x}} x^3} \, dx=\frac {4\,\left (\ln \left ({\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-\frac {1}{4}}\,{\mathrm {e}}^{-\frac {5}{x}}+1\right )+1\right )}{x} \]