Integrand size = 99, antiderivative size = 30 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=-3+\frac {2}{-e^{3+x-x^2 (-2+x (-x+\log (x)))}+x} \] Output:
2/(x-exp(x-x^2*(x*(ln(x)-x)-2)+3))-3
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=-\frac {2 x^{x^3}}{e^{3+x+2 x^2+x^4}-x^{1+x^3}} \] Input:
Integrate[(-2 + E^(3 + x + 2*x^2 + x^4 - x^3*Log[x])*(2 + 8*x - 2*x^2 + 8* x^3 - 6*x^2*Log[x]))/(E^(6 + 2*x + 4*x^2 + 2*x^4 - 2*x^3*Log[x]) - 2*E^(3 + x + 2*x^2 + x^4 - x^3*Log[x])*x + x^2),x]
Output:
(-2*x^x^3)/(E^(3 + x + 2*x^2 + x^4) - x^(1 + x^3))
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.
\(\displaystyle \int \frac {e^{x^4-x^3 \log (x)+2 x^2+x+3} \left (8 x^3-2 x^2-6 x^2 \log (x)+8 x+2\right )-2}{x^2-2 x e^{x^4-x^3 \log (x)+2 x^2+x+3}+e^{2 x^4-2 x^3 \log (x)+4 x^2+2 x+6}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^{2 x^3} \left (e^{x^4-x^3 \log (x)+2 x^2+x+3} \left (8 x^3-2 x^2-6 x^2 \log (x)+8 x+2\right )-2\right )}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 x^{2 x^3}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}+\frac {8 e^{x^4+2 x^2+x+3} x^{x^3+1}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}-\frac {2 e^{x^4+2 x^2+x+3} x^{x^3+2}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}+\frac {8 e^{x^4+2 x^2+x+3} x^{x^3+3}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}-\frac {6 e^{x^4+2 x^2+x+3} x^{x^3+2} \log (x)}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}+\frac {2 e^{x^4+2 x^2+x+3} x^{x^3}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {x^{2 x^3}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx+8 \int \frac {e^{x^4+2 x^2+x+3} x^{x^3+1}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx-2 \int \frac {e^{x^4+2 x^2+x+3} x^{x^3+2}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx+8 \int \frac {e^{x^4+2 x^2+x+3} x^{x^3+3}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx+6 \int \frac {\int \frac {e^{x^4+2 x^2+x+3} x^{x^3+2}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx}{x}dx-6 \log (x) \int \frac {e^{x^4+2 x^2+x+3} x^{x^3+2}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx+2 \int \frac {e^{x^4+2 x^2+x+3} x^{x^3}}{\left (e^{x^4+2 x^2+x+3}-x^{x^3+1}\right )^2}dx\) |
Input:
Int[(-2 + E^(3 + x + 2*x^2 + x^4 - x^3*Log[x])*(2 + 8*x - 2*x^2 + 8*x^3 - 6*x^2*Log[x]))/(E^(6 + 2*x + 4*x^2 + 2*x^4 - 2*x^3*Log[x]) - 2*E^(3 + x + 2*x^2 + x^4 - x^3*Log[x])*x + x^2),x]
Output:
$Aborted
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {2}{x -x^{-x^{3}} {\mathrm e}^{x^{4}+2 x^{2}+x +3}}\) | \(28\) |
parallelrisch | \(\frac {2}{x -{\mathrm e}^{-x^{3} \ln \left (x \right )+x^{4}+2 x^{2}+x +3}}\) | \(28\) |
Input:
int(((-6*x^2*ln(x)+8*x^3-2*x^2+8*x+2)*exp(-x^3*ln(x)+x^4+2*x^2+x+3)-2)/(ex p(-x^3*ln(x)+x^4+2*x^2+x+3)^2-2*x*exp(-x^3*ln(x)+x^4+2*x^2+x+3)+x^2),x,met hod=_RETURNVERBOSE)
Output:
2/(x-x^(-x^3)*exp(x^4+2*x^2+x+3))
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=\frac {2}{x - e^{\left (x^{4} - x^{3} \log \left (x\right ) + 2 \, x^{2} + x + 3\right )}} \] Input:
integrate(((-6*x^2*log(x)+8*x^3-2*x^2+8*x+2)*exp(-x^3*log(x)+x^4+2*x^2+x+3 )-2)/(exp(-x^3*log(x)+x^4+2*x^2+x+3)^2-2*x*exp(-x^3*log(x)+x^4+2*x^2+x+3)+ x^2),x, algorithm="fricas")
Output:
2/(x - e^(x^4 - x^3*log(x) + 2*x^2 + x + 3))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=- \frac {2}{- x + e^{x^{4} - x^{3} \log {\left (x \right )} + 2 x^{2} + x + 3}} \] Input:
integrate(((-6*x**2*ln(x)+8*x**3-2*x**2+8*x+2)*exp(-x**3*ln(x)+x**4+2*x**2 +x+3)-2)/(exp(-x**3*ln(x)+x**4+2*x**2+x+3)**2-2*x*exp(-x**3*ln(x)+x**4+2*x **2+x+3)+x**2),x)
Output:
-2/(-x + exp(x**4 - x**3*log(x) + 2*x**2 + x + 3))
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=\frac {2 \, x^{\left (x^{3}\right )}}{x x^{\left (x^{3}\right )} - e^{\left (x^{4} + 2 \, x^{2} + x + 3\right )}} \] Input:
integrate(((-6*x^2*log(x)+8*x^3-2*x^2+8*x+2)*exp(-x^3*log(x)+x^4+2*x^2+x+3 )-2)/(exp(-x^3*log(x)+x^4+2*x^2+x+3)^2-2*x*exp(-x^3*log(x)+x^4+2*x^2+x+3)+ x^2),x, algorithm="maxima")
Output:
2*x^(x^3)/(x*x^(x^3) - e^(x^4 + 2*x^2 + x + 3))
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=\frac {2}{x - e^{\left (x^{4} - x^{3} \log \left (x\right ) + 2 \, x^{2} + x + 3\right )}} \] Input:
integrate(((-6*x^2*log(x)+8*x^3-2*x^2+8*x+2)*exp(-x^3*log(x)+x^4+2*x^2+x+3 )-2)/(exp(-x^3*log(x)+x^4+2*x^2+x+3)^2-2*x*exp(-x^3*log(x)+x^4+2*x^2+x+3)+ x^2),x, algorithm="giac")
Output:
2/(x - e^(x^4 - x^3*log(x) + 2*x^2 + x + 3))
Time = 3.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=\frac {2}{x-\frac {{\mathrm {e}}^{x^4}\,{\mathrm {e}}^3\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^x}{x^{x^3}}} \] Input:
int((exp(x - x^3*log(x) + 2*x^2 + x^4 + 3)*(8*x - 6*x^2*log(x) - 2*x^2 + 8 *x^3 + 2) - 2)/(exp(2*x - 2*x^3*log(x) + 4*x^2 + 2*x^4 + 6) - 2*x*exp(x - x^3*log(x) + 2*x^2 + x^4 + 3) + x^2),x)
Output:
2/(x - (exp(x^4)*exp(3)*exp(2*x^2)*exp(x))/x^(x^3))
Time = 1.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-2+e^{3+x+2 x^2+x^4-x^3 \log (x)} \left (2+8 x-2 x^2+8 x^3-6 x^2 \log (x)\right )}{e^{6+2 x+4 x^2+2 x^4-2 x^3 \log (x)}-2 e^{3+x+2 x^2+x^4-x^3 \log (x)} x+x^2} \, dx=-\frac {2 x^{x^{3}}}{e^{x^{4}+2 x^{2}+x} e^{3}-x^{x^{3}} x} \] Input:
int(((-6*x^2*log(x)+8*x^3-2*x^2+8*x+2)*exp(-x^3*log(x)+x^4+2*x^2+x+3)-2)/( exp(-x^3*log(x)+x^4+2*x^2+x+3)^2-2*x*exp(-x^3*log(x)+x^4+2*x^2+x+3)+x^2),x )
Output:
( - 2*x**(x**3))/(e**(x**4 + 2*x**2 + x)*e**3 - x**(x**3)*x)