Integrand size = 147, antiderivative size = 23 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=x \log \left (-1+e^{\frac {5 x}{-2 x+e^2 x^5}}\right ) \] Output:
ln(exp(5/(exp(2)*x^5-2*x)*x)-1)*x
Time = 0.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=x \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \] Input:
Integrate[(-20*E^(2 + 5/(-2 + E^2*x^4))*x^4 + (-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4 + E^4*x^8))*Log[-1 + E^(5/(-2 + E^2*x^ 4))])/(-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4 + E^ 4*x^8)),x]
Output:
x*Log[-1 + E^(5/(-2 + E^2*x^4))]
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.
\(\displaystyle \int \frac {\left (-e^4 x^8+4 e^2 x^4+e^{\frac {5}{e^2 x^4-2}} \left (e^4 x^8-4 e^2 x^4+4\right )-4\right ) \log \left (e^{\frac {5}{e^2 x^4-2}}-1\right )-20 e^{\frac {5}{e^2 x^4-2}+2} x^4}{-e^4 x^8+4 e^2 x^4+e^{\frac {5}{e^2 x^4-2}} \left (e^4 x^8-4 e^2 x^4+4\right )-4} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\log \left (e^{\frac {5}{e^2 x^4-2}}-1\right )-\frac {20 e^{\frac {5}{e^2 x^4-2}+2} x^4}{\left (e^{\frac {5}{e^2 x^4-2}}-1\right ) \left (e^2 x^4-2\right )^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\log \left (e^{\frac {5}{e^2 x^4-2}}-1\right )-\frac {20 e^{\frac {5}{e^2 x^4-2}+2} x^4}{\left (e^{\frac {5}{e^2 x^4-2}}-1\right ) \left (e^2 x^4-2\right )^2}\right )dx\) |
Input:
Int[(-20*E^(2 + 5/(-2 + E^2*x^4))*x^4 + (-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/( -2 + E^2*x^4))*(4 - 4*E^2*x^4 + E^4*x^8))*Log[-1 + E^(5/(-2 + E^2*x^4))])/ (-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4 + E^4*x^8) ),x]
Output:
$Aborted
Time = 59.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
risch | \(x \ln \left ({\mathrm e}^{\frac {5}{x^{4} {\mathrm e}^{2}-2}}-1\right )\) | \(19\) |
parallelrisch | \(x \ln \left ({\mathrm e}^{\frac {5}{x^{4} {\mathrm e}^{2}-2}}-1\right )\) | \(19\) |
Input:
int((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x ^4*exp(2)-4)*ln(exp(5/(x^4*exp(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*exp(2)-2 )))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x^ 4*exp(2)-4),x,method=_RETURNVERBOSE)
Output:
x*ln(exp(5/(x^4*exp(2)-2))-1)
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=x \log \left (-{\left (e^{2} - e^{\left (\frac {2 \, x^{4} e^{2} + 1}{x^{4} e^{2} - 2}\right )}\right )} e^{\left (-2\right )}\right ) \] Input:
integrate((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2) ^2+4*x^4*exp(2)-4)*log(exp(5/(x^4*exp(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*e xp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2) ^2+4*x^4*exp(2)-4),x, algorithm="fricas")
Output:
x*log(-(e^2 - e^((2*x^4*e^2 + 1)/(x^4*e^2 - 2)))*e^(-2))
Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=x \log {\left (e^{\frac {5}{x^{4} e^{2} - 2}} - 1 \right )} \] Input:
integrate((((x**8*exp(2)**2-4*x**4*exp(2)+4)*exp(5/(x**4*exp(2)-2))-x**8*e xp(2)**2+4*x**4*exp(2)-4)*ln(exp(5/(x**4*exp(2)-2))-1)-20*x**4*exp(2)*exp( 5/(x**4*exp(2)-2)))/((x**8*exp(2)**2-4*x**4*exp(2)+4)*exp(5/(x**4*exp(2)-2 ))-x**8*exp(2)**2+4*x**4*exp(2)-4),x)
Output:
x*log(exp(5/(x**4*exp(2) - 2)) - 1)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=x \log \left (e^{\left (\frac {4}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {3}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {2}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {1}{x^{4} e^{2} - 2}\right )} + 1\right ) + x \log \left (e^{\left (\frac {1}{x^{4} e^{2} - 2}\right )} - 1\right ) \] Input:
integrate((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2) ^2+4*x^4*exp(2)-4)*log(exp(5/(x^4*exp(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*e xp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2) ^2+4*x^4*exp(2)-4),x, algorithm="maxima")
Output:
x*log(e^(4/(x^4*e^2 - 2)) + e^(3/(x^4*e^2 - 2)) + e^(2/(x^4*e^2 - 2)) + e^ (1/(x^4*e^2 - 2)) + 1) + x*log(e^(1/(x^4*e^2 - 2)) - 1)
Exception generated. \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=\text {Exception raised: AttributeError} \] Input:
integrate((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2) ^2+4*x^4*exp(2)-4)*log(exp(5/(x^4*exp(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*e xp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2) ^2+4*x^4*exp(2)-4),x, algorithm="giac")
Output:
Exception raised: AttributeError >> type
Time = 0.54 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=x\,\ln \left ({\mathrm {e}}^{\frac {5}{x^4\,{\mathrm {e}}^2-2}}-1\right ) \] Input:
int((log(exp(5/(x^4*exp(2) - 2)) - 1)*(exp(5/(x^4*exp(2) - 2))*(x^8*exp(4) - 4*x^4*exp(2) + 4) + 4*x^4*exp(2) - x^8*exp(4) - 4) - 20*x^4*exp(5/(x^4* exp(2) - 2))*exp(2))/(exp(5/(x^4*exp(2) - 2))*(x^8*exp(4) - 4*x^4*exp(2) + 4) + 4*x^4*exp(2) - x^8*exp(4) - 4),x)
Output:
x*log(exp(5/(x^4*exp(2) - 2)) - 1)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+\left (-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )\right ) \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} \left (4-4 e^2 x^4+e^4 x^8\right )} \, dx=\mathrm {log}\left (e^{\frac {5}{e^{2} x^{4}-2}}-1\right ) x \] Input:
int((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x ^4*exp(2)-4)*log(exp(5/(x^4*exp(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*exp(2)- 2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x ^4*exp(2)-4),x)
Output:
log(e**(5/(e**2*x**4 - 2)) - 1)*x