Integrand size = 93, antiderivative size = 29 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=\log \left (\frac {x^2 \log (5)}{5 e^x-\frac {2}{\left (e^4-x\right ) x}}\right ) \] Output:
ln(x^2*ln(5)/(5*exp(x)-2/x/(exp(2)^2-x)))
Time = 4.98 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=2 \log (x)+\log \left (x \left (-e^4+x\right )\right )-\log \left (2-5 e^{4+x} x+5 e^x x^2\right ) \] Input:
Integrate[(-6*E^4 + 8*x + E^x*(10*x^3 - 5*x^4 + E^8*(10*x - 5*x^2) + E^4*( -20*x^2 + 10*x^3)))/(-2*E^4*x + 2*x^2 + E^x*(5*E^8*x^2 - 10*E^4*x^3 + 5*x^ 4)),x]
Output:
2*Log[x] + Log[x*(-E^4 + x)] - Log[2 - 5*E^(4 + x)*x + 5*E^x*x^2]
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.
\(\displaystyle \int \frac {e^x \left (-5 x^4+10 x^3+e^8 \left (10 x-5 x^2\right )+e^4 \left (10 x^3-20 x^2\right )\right )+8 x-6 e^4}{2 x^2+e^x \left (5 x^4-10 e^4 x^3+5 e^8 x^2\right )-2 e^4 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^x \left (-5 x^4+10 x^3+e^8 \left (10 x-5 x^2\right )+e^4 \left (10 x^3-20 x^2\right )\right )-8 x+6 e^4}{\left (e^4-x\right ) x \left (5 e^x x^2-5 e^{x+4} x+2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (-x^2-\left (2-e^4\right ) x+e^4\right )}{\left (e^4-x\right ) x \left (5 e^x x^2-5 e^{x+4} x+2\right )}+\frac {2-x}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{-5 e^x x^2+5 e^{x+4} x-2}dx+2 \int \frac {1}{\left (e^4-x\right ) \left (-5 e^x x^2+5 e^{x+4} x-2\right )}dx+2 \int \frac {1}{x \left (5 e^x x^2-5 e^{x+4} x+2\right )}dx-x+2 \log (x)\) |
Input:
Int[(-6*E^4 + 8*x + E^x*(10*x^3 - 5*x^4 + E^8*(10*x - 5*x^2) + E^4*(-20*x^ 2 + 10*x^3)))/(-2*E^4*x + 2*x^2 + E^x*(5*E^8*x^2 - 10*E^4*x^3 + 5*x^4)),x]
Output:
$Aborted
Time = 0.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
risch | \(2 \ln \left (x \right )-\ln \left ({\mathrm e}^{x}-\frac {2}{5 x \left ({\mathrm e}^{4}-x \right )}\right )\) | \(25\) |
parallelrisch | \(3 \ln \left (x \right )+\ln \left (x -{\mathrm e}^{4}\right )-\ln \left (-{\mathrm e}^{4} {\mathrm e}^{x} x +{\mathrm e}^{x} x^{2}+\frac {2}{5}\right )\) | \(35\) |
norman | \(3 \ln \left (x \right )-\ln \left (5 \,{\mathrm e}^{4} {\mathrm e}^{x} x -5 \,{\mathrm e}^{x} x^{2}-2\right )+\ln \left ({\mathrm e}^{4}-x \right )\) | \(36\) |
Input:
int((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)*exp(x) -6*exp(2)^2+8*x)/((5*x^2*exp(2)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x*exp(2) ^2+2*x^2),x,method=_RETURNVERBOSE)
Output:
2*ln(x)-ln(exp(x)-2/5/x/(exp(4)-x))
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=2 \, \log \left (x\right ) - \log \left (-\frac {5 \, {\left (x^{2} - x e^{4}\right )} e^{x} + 2}{x^{2} - x e^{4}}\right ) \] Input:
integrate((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)* exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x* exp(2)^2+2*x^2),x, algorithm="fricas")
Output:
2*log(x) - log(-(5*(x^2 - x*e^4)*e^x + 2)/(x^2 - x*e^4))
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=2 \log {\left (x \right )} - \log {\left (e^{x} + \frac {2}{5 x^{2} - 5 x e^{4}} \right )} \] Input:
integrate((((-5*x**2+10*x)*exp(2)**4+(10*x**3-20*x**2)*exp(2)**2-5*x**4+10 *x**3)*exp(x)-6*exp(2)**2+8*x)/((5*x**2*exp(2)**4-10*x**3*exp(2)**2+5*x**4 )*exp(x)-2*x*exp(2)**2+2*x**2),x)
Output:
2*log(x) - log(exp(x) + 2/(5*x**2 - 5*x*exp(4)))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=2 \, \log \left (x\right ) - \log \left (\frac {5 \, {\left (x^{2} - x e^{4}\right )} e^{x} + 2}{5 \, {\left (x^{2} - x e^{4}\right )}}\right ) \] Input:
integrate((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)* exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x* exp(2)^2+2*x^2),x, algorithm="maxima")
Output:
2*log(x) - log(1/5*(5*(x^2 - x*e^4)*e^x + 2)/(x^2 - x*e^4))
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=-\log \left (-5 \, x^{2} e^{x} + 5 \, x e^{\left (x + 4\right )} - 2\right ) + \log \left (x - e^{4}\right ) + 3 \, \log \left (x\right ) \] Input:
integrate((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)* exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x* exp(2)^2+2*x^2),x, algorithm="giac")
Output:
-log(-5*x^2*e^x + 5*x*e^(x + 4) - 2) + log(x - e^4) + 3*log(x)
Time = 1.82 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=\ln \left (x-{\mathrm {e}}^4\right )+3\,\ln \left (x\right )-\ln \left (5\,x^2\,{\mathrm {e}}^x-5\,x\,{\mathrm {e}}^4\,{\mathrm {e}}^x+2\right ) \] Input:
int((8*x - 6*exp(4) + exp(x)*(exp(8)*(10*x - 5*x^2) - exp(4)*(20*x^2 - 10* x^3) + 10*x^3 - 5*x^4))/(exp(x)*(5*x^2*exp(8) - 10*x^3*exp(4) + 5*x^4) - 2 *x*exp(4) + 2*x^2),x)
Output:
log(x - exp(4)) + 3*log(x) - log(5*x^2*exp(x) - 5*x*exp(4)*exp(x) + 2)
Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx=\mathrm {log}\left (e^{4}-x \right )-\mathrm {log}\left (5 e^{x} e^{4} x -5 e^{x} x^{2}-2\right )+3 \,\mathrm {log}\left (x \right ) \] Input:
int((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)*exp(x) -6*exp(2)^2+8*x)/((5*x^2*exp(2)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x*exp(2) ^2+2*x^2),x)
Output:
log(e**4 - x) - log(5*e**x*e**4*x - 5*e**x*x**2 - 2) + 3*log(x)