Integrand size = 92, antiderivative size = 25 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {5-x+\log (3)}{1+e^{e^x+x+x^4}-x} \] Output:
(ln(3)-x+5)/(exp(exp(x)+x^4+x)-x+1)
Time = 0.61 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {5-x+\log (3)}{1+e^{e^x+x+x^4}-x} \] Input:
Integrate[(4 + Log[3] + E^(E^x + x + x^4)*(-6 + x - 20*x^3 + 4*x^4 + E^x*( -5 + x - Log[3]) + (-1 - 4*x^3)*Log[3]))/(1 + E^(2*E^x + 2*x + 2*x^4) + E^ (E^x + x + x^4)*(2 - 2*x) - 2*x + x^2),x]
Output:
(5 - x + Log[3])/(1 + E^(E^x + x + x^4) - 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.
\(\displaystyle \int \frac {e^{x^4+x+e^x} \left (4 x^4-20 x^3+\left (-4 x^3-1\right ) \log (3)+x+e^x (x-5-\log (3))-6\right )+4+\log (3)}{e^{2 x^4+2 x+2 e^x}+e^{x^4+x+e^x} (2-2 x)+x^2-2 x+1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x^4+x+e^x} \left (4 x^4-20 x^3+\left (-4 x^3-1\right ) \log (3)+x+e^x (x-5-\log (3))-6\right )+4 \left (1+\frac {\log (3)}{4}\right )}{\left (e^{x^4+x+e^x}-x+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 x^4-20 x^3 \left (1+\frac {\log (3)}{5}\right )+e^x x+x-5 e^x \left (1+\frac {\log (3)}{5}\right )-6 \left (1+\frac {\log (3)}{6}\right )}{e^{x^4+x+e^x}-x+1}+\frac {\left (4 x^4-4 x^3+e^x x+x-e^x-2\right ) (x-5-\log (3))}{\left (e^{x^4+x+e^x}-x+1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {x}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx-\int \frac {e^x x}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx+\int \frac {x}{-x+e^{x^4+x+e^x}+1}dx+\int \frac {e^x x}{-x+e^{x^4+x+e^x}+1}dx-4 \int \frac {x^4}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx+4 \int \frac {x^4}{-x+e^{x^4+x+e^x}+1}dx+2 (5+\log (3)) \int \frac {1}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx+(5+\log (3)) \int \frac {e^x}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx-(6+\log (3)) \int \frac {1}{-x+e^{x^4+x+e^x}+1}dx-(5+\log (3)) \int \frac {e^x}{-x+e^{x^4+x+e^x}+1}dx-(5+\log (3)) \int \frac {x}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx-(5+\log (3)) \int \frac {e^x x}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx-4 (5+\log (3)) \int \frac {x^4}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx+4 \int \frac {x^5}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx+4 (5+\log (3)) \int \frac {x^3}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx-4 (5+\log (3)) \int \frac {x^3}{-x+e^{x^4+x+e^x}+1}dx+\int \frac {x^2}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx+\int \frac {e^x x^2}{\left (-x+e^{x^4+x+e^x}+1\right )^2}dx\) |
Input:
Int[(4 + Log[3] + E^(E^x + x + x^4)*(-6 + x - 20*x^3 + 4*x^4 + E^x*(-5 + x - Log[3]) + (-1 - 4*x^3)*Log[3]))/(1 + E^(2*E^x + 2*x + 2*x^4) + E^(E^x + x + x^4)*(2 - 2*x) - 2*x + x^2),x]
Output:
$Aborted
Time = 1.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {\ln \left (3\right )-x +5}{x -{\mathrm e}^{{\mathrm e}^{x}+x^{4}+x}-1}\) | \(25\) |
parallelrisch | \(-\frac {\ln \left (3\right )-x +5}{x -{\mathrm e}^{{\mathrm e}^{x}+x^{4}+x}-1}\) | \(25\) |
norman | \(\frac {{\mathrm e}^{{\mathrm e}^{x}+x^{4}+x}-4-\ln \left (3\right )}{x -{\mathrm e}^{{\mathrm e}^{x}+x^{4}+x}-1}\) | \(31\) |
Input:
int((((-ln(3)+x-5)*exp(x)+(-4*x^3-1)*ln(3)+4*x^4-20*x^3+x-6)*exp(exp(x)+x^ 4+x)+4+ln(3))/(exp(exp(x)+x^4+x)^2+(2-2*x)*exp(exp(x)+x^4+x)+x^2-2*x+1),x, method=_RETURNVERBOSE)
Output:
-(ln(3)-x+5)/(x-exp(exp(x)+x^4+x)-1)
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {x - \log \left (3\right ) - 5}{x - e^{\left (x^{4} + x + e^{x}\right )} - 1} \] Input:
integrate((((-log(3)+x-5)*exp(x)+(-4*x^3-1)*log(3)+4*x^4-20*x^3+x-6)*exp(e xp(x)+x^4+x)+4+log(3))/(exp(exp(x)+x^4+x)^2+(2-2*x)*exp(exp(x)+x^4+x)+x^2- 2*x+1),x, algorithm="fricas")
Output:
(x - log(3) - 5)/(x - e^(x^4 + x + e^x) - 1)
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {- x + \log {\left (3 \right )} + 5}{- x + e^{x^{4} + x + e^{x}} + 1} \] Input:
integrate((((-ln(3)+x-5)*exp(x)+(-4*x**3-1)*ln(3)+4*x**4-20*x**3+x-6)*exp( exp(x)+x**4+x)+4+ln(3))/(exp(exp(x)+x**4+x)**2+(2-2*x)*exp(exp(x)+x**4+x)+ x**2-2*x+1),x)
Output:
(-x + log(3) + 5)/(-x + exp(x**4 + x + exp(x)) + 1)
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {x - \log \left (3\right ) - 5}{x - e^{\left (x^{4} + x + e^{x}\right )} - 1} \] Input:
integrate((((-log(3)+x-5)*exp(x)+(-4*x^3-1)*log(3)+4*x^4-20*x^3+x-6)*exp(e xp(x)+x^4+x)+4+log(3))/(exp(exp(x)+x^4+x)^2+(2-2*x)*exp(exp(x)+x^4+x)+x^2- 2*x+1),x, algorithm="maxima")
Output:
(x - log(3) - 5)/(x - e^(x^4 + x + e^x) - 1)
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {x - \log \left (3\right ) - 5}{x - e^{\left (x^{4} + x + e^{x}\right )} - 1} \] Input:
integrate((((-log(3)+x-5)*exp(x)+(-4*x^3-1)*log(3)+4*x^4-20*x^3+x-6)*exp(e xp(x)+x^4+x)+4+log(3))/(exp(exp(x)+x^4+x)^2+(2-2*x)*exp(exp(x)+x^4+x)+x^2- 2*x+1),x, algorithm="giac")
Output:
(x - log(3) - 5)/(x - e^(x^4 + x + e^x) - 1)
Time = 2.85 (sec) , antiderivative size = 113, normalized size of antiderivative = 4.52 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=-\frac {2\,\ln \left (3\right )-7\,x+5\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^x-x\,\ln \left (3\right )+4\,x^3\,\ln \left (3\right )-4\,x^4\,\ln \left (3\right )+{\mathrm {e}}^x\,\ln \left (3\right )-6\,x\,{\mathrm {e}}^x+x^2+20\,x^3-24\,x^4+4\,x^5-x\,{\mathrm {e}}^x\,\ln \left (3\right )+10}{\left ({\mathrm {e}}^{x+{\mathrm {e}}^x+x^4}-x+1\right )\,\left (x-{\mathrm {e}}^x+x\,{\mathrm {e}}^x-4\,x^3+4\,x^4-2\right )} \] Input:
int((log(3) - exp(x + exp(x) + x^4)*(log(3)*(4*x^3 + 1) - x + 20*x^3 - 4*x ^4 + exp(x)*(log(3) - x + 5) + 6) + 4)/(exp(2*x + 2*exp(x) + 2*x^4) - 2*x - exp(x + exp(x) + x^4)*(2*x - 2) + x^2 + 1),x)
Output:
-(2*log(3) - 7*x + 5*exp(x) + x^2*exp(x) - x*log(3) + 4*x^3*log(3) - 4*x^4 *log(3) + exp(x)*log(3) - 6*x*exp(x) + x^2 + 20*x^3 - 24*x^4 + 4*x^5 - x*e xp(x)*log(3) + 10)/((exp(x + exp(x) + x^4) - x + 1)*(x - exp(x) + x*exp(x) - 4*x^3 + 4*x^4 - 2))
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {4+\log (3)+e^{e^x+x+x^4} \left (-6+x-20 x^3+4 x^4+e^x (-5+x-\log (3))+\left (-1-4 x^3\right ) \log (3)\right )}{1+e^{2 e^x+2 x+2 x^4}+e^{e^x+x+x^4} (2-2 x)-2 x+x^2} \, dx=\frac {-e^{e^{x}+x^{4}+x} \mathrm {log}\left (3\right )-5 e^{e^{x}+x^{4}+x}+\mathrm {log}\left (3\right ) x +4 x}{e^{e^{x}+x^{4}+x}-x +1} \] Input:
int((((-log(3)+x-5)*exp(x)+(-4*x^3-1)*log(3)+4*x^4-20*x^3+x-6)*exp(exp(x)+ x^4+x)+4+log(3))/(exp(exp(x)+x^4+x)^2+(2-2*x)*exp(exp(x)+x^4+x)+x^2-2*x+1) ,x)
Output:
( - e**(e**x + x**4 + x)*log(3) - 5*e**(e**x + x**4 + x) + log(3)*x + 4*x) /(e**(e**x + x**4 + x) - x + 1)