Integrand size = 117, antiderivative size = 31 \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=5-\frac {4}{-5+e^{e^{18-2 e^x-2 x}-x (-4+\log (3))}+x} \] Output:
5-4/(exp(exp(-exp(x)+9-x)^2-x*(-4+ln(3)))-5+x)
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=-\frac {4\ 3^x}{e^{e^{-2 \left (-9+e^x+x\right )}+4 x}+3^x (-5+x)} \] Input:
Integrate[(4 + E^(E^(18 - 2*E^x - 2*x) + 4*x - x*Log[3])*(16 + E^(18 - 2*E ^x - 2*x)*(-8 - 8*E^x) - 4*Log[3]))/(25 + E^(2*E^(18 - 2*E^x - 2*x) + 8*x - 2*x*Log[3]) - 10*x + x^2 + E^(E^(18 - 2*E^x - 2*x) + 4*x - x*Log[3])*(-1 0 + 2*x)),x]
Output:
(-4*3^x)/(E^(E^(-2*(-9 + E^x + x)) + 4*x) + 3^x*(-5 + 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.
| \(\displaystyle \int \frac {e^{4 x+e^{-2 x-2 e^x+18}+x (-\log (3))} \left (e^{-2 x-2 e^x+18} \left (-8 e^x-8\right )+16-4 \log (3)\right )+4}{\exp \left (8 x+2 e^{-2 x-2 e^x+18}-2 x \log (3)\right )+x^2-10 x+(2 x-10) e^{4 x+e^{-2 x-2 e^x+18}+x (-\log (3))}+25} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {3^{2 x} \left (e^{4 x+e^{-2 x-2 e^x+18}+x (-\log (3))} \left (e^{-2 x-2 e^x+18} \left (-8 e^x-8\right )+16-4 \log (3)\right )+4\right )}{\left (-3^x x+5\ 3^x-e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {4\ 3^{2 x}}{\left (3^x x-5\ 3^x+e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}+\frac {4\ 3^x e^{2 x+e^{-2 x-2 e^x+18}-2 e^x} \left (-2 e^{x+18}+4 e^{2 x+2 e^x} \left (1-\frac {\log (3)}{4}\right )-2 e^{18}\right )}{\left (-3^x x+5\ 3^x-e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 (4-\log (3)) \int \frac {3^x \exp \left (e^{-2 \left (x+e^x\right )} \left (4 e^{2 x+2 e^x} x+e^{18}\right )\right )}{\left (-3^x x+5\ 3^x-e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}dx+4 \int \frac {3^{2 x}}{\left (3^x x-5\ 3^x+e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}dx-8 \int \frac {3^x e^{2 x+e^{-2 x-2 e^x+18}-2 e^x+18}}{\left (3^x x-5\ 3^x+e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}dx-8 \int \frac {3^x e^{3 x+e^{-2 x-2 e^x+18}-2 e^x+18}}{\left (3^x x-5\ 3^x+e^{4 x+e^{-2 x-2 e^x+18}}\right )^2}dx\) |
Input:
Int[(4 + E^(E^(18 - 2*E^x - 2*x) + 4*x - x*Log[3])*(16 + E^(18 - 2*E^x - 2 *x)*(-8 - 8*E^x) - 4*Log[3]))/(25 + E^(2*E^(18 - 2*E^x - 2*x) + 8*x - 2*x* Log[3]) - 10*x + x^2 + E^(E^(18 - 2*E^x - 2*x) + 4*x - x*Log[3])*(-10 + 2* x)),x]
Output:
$Aborted
Time = 1.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
| method | result | size |
| risch | \(-\frac {4}{-5+\left (\frac {1}{3}\right )^{x} {\mathrm e}^{{\mathrm e}^{-2 \,{\mathrm e}^{x}+18-2 x}+4 x}+x}\) | \(27\) |
| parallelrisch | \(-\frac {4}{-5+{\mathrm e}^{{\mathrm e}^{-2 \,{\mathrm e}^{x}+18-2 x}-x \ln \left (3\right )+4 x}+x}\) | \(30\) |
Input:
int((((-8*exp(x)-8)*exp(-exp(x)+9-x)^2-4*ln(3)+16)*exp(exp(-exp(x)+9-x)^2- x*ln(3)+4*x)+4)/(exp(exp(-exp(x)+9-x)^2-x*ln(3)+4*x)^2+(2*x-10)*exp(exp(-e xp(x)+9-x)^2-x*ln(3)+4*x)+x^2-10*x+25),x,method=_RETURNVERBOSE)
Output:
-4/(-5+(1/3)^x*exp(exp(-2*exp(x)+18-2*x)+4*x)+x)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=-\frac {4}{x + e^{\left (-x \log \left (3\right ) + 4 \, x + e^{\left (-2 \, x - 2 \, e^{x} + 18\right )}\right )} - 5} \] Input:
integrate((((-8*exp(x)-8)*exp(-exp(x)+9-x)^2-4*log(3)+16)*exp(exp(-exp(x)+ 9-x)^2-x*log(3)+4*x)+4)/(exp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)^2+(2*x-10)*e xp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)+x^2-10*x+25),x, algorithm="fricas")
Output:
-4/(x + e^(-x*log(3) + 4*x + e^(-2*x - 2*e^x + 18)) - 5)
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=- \frac {4}{x + e^{- x \log {\left (3 \right )} + 4 x + e^{- 2 x - 2 e^{x} + 18}} - 5} \] Input:
integrate((((-8*exp(x)-8)*exp(-exp(x)+9-x)**2-4*ln(3)+16)*exp(exp(-exp(x)+ 9-x)**2-x*ln(3)+4*x)+4)/(exp(exp(-exp(x)+9-x)**2-x*ln(3)+4*x)**2+(2*x-10)* exp(exp(-exp(x)+9-x)**2-x*ln(3)+4*x)+x**2-10*x+25),x)
Output:
-4/(x + exp(-x*log(3) + 4*x + exp(-2*x - 2*exp(x) + 18)) - 5)
Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=-\frac {4 \cdot 3^{x}}{3^{x} {\left (x - 5\right )} + e^{\left (4 \, x + e^{\left (-2 \, x - 2 \, e^{x} + 18\right )}\right )}} \] Input:
integrate((((-8*exp(x)-8)*exp(-exp(x)+9-x)^2-4*log(3)+16)*exp(exp(-exp(x)+ 9-x)^2-x*log(3)+4*x)+4)/(exp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)^2+(2*x-10)*e xp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)+x^2-10*x+25),x, algorithm="maxima")
Output:
-4*3^x/(3^x*(x - 5) + e^(4*x + e^(-2*x - 2*e^x + 18)))
Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=-\frac {4}{x + e^{\left (-x \log \left (3\right ) + 4 \, x + e^{\left (-2 \, x - 2 \, e^{x} + 18\right )}\right )} - 5} \] Input:
integrate((((-8*exp(x)-8)*exp(-exp(x)+9-x)^2-4*log(3)+16)*exp(exp(-exp(x)+ 9-x)^2-x*log(3)+4*x)+4)/(exp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)^2+(2*x-10)*e xp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)+x^2-10*x+25),x, algorithm="giac")
Output:
-4/(x + e^(-x*log(3) + 4*x + e^(-2*x - 2*e^x + 18)) - 5)
Timed out. \[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=\int -\frac {{\mathrm {e}}^{4\,x+{\mathrm {e}}^{18-2\,{\mathrm {e}}^x-2\,x}-x\,\ln \left (3\right )}\,\left (4\,\ln \left (3\right )+{\mathrm {e}}^{18-2\,{\mathrm {e}}^x-2\,x}\,\left (8\,{\mathrm {e}}^x+8\right )-16\right )-4}{{\mathrm {e}}^{8\,x+2\,{\mathrm {e}}^{18-2\,{\mathrm {e}}^x-2\,x}-2\,x\,\ln \left (3\right )}-10\,x+{\mathrm {e}}^{4\,x+{\mathrm {e}}^{18-2\,{\mathrm {e}}^x-2\,x}-x\,\ln \left (3\right )}\,\left (2\,x-10\right )+x^2+25} \,d x \] Input:
int(-(exp(4*x + exp(18 - 2*exp(x) - 2*x) - x*log(3))*(4*log(3) + exp(18 - 2*exp(x) - 2*x)*(8*exp(x) + 8) - 16) - 4)/(exp(8*x + 2*exp(18 - 2*exp(x) - 2*x) - 2*x*log(3)) - 10*x + exp(4*x + exp(18 - 2*exp(x) - 2*x) - x*log(3) )*(2*x - 10) + x^2 + 25),x)
Output:
int(-(exp(4*x + exp(18 - 2*exp(x) - 2*x) - x*log(3))*(4*log(3) + exp(18 - 2*exp(x) - 2*x)*(8*exp(x) + 8) - 16) - 4)/(exp(8*x + 2*exp(18 - 2*exp(x) - 2*x) - 2*x*log(3)) - 10*x + exp(4*x + exp(18 - 2*exp(x) - 2*x) - x*log(3) )*(2*x - 10) + x^2 + 25), x)
\[ \int \frac {4+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} \left (16+e^{18-2 e^x-2 x} \left (-8-8 e^x\right )-4 \log (3)\right )}{25+e^{2 e^{18-2 e^x-2 x}+8 x-2 x \log (3)}-10 x+x^2+e^{e^{18-2 e^x-2 x}+4 x-x \log (3)} (-10+2 x)} \, dx=4 \left (\int \frac {3^{2 x}}{e^{\frac {8 e^{2 e^{x}+2 x} x +2 e^{18}}{e^{2 e^{x}+2 x}}}+2 e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x} x -10 e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}+3^{2 x} x^{2}-10 \,3^{2 x} x +25 \,3^{2 x}}d x \right )-4 \left (\int \frac {e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}}{e^{\frac {8 e^{2 e^{x}+2 x} x +2 e^{18}}{e^{2 e^{x}+2 x}}}+2 e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x} x -10 e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}+3^{2 x} x^{2}-10 \,3^{2 x} x +25 \,3^{2 x}}d x \right ) \mathrm {log}\left (3\right )+16 \left (\int \frac {e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}}{e^{\frac {8 e^{2 e^{x}+2 x} x +2 e^{18}}{e^{2 e^{x}+2 x}}}+2 e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x} x -10 e^{\frac {4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}+3^{2 x} x^{2}-10 \,3^{2 x} x +25 \,3^{2 x}}d x \right )-8 \left (\int \frac {e^{\frac {3 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}}{e^{\frac {2 e^{2 e^{x}+3 x}+8 e^{2 e^{x}+2 x} x +2 e^{18}}{e^{2 e^{x}+2 x}}}+2 e^{\frac {2 e^{2 e^{x}+3 x}+4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x} x -10 e^{\frac {2 e^{2 e^{x}+3 x}+4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}+e^{2 e^{x}} 3^{2 x} x^{2}-10 e^{2 e^{x}} 3^{2 x} x +25 e^{2 e^{x}} 3^{2 x}}d x \right ) e^{18}-8 \left (\int \frac {e^{\frac {2 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}}{e^{\frac {2 e^{2 e^{x}+3 x}+8 e^{2 e^{x}+2 x} x +2 e^{18}}{e^{2 e^{x}+2 x}}}+2 e^{\frac {2 e^{2 e^{x}+3 x}+4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x} x -10 e^{\frac {2 e^{2 e^{x}+3 x}+4 e^{2 e^{x}+2 x} x +e^{18}}{e^{2 e^{x}+2 x}}} 3^{x}+e^{2 e^{x}} 3^{2 x} x^{2}-10 e^{2 e^{x}} 3^{2 x} x +25 e^{2 e^{x}} 3^{2 x}}d x \right ) e^{18} \] Input:
int((((-8*exp(x)-8)*exp(-exp(x)+9-x)^2-4*log(3)+16)*exp(exp(-exp(x)+9-x)^2 -x*log(3)+4*x)+4)/(exp(exp(-exp(x)+9-x)^2-x*log(3)+4*x)^2+(2*x-10)*exp(exp (-exp(x)+9-x)^2-x*log(3)+4*x)+x^2-10*x+25),x)
Output:
4*(int(3**(2*x)/(e**((8*e**(2*e**x + 2*x)*x + 2*e**18)/e**(2*e**x + 2*x)) + 2*e**((4*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x*x - 10*e** ((4*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x + 3**(2*x)*x**2 - 10*3**(2*x)*x + 25*3**(2*x)),x) - int((e**((4*e**(2*e**x + 2*x)*x + e**18 )/e**(2*e**x + 2*x))*3**x)/(e**((8*e**(2*e**x + 2*x)*x + 2*e**18)/e**(2*e* *x + 2*x)) + 2*e**((4*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x *x - 10*e**((4*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x + 3**( 2*x)*x**2 - 10*3**(2*x)*x + 25*3**(2*x)),x)*log(3) + 4*int((e**((4*e**(2*e **x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x)/(e**((8*e**(2*e**x + 2*x)*x + 2*e**18)/e**(2*e**x + 2*x)) + 2*e**((4*e**(2*e**x + 2*x)*x + e**18)/e** (2*e**x + 2*x))*3**x*x - 10*e**((4*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x + 3**(2*x)*x**2 - 10*3**(2*x)*x + 25*3**(2*x)),x) - 2*int((e **((3*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x)/(e**((2*e**(2* e**x + 3*x) + 8*e**(2*e**x + 2*x)*x + 2*e**18)/e**(2*e**x + 2*x)) + 2*e**( (2*e**(2*e**x + 3*x) + 4*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3 **x*x - 10*e**((2*e**(2*e**x + 3*x) + 4*e**(2*e**x + 2*x)*x + e**18)/e**(2 *e**x + 2*x))*3**x + e**(2*e**x)*3**(2*x)*x**2 - 10*e**(2*e**x)*3**(2*x)*x + 25*e**(2*e**x)*3**(2*x)),x)*e**18 - 2*int((e**((2*e**(2*e**x + 2*x)*x + e**18)/e**(2*e**x + 2*x))*3**x)/(e**((2*e**(2*e**x + 3*x) + 8*e**(2*e**x + 2*x)*x + 2*e**18)/e**(2*e**x + 2*x)) + 2*e**((2*e**(2*e**x + 3*x) + 4...