Integrand size = 102, antiderivative size = 31 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{-2+\frac {-1+e^2-e^4-\log (3)}{5+e^{4 x}-x}} \] Output:
exp((-ln(3)-exp(4)+exp(2)-1)/(exp(4*x)+5-x)-2)
Time = 2.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=3^{-\frac {1}{5+e^{4 x}-x}} e^{-2+\frac {-1+e^2-e^4}{5+e^{4 x}-x}} \] Input:
Integrate[(E^((-11 + E^2 - E^4 - 2*E^(4*x) + 2*x - Log[3])/(5 + E^(4*x) - x))*(-1 + E^2 - E^4 - Log[3] + E^(4*x)*(4 - 4*E^2 + 4*E^4 + 4*Log[3])))/(2 5 + E^(8*x) + E^(4*x)*(10 - 2*x) - 10*x + x^2),x]
Output:
E^(-2 + (-1 + E^2 - E^4)/(5 + E^(4*x) - x))/3^(5 + E^(4*x) - x)^(-1)
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.
| \(\displaystyle \int \frac {\left (e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )-1+e^2-e^4-\log (3)\right ) \exp \left (\frac {2 x-2 e^{4 x}-e^4+e^2-11-\log (3)}{-x+e^{4 x}+5}\right )}{x^2-10 x+e^{8 x}+e^{4 x} (10-2 x)+25} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (1-4 e^{4 x}\right ) \left (-1+e^2-e^4-\log (3)\right ) \exp \left (\frac {2 x-2 e^{4 x}-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right )}{\left (-x+e^{4 x}+5\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\left (\left (1-e^2+e^4+\log (3)\right ) \int \frac {\exp \left (-\frac {-2 x+2 e^{4 x}+\log (3)+e^4-e^2+11}{-x+e^{4 x}+5}\right ) \left (1-4 e^{4 x}\right )}{\left (-x+e^{4 x}+5\right )^2}dx\right )\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\left (\left (1-e^2+e^4+\log (3)\right ) \int \frac {\exp \left (-\frac {-2 x+2 e^{4 x}+11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right ) \left (1-4 e^{4 x}\right )}{\left (-x+e^{4 x}+5\right )^2}dx\right )\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\left (\left (1-e^2+e^4+\log (3)\right ) \int \left (-\frac {\exp \left (-\frac {-2 x+2 e^{4 x}+11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right ) (4 x-21)}{\left (x-e^{4 x}-5\right )^2}-\frac {4 \exp \left (-\frac {-2 x+2 e^{4 x}+11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right )}{-x+e^{4 x}+5}\right )dx\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\left (\left (1-e^2+e^4+\log (3)\right ) \left (21 \int \frac {\exp \left (-\frac {-2 x+2 e^{4 x}+11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right )}{\left (-x+e^{4 x}+5\right )^2}dx-4 \int \frac {\exp \left (-\frac {-2 x+2 e^{4 x}+11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right )}{-x+e^{4 x}+5}dx-4 \int \frac {\exp \left (-\frac {-2 x+2 e^{4 x}+11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{-x+e^{4 x}+5}\right ) x}{\left (-x+e^{4 x}+5\right )^2}dx\right )\right )\) |
Input:
Int[(E^((-11 + E^2 - E^4 - 2*E^(4*x) + 2*x - Log[3])/(5 + E^(4*x) - x))*(- 1 + E^2 - E^4 - Log[3] + E^(4*x)*(4 - 4*E^2 + 4*E^4 + 4*Log[3])))/(25 + E^ (8*x) + E^(4*x)*(10 - 2*x) - 10*x + x^2),x]
Output:
$Aborted
Time = 0.66 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10
| method | result | size |
| parallelrisch | \({\mathrm e}^{-\frac {2 \,{\mathrm e}^{4 x}+\ln \left (3\right )+{\mathrm e}^{4}-{\mathrm e}^{2}-2 x +11}{{\mathrm e}^{4 x}+5-x}}\) | \(34\) |
| risch | \({\mathrm e}^{-\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{-{\mathrm e}^{4 x}-5+x}}\) | \(36\) |
| norman | \(\frac {x \,{\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}-{\mathrm e}^{4 x} {\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}-5 \,{\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \left (3\right )-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}}{-{\mathrm e}^{4 x}-5+x}\) | \(126\) |
Input:
int(((4*ln(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-ln(3)-exp(4)+exp(2)-1)*exp((-2 *exp(4*x)-ln(3)-exp(4)+exp(2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2+(-2*x+10 )*exp(4*x)+x^2-10*x+25),x,method=_RETURNVERBOSE)
Output:
exp(-(2*exp(4*x)+ln(3)+exp(4)-exp(2)-2*x+11)/(exp(4*x)+5-x))
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{\left (-\frac {2 \, x - e^{4} + e^{2} - 2 \, e^{\left (4 \, x\right )} - \log \left (3\right ) - 11}{x - e^{\left (4 \, x\right )} - 5}\right )} \] Input:
integrate(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1) *exp((-2*exp(4*x)-log(3)-exp(4)+exp(2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2 +(-2*x+10)*exp(4*x)+x^2-10*x+25),x, algorithm="fricas")
Output:
e^(-(2*x - e^4 + e^2 - 2*e^(4*x) - log(3) - 11)/(x - e^(4*x) - 5))
Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{\frac {2 x - 2 e^{4 x} - e^{4} - 11 - \log {\left (3 \right )} + e^{2}}{- x + e^{4 x} + 5}} \] Input:
integrate(((4*ln(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-ln(3)-exp(4)+exp(2)-1)*e xp((-2*exp(4*x)-ln(3)-exp(4)+exp(2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)**2+( -2*x+10)*exp(4*x)+x**2-10*x+25),x)
Output:
exp((2*x - 2*exp(4*x) - exp(4) - 11 - log(3) + exp(2))/(-x + exp(4*x) + 5) )
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.84 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=e^{\left (\frac {e^{4}}{x - e^{\left (4 \, x\right )} - 5} - \frac {e^{2}}{x - e^{\left (4 \, x\right )} - 5} + \frac {\log \left (3\right )}{x - e^{\left (4 \, x\right )} - 5} + \frac {1}{x - e^{\left (4 \, x\right )} - 5} - 2\right )} \] Input:
integrate(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1) *exp((-2*exp(4*x)-log(3)-exp(4)+exp(2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2 +(-2*x+10)*exp(4*x)+x^2-10*x+25),x, algorithm="maxima")
Output:
e^(e^4/(x - e^(4*x) - 5) - e^2/(x - e^(4*x) - 5) + log(3)/(x - e^(4*x) - 5 ) + 1/(x - e^(4*x) - 5) - 2)
\[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=\int { \frac {{\left (4 \, {\left (e^{4} - e^{2} + \log \left (3\right ) + 1\right )} e^{\left (4 \, x\right )} - e^{4} + e^{2} - \log \left (3\right ) - 1\right )} e^{\left (-\frac {2 \, x - e^{4} + e^{2} - 2 \, e^{\left (4 \, x\right )} - \log \left (3\right ) - 11}{x - e^{\left (4 \, x\right )} - 5}\right )}}{x^{2} - 2 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} - 10 \, x + e^{\left (8 \, x\right )} + 25} \,d x } \] Input:
integrate(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1) *exp((-2*exp(4*x)-log(3)-exp(4)+exp(2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2 +(-2*x+10)*exp(4*x)+x^2-10*x+25),x, algorithm="giac")
Output:
integrate((4*(e^4 - e^2 + log(3) + 1)*e^(4*x) - e^4 + e^2 - log(3) - 1)*e^ (-(2*x - e^4 + e^2 - 2*e^(4*x) - log(3) - 11)/(x - e^(4*x) - 5))/(x^2 - 2* (x - 5)*e^(4*x) - 10*x + e^(8*x) + 25), x)
Time = 3.60 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.03 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=\frac {{\mathrm {e}}^{\frac {2\,x}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {11}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^4}{{\mathrm {e}}^{4\,x}-x+5}}}{3^{\frac {1}{{\mathrm {e}}^{4\,x}-x+5}}} \] Input:
int(-(exp(-(2*exp(4*x) - 2*x - exp(2) + exp(4) + log(3) + 11)/(exp(4*x) - x + 5))*(exp(4) - exp(2) + log(3) - exp(4*x)*(4*exp(4) - 4*exp(2) + 4*log( 3) + 4) + 1))/(exp(8*x) - 10*x - exp(4*x)*(2*x - 10) + x^2 + 25),x)
Output:
(exp((2*x)/(exp(4*x) - x + 5))*exp(-11/(exp(4*x) - x + 5))*exp(-(2*exp(4*x ))/(exp(4*x) - x + 5))*exp(exp(2)/(exp(4*x) - x + 5))*exp(-exp(4)/(exp(4*x ) - x + 5)))/3^(1/(exp(4*x) - x + 5))
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx=\frac {e^{\frac {e^{2}}{e^{4 x}-x +5}}}{e^{\frac {\mathrm {log}\left (3\right )+e^{4}+1}{e^{4 x}-x +5}} e^{2}} \] Input:
int(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1)*exp(( -2*exp(4*x)-log(3)-exp(4)+exp(2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2+(-2*x +10)*exp(4*x)+x^2-10*x+25),x)
Output:
e**(e**2/(e**(4*x) - x + 5))/(e**((log(3) + e**4 + 1)/(e**(4*x) - x + 5))* e**2)