Integrand size = 111, antiderivative size = 24 \[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \] Output:
x/exp(-exp(3*x/(exp(2)+5+2*x))+x-5)
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \] Input:
Integrate[(E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)*(25 + E^4*(1 - x) - 5*x - 16*x^2 - 4*x^3 + E^((3*x)/(5 + E^2 + 2*x))*(15*x + 3*E^2*x) + E^2*(10 - 6 *x - 4*x^2)))/(25 + E^4 + 20*x + 4*x^2 + E^2*(10 + 4*x)),x]
Output:
E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)*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.
| \(\displaystyle \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5} \left (-4 x^3-16 x^2+e^2 \left (-4 x^2-6 x+10\right )-5 x+e^4 (1-x)+e^{\frac {3 x}{2 x+e^2+5}} \left (3 e^2 x+15 x\right )+25\right )}{4 x^2+20 x+e^2 (4 x+10)+e^4+25} \, dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5} \left (-4 x^3-16 x^2+e^2 \left (-4 x^2-6 x+10\right )-5 x+e^4 (1-x)+e^{\frac {3 x}{2 x+e^2+5}} \left (3 e^2 x+15 x\right )+25\right )}{\left (2 x+e^2+5\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {4 e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5} x^3}{\left (2 x+e^2+5\right )^2}-\frac {16 e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5} x^2}{\left (2 x+e^2+5\right )^2}-\frac {5 e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5} x}{\left (2 x+e^2+5\right )^2}+\frac {3 \left (5+e^2\right ) e^{\frac {3 x}{2 x+e^2+5}-x+e^{\frac {3 x}{2 x+e^2+5}}+5} x}{\left (2 x+e^2+5\right )^2}+\frac {25 e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{\left (2 x+e^2+5\right )^2}-\frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+9} (x-1)}{\left (2 x+e^2+5\right )^2}-\frac {2 e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+7} (x-1) (2 x+5)}{\left (2 x+e^2+5\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \left (5+e^2\right ) \int e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}dx-4 \int e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}dx-\int e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+7}dx-\int e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5} xdx+\frac {1}{2} \left (5+e^2\right )^3 \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{\left (2 x+e^2+5\right )^2}dx-4 \left (5+e^2\right )^2 \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{\left (2 x+e^2+5\right )^2}dx+\frac {5}{2} \left (5+e^2\right ) \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{\left (2 x+e^2+5\right )^2}dx+25 \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{\left (2 x+e^2+5\right )^2}dx-\frac {1}{2} \left (7+e^2\right ) \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+9}}{\left (2 x+e^2+5\right )^2}dx-\frac {3}{2} \left (5+e^2\right )^2 \int \frac {e^{\frac {3 x}{2 x+e^2+5}-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{\left (2 x+e^2+5\right )^2}dx-\frac {3}{2} \left (5+e^2\right )^2 \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{2 x+e^2+5}dx+8 \left (5+e^2\right ) \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{2 x+e^2+5}dx-\frac {5}{2} \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{2 x+e^2+5}dx+\left (7+2 e^2\right ) \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+7}}{2 x+e^2+5}dx-\frac {1}{2} \int \frac {e^{-x+e^{\frac {3 x}{2 x+e^2+5}}+9}}{2 x+e^2+5}dx+\frac {3}{2} \left (5+e^2\right ) \int \frac {e^{\frac {3 x}{2 x+e^2+5}-x+e^{\frac {3 x}{2 x+e^2+5}}+5}}{2 x+e^2+5}dx\) |
Input:
Int[(E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)*(25 + E^4*(1 - x) - 5*x - 16*x^ 2 - 4*x^3 + E^((3*x)/(5 + E^2 + 2*x))*(15*x + 3*E^2*x) + E^2*(10 - 6*x - 4 *x^2)))/(25 + E^4 + 20*x + 4*x^2 + E^2*(10 + 4*x)),x]
Output:
$Aborted
Time = 2.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(x \,{\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}\) | \(22\) |
| parallelrisch | \(x \,{\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}\) | \(24\) |
| norman | \(\frac {\left (\left ({\mathrm e}^{2}+5\right ) x +2 x^{2}\right ) {\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}}{{\mathrm e}^{2}+5+2 x}\) | \(44\) |
Input:
int(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)^2+(-4*x^2-6*x+ 10)*exp(2)-4*x^3-16*x^2-5*x+25)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x+25)/e xp(-exp(3*x/(exp(2)+5+2*x))+x-5),x,method=_RETURNVERBOSE)
Output:
x*exp(exp(3*x/(exp(2)+5+2*x))-x+5)
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=x e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )} \] Input:
integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)^2+(-4*x^ 2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+25)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x +25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x, algorithm="fricas")
Output:
x*e^(-x + e^(3*x/(2*x + e^2 + 5)) + 5)
Time = 14.84 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=x e^{- x + e^{\frac {3 x}{2 x + 5 + e^{2}}} + 5} \] Input:
integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)**2+(-4*x **2-6*x+10)*exp(2)-4*x**3-16*x**2-5*x+25)/(exp(2)**2+(4*x+10)*exp(2)+4*x** 2+20*x+25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x)
Output:
x*exp(-x + exp(3*x/(2*x + 5 + exp(2))) + 5)
\[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=\int { -\frac {{\left (4 \, x^{3} + 16 \, x^{2} + {\left (x - 1\right )} e^{4} + 2 \, {\left (2 \, x^{2} + 3 \, x - 5\right )} e^{2} - 3 \, {\left (x e^{2} + 5 \, x\right )} e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5 \, x - 25\right )} e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )}}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} e^{2} + 20 \, x + e^{4} + 25} \,d x } \] Input:
integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)^2+(-4*x^ 2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+25)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x +25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x, algorithm="maxima")
Output:
-integrate((4*x^3 + 16*x^2 + (x - 1)*e^4 + 2*(2*x^2 + 3*x - 5)*e^2 - 3*(x* e^2 + 5*x)*e^(3*x/(2*x + e^2 + 5)) + 5*x - 25)*e^(-x + e^(3*x/(2*x + e^2 + 5)) + 5)/(4*x^2 + 2*(2*x + 5)*e^2 + 20*x + e^4 + 25), x)
\[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=\int { -\frac {{\left (4 \, x^{3} + 16 \, x^{2} + {\left (x - 1\right )} e^{4} + 2 \, {\left (2 \, x^{2} + 3 \, x - 5\right )} e^{2} - 3 \, {\left (x e^{2} + 5 \, x\right )} e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5 \, x - 25\right )} e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )}}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} e^{2} + 20 \, x + e^{4} + 25} \,d x } \] Input:
integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)^2+(-4*x^ 2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+25)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x +25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x, algorithm="giac")
Output:
integrate(-(4*x^3 + 16*x^2 + (x - 1)*e^4 + 2*(2*x^2 + 3*x - 5)*e^2 - 3*(x* e^2 + 5*x)*e^(3*x/(2*x + e^2 + 5)) + 5*x - 25)*e^(-x + e^(3*x/(2*x + e^2 + 5)) + 5)/(4*x^2 + 2*(2*x + 5)*e^2 + 20*x + e^4 + 25), x)
Time = 4.47 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {3\,x}{2\,x+{\mathrm {e}}^2+5}}} \] Input:
int(-(exp(exp((3*x)/(2*x + exp(2) + 5)) - x + 5)*(5*x - exp((3*x)/(2*x + e xp(2) + 5))*(15*x + 3*x*exp(2)) + exp(2)*(6*x + 4*x^2 - 10) + exp(4)*(x - 1) + 16*x^2 + 4*x^3 - 25))/(20*x + exp(4) + 4*x^2 + exp(2)*(4*x + 10) + 25 ),x)
Output:
x*exp(-x)*exp(5)*exp(exp((3*x)/(2*x + exp(2) + 5)))
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx=\frac {e^{e^{\frac {3 x}{e^{2}+2 x +5}}} e^{5} x}{e^{x}} \] Input:
int(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)^2+(-4*x^2-6*x+ 10)*exp(2)-4*x^3-16*x^2-5*x+25)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x+25)/e xp(-exp(3*x/(exp(2)+5+2*x))+x-5),x)
Output:
(e**(e**((3*x)/(e**2 + 2*x + 5)))*e**5*x)/e**x