Integrand size = 92, antiderivative size = 26 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{-3+x}}{x \left (5-e^{-e^8 x^2} x\right )} \] Output:
exp(x)/exp(3)/x/(5-x/exp(x^2*exp(4)^2))
Time = 2.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{-3+x+e^8 x^2}}{\left (5 e^{e^8 x^2}-x\right ) x} \] Input:
Integrate[(E^(x + 2*E^8*x^2)*(-5 + 5*x) + E^(x + E^8*x^2)*(2*x - x^2 - 2*E ^8*x^3))/(25*E^(3 + 2*E^8*x^2)*x^2 - 10*E^(3 + E^8*x^2)*x^3 + E^3*x^4),x]
Output:
E^(-3 + x + E^8*x^2)/((5*E^(E^8*x^2) - 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.
\(\displaystyle \int \frac {e^{2 e^8 x^2+x} (5 x-5)+e^{e^8 x^2+x} \left (-2 e^8 x^3-x^2+2 x\right )}{e^3 x^4+25 e^{2 e^8 x^2+3} x^2-10 e^{e^8 x^2+3} x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 e^8 x^2+x} (5 x-5)+e^{e^8 x^2+x} \left (-2 e^8 x^3-x^2+2 x\right )}{e^3 \left (5 e^{e^8 x^2}-x\right )^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {5 e^{2 e^8 x^2+x} (1-x)-e^{e^8 x^2+x} \left (-2 e^8 x^3-x^2+2 x\right )}{\left (5 e^{e^8 x^2}-x\right )^2 x^2}dx}{e^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {5 e^{2 e^8 x^2+x} (1-x)-e^{e^8 x^2+x} \left (-2 e^8 x^3-x^2+2 x\right )}{\left (5 e^{e^8 x^2}-x\right )^2 x^2}dx}{e^3}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (-\frac {e^x (x-1)}{5 x^2}+\frac {e^x \left (2 e^8 x-1\right )}{5 \left (5 e^{e^8 x^2}-x\right )}+\frac {e^x \left (2 e^8 x^2-1\right )}{5 \left (5 e^{e^8 x^2}-x\right )^2}\right )dx}{e^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} \int \frac {e^x}{\left (5 e^{e^8 x^2}-x\right )^2}dx-\frac {1}{5} \int \frac {e^x}{5 e^{e^8 x^2}-x}dx+\frac {2}{5} \int \frac {e^{x+8} x}{5 e^{e^8 x^2}-x}dx+\frac {2}{5} \int \frac {e^{x+8} x^2}{\left (5 e^{e^8 x^2}-x\right )^2}dx-\frac {e^x}{5 x}}{e^3}\) |
Input:
Int[(E^(x + 2*E^8*x^2)*(-5 + 5*x) + E^(x + E^8*x^2)*(2*x - x^2 - 2*E^8*x^3 ))/(25*E^(3 + 2*E^8*x^2)*x^2 - 10*E^(3 + E^8*x^2)*x^3 + E^3*x^4),x]
Output:
$Aborted
Time = 0.49 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15
method | result | size |
risch | \(\frac {{\mathrm e}^{-3+x +x^{2} {\mathrm e}^{8}}}{\left (-x +5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right ) x}\) | \(30\) |
norman | \(-\frac {{\mathrm e}^{x} {\mathrm e}^{-3} {\mathrm e}^{x^{2} {\mathrm e}^{8}}}{x \left (x -5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right )}\) | \(36\) |
parallelrisch | \(-\frac {{\mathrm e}^{x} {\mathrm e}^{-3} {\mathrm e}^{x^{2} {\mathrm e}^{8}}}{x \left (x -5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right )}\) | \(36\) |
Input:
int(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*e xp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2 *exp(4)^2)+x^4*exp(3)),x,method=_RETURNVERBOSE)
Output:
1/(-x+5*exp(x^2*exp(8)))/x*exp(-3+x+x^2*exp(8))
Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=-\frac {e^{\left (2 \, x^{2} e^{8} + 2 \, x\right )}}{x^{2} e^{\left (x^{2} e^{8} + x + 3\right )} - 5 \, x e^{\left (2 \, x^{2} e^{8} + x + 3\right )}} \] Input:
integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*ex p(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*e xp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="fricas")
Output:
-e^(2*x^2*e^8 + 2*x)/(x^2*e^(x^2*e^8 + x + 3) - 5*x*e^(2*x^2*e^8 + x + 3))
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{x}}{- 5 x e^{3} + 25 e^{3} e^{x^{2} e^{8}}} + \frac {e^{x}}{5 x e^{3}} \] Input:
integrate(((5*x-5)*exp(x)*exp(x**2*exp(4)**2)**2+(-2*x**3*exp(4)**2-x**2+2 *x)*exp(x)*exp(x**2*exp(4)**2))/(25*x**2*exp(3)*exp(x**2*exp(4)**2)**2-10* x**3*exp(3)*exp(x**2*exp(4)**2)+x**4*exp(3)),x)
Output:
exp(x)/(-5*x*exp(3) + 25*exp(3)*exp(x**2*exp(8))) + exp(-3)*exp(x)/(5*x)
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=-\frac {e^{\left (x^{2} e^{8} + x\right )}}{x^{2} e^{3} - 5 \, x e^{\left (x^{2} e^{8} + 3\right )}} \] Input:
integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*ex p(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*e xp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="maxima")
Output:
-e^(x^2*e^8 + x)/(x^2*e^3 - 5*x*e^(x^2*e^8 + 3))
\[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\int { \frac {5 \, {\left (x - 1\right )} e^{\left (2 \, x^{2} e^{8} + x\right )} - {\left (2 \, x^{3} e^{8} + x^{2} - 2 \, x\right )} e^{\left (x^{2} e^{8} + x\right )}}{x^{4} e^{3} - 10 \, x^{3} e^{\left (x^{2} e^{8} + 3\right )} + 25 \, x^{2} e^{\left (2 \, x^{2} e^{8} + 3\right )}} \,d x } \] Input:
integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*ex p(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*e xp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="giac")
Output:
integrate((5*(x - 1)*e^(2*x^2*e^8 + x) - (2*x^3*e^8 + x^2 - 2*x)*e^(x^2*e^ 8 + x))/(x^4*e^3 - 10*x^3*e^(x^2*e^8 + 3) + 25*x^2*e^(2*x^2*e^8 + 3)), x)
Time = 4.14 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {{\mathrm {e}}^{x-3}}{5\,x}+\frac {{\mathrm {e}}^{-3}\,\left ({\mathrm {e}}^x-2\,x^2\,{\mathrm {e}}^{x+8}\right )}{5\,\left (x-5\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^8}\right )\,\left (2\,x^2\,{\mathrm {e}}^8-1\right )} \] Input:
int((exp(2*x^2*exp(8))*exp(x)*(5*x - 5) - exp(x^2*exp(8))*exp(x)*(2*x^3*ex p(8) - 2*x + x^2))/(x^4*exp(3) - 10*x^3*exp(x^2*exp(8))*exp(3) + 25*x^2*ex p(2*x^2*exp(8))*exp(3)),x)
Output:
exp(x - 3)/(5*x) + (exp(-3)*(exp(x) - 2*x^2*exp(x + 8)))/(5*(x - 5*exp(x^2 *exp(8)))*(2*x^2*exp(8) - 1))
Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{e^{8} x^{2}+x}}{e^{3} x \left (5 e^{e^{8} x^{2}}-x \right )} \] Input:
int(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*e xp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2 *exp(4)^2)+x^4*exp(3)),x)
Output:
e**(e**8*x**2 + x)/(e**3*x*(5*e**(e**8*x**2) - x))