Integrand size = 89, antiderivative size = 27 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\frac {5}{e^{e^2+x}-x+\frac {x (16+x)}{1-x}} \] Output:
5/(x*(x+16)/(1-x)+exp(x+exp(2))-x)
Time = 2.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 (-1+x)}{-e^{e^2+x} (-1+x)+x (15+2 x)} \] Input:
Integrate[(-75 - 20*x + 10*x^2 + E^(E^2 + x)*(-5 + 10*x - 5*x^2))/(225*x^2 + 60*x^3 + 4*x^4 + E^(2*E^2 + 2*x)*(1 - 2*x + x^2) + E^(E^2 + x)*(30*x - 26*x^2 - 4*x^3)),x]
Output:
(-5*(-1 + x))/(-(E^(E^2 + x)*(-1 + x)) + x*(15 + 2*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.
\(\displaystyle \int \frac {10 x^2+e^{x+e^2} \left (-5 x^2+10 x-5\right )-20 x-75}{4 x^4+60 x^3+225 x^2+e^{2 x+2 e^2} \left (x^2-2 x+1\right )+e^{x+e^2} \left (-4 x^3-26 x^2+30 x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 \left (2 x^2-e^{x+e^2} (x-1)^2-4 x-15\right )}{\left (e^{x+e^2} (x-1)-x (2 x+15)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {e^{x+e^2} (1-x)^2-2 x^2+4 x+15}{\left (e^{x+e^2} (1-x)+x (2 x+15)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {e^{x+e^2} (1-x)^2-2 x^2+4 x+15}{\left (e^{x+e^2} (1-x)+x (2 x+15)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {2 x^3+11 x^2-11 x+15}{\left (-2 x^2+e^{x+e^2} x-15 x-e^{x+e^2}\right )^2}-\frac {x-1}{2 x^2-e^{x+e^2} x+15 x+e^{x+e^2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (15 \int \frac {1}{\left (-2 x^2+e^{x+e^2} x-15 x-e^{x+e^2}\right )^2}dx-\int \frac {1}{-2 x^2+e^{x+e^2} x-15 x-e^{x+e^2}}dx-11 \int \frac {x}{\left (2 x^2-e^{x+e^2} x+15 x+e^{x+e^2}\right )^2}dx+11 \int \frac {x^2}{\left (2 x^2-e^{x+e^2} x+15 x+e^{x+e^2}\right )^2}dx-\int \frac {x}{2 x^2-e^{x+e^2} x+15 x+e^{x+e^2}}dx+2 \int \frac {x^3}{\left (2 x^2-e^{x+e^2} x+15 x+e^{x+e^2}\right )^2}dx\right )\) |
Input:
Int[(-75 - 20*x + 10*x^2 + E^(E^2 + x)*(-5 + 10*x - 5*x^2))/(225*x^2 + 60* x^3 + 4*x^4 + E^(2*E^2 + 2*x)*(1 - 2*x + x^2) + E^(E^2 + x)*(30*x - 26*x^2 - 4*x^3)),x]
Output:
$Aborted
Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
risch | \(-\frac {5 \left (-1+x \right )}{2 x^{2}-{\mathrm e}^{x +{\mathrm e}^{2}} x +15 x +{\mathrm e}^{x +{\mathrm e}^{2}}}\) | \(30\) |
norman | \(\frac {-5 x +5}{2 x^{2}-{\mathrm e}^{x +{\mathrm e}^{2}} x +15 x +{\mathrm e}^{x +{\mathrm e}^{2}}}\) | \(31\) |
parallelrisch | \(-\frac {5 x -5}{2 x^{2}-{\mathrm e}^{x +{\mathrm e}^{2}} x +15 x +{\mathrm e}^{x +{\mathrm e}^{2}}}\) | \(32\) |
Input:
int(((-5*x^2+10*x-5)*exp(x+exp(2))+10*x^2-20*x-75)/((x^2-2*x+1)*exp(x+exp( 2))^2+(-4*x^3-26*x^2+30*x)*exp(x+exp(2))+4*x^4+60*x^3+225*x^2),x,method=_R ETURNVERBOSE)
Output:
-5*(-1+x)/(2*x^2-exp(x+exp(2))*x+15*x+exp(x+exp(2)))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 \, {\left (x - 1\right )}}{2 \, x^{2} - {\left (x - 1\right )} e^{\left (x + e^{2}\right )} + 15 \, x} \] Input:
integrate(((-5*x^2+10*x-5)*exp(x+exp(2))+10*x^2-20*x-75)/((x^2-2*x+1)*exp( x+exp(2))^2+(-4*x^3-26*x^2+30*x)*exp(x+exp(2))+4*x^4+60*x^3+225*x^2),x, al gorithm="fricas")
Output:
-5*(x - 1)/(2*x^2 - (x - 1)*e^(x + e^2) + 15*x)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\frac {5 x - 5}{- 2 x^{2} - 15 x + \left (x - 1\right ) e^{x + e^{2}}} \] Input:
integrate(((-5*x**2+10*x-5)*exp(x+exp(2))+10*x**2-20*x-75)/((x**2-2*x+1)*e xp(x+exp(2))**2+(-4*x**3-26*x**2+30*x)*exp(x+exp(2))+4*x**4+60*x**3+225*x* *2),x)
Output:
(5*x - 5)/(-2*x**2 - 15*x + (x - 1)*exp(x + exp(2)))
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 \, {\left (x - 1\right )}}{2 \, x^{2} - {\left (x e^{\left (e^{2}\right )} - e^{\left (e^{2}\right )}\right )} e^{x} + 15 \, x} \] Input:
integrate(((-5*x^2+10*x-5)*exp(x+exp(2))+10*x^2-20*x-75)/((x^2-2*x+1)*exp( x+exp(2))^2+(-4*x^3-26*x^2+30*x)*exp(x+exp(2))+4*x^4+60*x^3+225*x^2),x, al gorithm="maxima")
Output:
-5*(x - 1)/(2*x^2 - (x*e^(e^2) - e^(e^2))*e^x + 15*x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (23) = 46\).
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 \, {\left (x - 1\right )}}{2 \, {\left (x + e^{2}\right )}^{2} - 4 \, {\left (x + e^{2}\right )} e^{2} - {\left (x + e^{2}\right )} e^{\left (x + e^{2}\right )} + 15 \, x + 2 \, e^{4} + e^{\left (x + e^{2} + 2\right )} + e^{\left (x + e^{2}\right )}} \] Input:
integrate(((-5*x^2+10*x-5)*exp(x+exp(2))+10*x^2-20*x-75)/((x^2-2*x+1)*exp( x+exp(2))^2+(-4*x^3-26*x^2+30*x)*exp(x+exp(2))+4*x^4+60*x^3+225*x^2),x, al gorithm="giac")
Output:
-5*(x - 1)/(2*(x + e^2)^2 - 4*(x + e^2)*e^2 - (x + e^2)*e^(x + e^2) + 15*x + 2*e^4 + e^(x + e^2 + 2) + e^(x + e^2))
Timed out. \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\int -\frac {20\,x+{\mathrm {e}}^{x+{\mathrm {e}}^2}\,\left (5\,x^2-10\,x+5\right )-10\,x^2+75}{{\mathrm {e}}^{2\,x+2\,{\mathrm {e}}^2}\,\left (x^2-2\,x+1\right )-{\mathrm {e}}^{x+{\mathrm {e}}^2}\,\left (4\,x^3+26\,x^2-30\,x\right )+225\,x^2+60\,x^3+4\,x^4} \,d x \] Input:
int(-(20*x + exp(x + exp(2))*(5*x^2 - 10*x + 5) - 10*x^2 + 75)/(exp(2*x + 2*exp(2))*(x^2 - 2*x + 1) - exp(x + exp(2))*(26*x^2 - 30*x + 4*x^3) + 225* x^2 + 60*x^3 + 4*x^4),x)
Output:
int(-(20*x + exp(x + exp(2))*(5*x^2 - 10*x + 5) - 10*x^2 + 75)/(exp(2*x + 2*exp(2))*(x^2 - 2*x + 1) - exp(x + exp(2))*(26*x^2 - 30*x + 4*x^3) + 225* x^2 + 60*x^3 + 4*x^4), x)
Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\frac {e^{e^{2}+x} x -e^{e^{2}+x}-2 x^{2}-15}{3 e^{e^{2}+x} x -3 e^{e^{2}+x}-6 x^{2}-45 x} \] Input:
int(((-5*x^2+10*x-5)*exp(x+exp(2))+10*x^2-20*x-75)/((x^2-2*x+1)*exp(x+exp( 2))^2+(-4*x^3-26*x^2+30*x)*exp(x+exp(2))+4*x^4+60*x^3+225*x^2),x)
Output:
(e**(e**2 + x)*x - e**(e**2 + x) - 2*x**2 - 15)/(3*(e**(e**2 + x)*x - e**( e**2 + x) - 2*x**2 - 15*x))