\(\int \frac {-75-20 x+10 x^2+e^{e^2+x} (-5+10 x-5 x^2)}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} (1-2 x+x^2)+e^{e^2+x} (30 x-26 x^2-4 x^3)} \, dx\) [9043]

Optimal result

Integrand size = 89, antiderivative size = 27 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\frac {5}{e^{e^2+x}-x+\frac {x (16+x)}{1-x}} \]

[Out]

5/(x*(x+16)/(1-x)+exp(x+exp(2))-x)

Rubi [F]

\[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx \]

[In]

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]

[Out]

-75*Defer[Int][(-E^(E^2 + x) - 15*x + E^(E^2 + x)*x - 2*x^2)^(-2), x] + 5*Defer[Int][(-E^(E^2 + x) - 15*x + E^
(E^2 + x)*x - 2*x^2)^(-1), x] + 55*Defer[Int][x/(E^(E^2 + x) + 15*x - E^(E^2 + x)*x + 2*x^2)^2, x] - 55*Defer[
Int][x^2/(E^(E^2 + x) + 15*x - E^(E^2 + x)*x + 2*x^2)^2, x] - 10*Defer[Int][x^3/(E^(E^2 + x) + 15*x - E^(E^2 +
x)*x + 2*x^2)^2, x] + 5*Defer[Int][x/(E^(E^2 + x) + 15*x - E^(E^2 + x)*x + 2*x^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (-15-e^{e^2+x} (-1+x)^2-4 x+2 x^2\right )}{\left (e^{e^2+x} (-1+x)-x (15+2 x)\right )^2} \, dx \\ & = 5 \int \frac {-15-e^{e^2+x} (-1+x)^2-4 x+2 x^2}{\left (e^{e^2+x} (-1+x)-x (15+2 x)\right )^2} \, dx \\ & = 5 \int \left (\frac {-1+x}{e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2}-\frac {15-11 x+11 x^2+2 x^3}{\left (-e^{e^2+x}-15 x+e^{e^2+x} x-2 x^2\right )^2}\right ) \, dx \\ & = 5 \int \frac {-1+x}{e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2} \, dx-5 \int \frac {15-11 x+11 x^2+2 x^3}{\left (-e^{e^2+x}-15 x+e^{e^2+x} x-2 x^2\right )^2} \, dx \\ & = -\left (5 \int \left (\frac {15}{\left (-e^{e^2+x}-15 x+e^{e^2+x} x-2 x^2\right )^2}-\frac {11 x}{\left (e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2\right )^2}+\frac {11 x^2}{\left (e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2\right )^2}+\frac {2 x^3}{\left (e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2\right )^2}\right ) \, dx\right )+5 \int \left (\frac {1}{-e^{e^2+x}-15 x+e^{e^2+x} x-2 x^2}+\frac {x}{e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2}\right ) \, dx \\ & = 5 \int \frac {1}{-e^{e^2+x}-15 x+e^{e^2+x} x-2 x^2} \, dx+5 \int \frac {x}{e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2} \, dx-10 \int \frac {x^3}{\left (e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2\right )^2} \, dx+55 \int \frac {x}{\left (e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2\right )^2} \, dx-55 \int \frac {x^2}{\left (e^{e^2+x}+15 x-e^{e^2+x} x+2 x^2\right )^2} \, dx-75 \int \frac {1}{\left (-e^{e^2+x}-15 x+e^{e^2+x} x-2 x^2\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 (-1+x)}{-e^{e^2+x} (-1+x)+x (15+2 x)} \]

[In]

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]

[Out]

(-5*(-1 + x))/(-(E^(E^2 + x)*(-1 + x)) + x*(15 + 2*x))

Maple [A] (verified)

Time = 0.80 (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	\(-\frac {5 x -5}{2 x^{2}-{\mathrm e}^{x +{\mathrm e}^{2}} x +15 x +{\mathrm e}^{x +{\mathrm e}^{2}}}\)	\(32\)

[In]

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=_RETURNVERBOSE)

[Out]

-5*(-1+x)/(2*x^2-exp(x+exp(2))*x+15*x+exp(x+exp(2)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 \, {\left (x - 1\right )}}{2 \, x^{2} - {\left (x - 1\right )} e^{\left (x + e^{2}\right )} + 15 \, x} \]

[In]

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, algorithm="fricas")

[Out]

-5*(x - 1)/(2*x^2 - (x - 1)*e^(x + e^2) + 15*x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\frac {5 x - 5}{- 2 x^{2} - 15 x + \left (x - 1\right ) e^{x + e^{2}}} \]

[In]

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)

[Out]

(5*x - 5)/(-2*x**2 - 15*x + (x - 1)*exp(x + exp(2)))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 \, {\left (x - 1\right )}}{2 \, x^{2} - {\left (x e^{\left (e^{2}\right )} - e^{\left (e^{2}\right )}\right )} e^{x} + 15 \, x} \]

[In]

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, algorithm="maxima")

[Out]

-5*(x - 1)/(2*x^2 - (x*e^(e^2) - e^(e^2))*e^x + 15*x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (23) = 46\).

Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=-\frac {5 \, {\left (x - 1\right )}}{2 \, {\left (x + e^{2}\right )}^{2} - 4 \, {\left (x + e^{2}\right )} e^{2} - {\left (x + e^{2}\right )} e^{\left (x + e^{2}\right )} + 15 \, x + 2 \, e^{4} + e^{\left (x + e^{2} + 2\right )} + e^{\left (x + e^{2}\right )}} \]

[In]

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, algorithm="giac")

[Out]

-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))

Mupad [F(-1)]

Timed out. \[ \int \frac {-75-20 x+10 x^2+e^{e^2+x} \left (-5+10 x-5 x^2\right )}{225 x^2+60 x^3+4 x^4+e^{2 e^2+2 x} \left (1-2 x+x^2\right )+e^{e^2+x} \left (30 x-26 x^2-4 x^3\right )} \, dx=\int -\frac {20\,x+{\mathrm {e}}^{x+{\mathrm {e}}^2}\,\left (5\,x^2-10\,x+5\right )-10\,x^2+75}{{\mathrm {e}}^{2\,x+2\,{\mathrm {e}}^2}\,\left (x^2-2\,x+1\right )-{\mathrm {e}}^{x+{\mathrm {e}}^2}\,\left (4\,x^3+26\,x^2-30\,x\right )+225\,x^2+60\,x^3+4\,x^4} \,d x \]

[In]

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)

[Out]

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)