\(\int \frac {-25 e^2 x^2+e^4 (-200 x+50 x^2)}{4 x^2-4 x^3+x^4+e^4 (64-64 x+16 x^2)+e^2 (32 x-32 x^2+8 x^3)} \, dx\) [7479]

Optimal result

Integrand size = 71, antiderivative size = 24 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25 x^2}{2 (2-x) \left (-4-\frac {x}{e^2}\right )} \]

[Out]

x^2/(-x/exp(2)-4)/(4/25-2/25*x)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1694, 12, 1828, 8} \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 e^2 \left (\left (1-2 e^2\right ) x+4 e^2\right )}{-x^2+2 \left (1-2 e^2\right ) x+8 e^2} \]

[In]

Int[(-25*E^2*x^2 + E^4*(-200*x + 50*x^2))/(4*x^2 - 4*x^3 + x^4 + E^4*(64 - 64*x + 16*x^2) + E^2*(32*x - 32*x^2
+ 8*x^3)),x]

[Out]

(-25*E^2*(4*E^2 + (1 - 2*E^2)*x))/(8*E^2 + 2*(1 - 2*E^2)*x - x^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {25 \left (-e^2 \left (1-2 e^2\right ) \left (1+2 e^2\right )^2-2 e^2 \left (1+4 e^4\right ) x-e^2 \left (1-2 e^2\right ) x^2\right )}{\left (1+4 e^2+4 e^4-x^2\right )^2} \, dx,x,\frac {1}{4} \left (-4+8 e^2\right )+x\right ) \\ & = 25 \text {Subst}\left (\int \frac {-e^2 \left (1-2 e^2\right ) \left (1+2 e^2\right )^2-2 e^2 \left (1+4 e^4\right ) x-e^2 \left (1-2 e^2\right ) x^2}{\left (1+4 e^2+4 e^4-x^2\right )^2} \, dx,x,\frac {1}{4} \left (-4+8 e^2\right )+x\right ) \\ & = -\frac {25 e^2 \left (4 e^2+\left (1-2 e^2\right ) x\right )}{8 e^2+2 \left (1-2 e^2\right ) x-x^2}-\frac {25 \text {Subst}\left (\int 0 \, dx,x,\frac {1}{4} \left (-4+8 e^2\right )+x\right )}{2 \left (1+2 e^2\right )^2} \\ & = -\frac {25 e^2 \left (4 e^2+\left (1-2 e^2\right ) x\right )}{8 e^2+2 \left (1-2 e^2\right ) x-x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25 e^2 \left (-2 e^2 (-2+x)+x\right )}{(-2+x) \left (4 e^2+x\right )} \]

[In]

Integrate[(-25*E^2*x^2 + E^4*(-200*x + 50*x^2))/(4*x^2 - 4*x^3 + x^4 + E^4*(64 - 64*x + 16*x^2) + E^2*(32*x -
32*x^2 + 8*x^3)),x]

[Out]

(25*E^2*(-2*E^2*(-2 + x) + x))/((-2 + x)*(4*E^2 + x))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46

method	result	size
norman	\(\frac {\left (-50 \,{\mathrm e}^{4}+25 \,{\mathrm e}^{2}\right ) x +100 \,{\mathrm e}^{4}}{\left (-2+x \right ) \left (x +4 \,{\mathrm e}^{2}\right )}\)	\(35\)
gosper	\(-\frac {25 \left (2 \,{\mathrm e}^{2} x -4 \,{\mathrm e}^{2}-x \right ) {\mathrm e}^{2}}{4 \,{\mathrm e}^{2} x +x^{2}-8 \,{\mathrm e}^{2}-2 x}\)	\(36\)
risch	\(\frac {\left (-\frac {25 \,{\mathrm e}^{4}}{2}+\frac {25 \,{\mathrm e}^{2}}{4}\right ) x +25 \,{\mathrm e}^{4}}{{\mathrm e}^{2} x +\frac {x^{2}}{4}-2 \,{\mathrm e}^{2}-\frac {x}{2}}\)	\(37\)
parallelrisch	\(-\frac {50 x \,{\mathrm e}^{4}-100 \,{\mathrm e}^{4}-25 \,{\mathrm e}^{2} x}{4 \,{\mathrm e}^{2} x +x^{2}-8 \,{\mathrm e}^{2}-2 x}\)	\(40\)

[In]

int(((50*x^2-200*x)*exp(2)^2-25*x^2*exp(2))/((16*x^2-64*x+64)*exp(2)^2+(8*x^3-32*x^2+32*x)*exp(2)+x^4-4*x^3+4*
x^2),x,method=_RETURNVERBOSE)

[Out]

((-50*exp(2)^2+25*exp(2))*x+100*exp(2)^2)/(-2+x)/(x+4*exp(2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 \, {\left (2 \, {\left (x - 2\right )} e^{4} - x e^{2}\right )}}{x^{2} + 4 \, {\left (x - 2\right )} e^{2} - 2 \, x} \]

[In]

integrate(((50*x^2-200*x)*exp(2)^2-25*x^2*exp(2))/((16*x^2-64*x+64)*exp(2)^2+(8*x^3-32*x^2+32*x)*exp(2)+x^4-4*
x^3+4*x^2),x, algorithm="fricas")

[Out]

-25*(2*(x - 2)*e^4 - x*e^2)/(x^2 + 4*(x - 2)*e^2 - 2*x)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=- \frac {x \left (- 25 e^{2} + 50 e^{4}\right ) - 100 e^{4}}{x^{2} + x \left (-2 + 4 e^{2}\right ) - 8 e^{2}} \]

[In]

integrate(((50*x**2-200*x)*exp(2)**2-25*x**2*exp(2))/((16*x**2-64*x+64)*exp(2)**2+(8*x**3-32*x**2+32*x)*exp(2)
+x**4-4*x**3+4*x**2),x)

[Out]

-(x*(-25*exp(2) + 50*exp(4)) - 100*exp(4))/(x**2 + x*(-2 + 4*exp(2)) - 8*exp(2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 \, {\left (x {\left (2 \, e^{4} - e^{2}\right )} - 4 \, e^{4}\right )}}{x^{2} + 2 \, x {\left (2 \, e^{2} - 1\right )} - 8 \, e^{2}} \]

[In]

integrate(((50*x^2-200*x)*exp(2)^2-25*x^2*exp(2))/((16*x^2-64*x+64)*exp(2)^2+(8*x^3-32*x^2+32*x)*exp(2)+x^4-4*
x^3+4*x^2),x, algorithm="maxima")

[Out]

-25*(x*(2*e^4 - e^2) - 4*e^4)/(x^2 + 2*x*(2*e^2 - 1) - 8*e^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 \, {\left (2 \, x e^{4} - x e^{2} - 4 \, e^{4}\right )}}{x^{2} + 4 \, x e^{2} - 2 \, x - 8 \, e^{2}} \]

[In]

integrate(((50*x^2-200*x)*exp(2)^2-25*x^2*exp(2))/((16*x^2-64*x+64)*exp(2)^2+(8*x^3-32*x^2+32*x)*exp(2)+x^4-4*
x^3+4*x^2),x, algorithm="giac")

[Out]

-25*(2*x*e^4 - x*e^2 - 4*e^4)/(x^2 + 4*x*e^2 - 2*x - 8*e^2)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25\,{\mathrm {e}}^2\,\left (x+4\,{\mathrm {e}}^2-2\,x\,{\mathrm {e}}^2\right )}{\left (x+4\,{\mathrm {e}}^2\right )\,\left (x-2\right )} \]

[In]

int(-(exp(4)*(200*x - 50*x^2) + 25*x^2*exp(2))/(exp(4)*(16*x^2 - 64*x + 64) + exp(2)*(32*x - 32*x^2 + 8*x^3) +
4*x^2 - 4*x^3 + x^4),x)

[Out]

(25*exp(2)*(x + 4*exp(2) - 2*x*exp(2)))/((x + 4*exp(2))*(x - 2))