Integrand size = 71, antiderivative size = 24 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25 x^2}{2 (2-x) \left (-4-\frac {x}{e^2}\right )} \] Output:
x^2/(-x/exp(2)-4)/(4/25-2/25*x)
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25 e^2 \left (-2 e^2 (-2+x)+x\right )}{(-2+x) \left (4 e^2+x\right )} \] Input:
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]
Output:
(25*E^2*(-2*E^2*(-2 + x) + x))/((-2 + x)*(4*E^2 + x))
Leaf count is larger than twice the leaf count of optimal. \(92\) vs. \(2(24)=48\).
Time = 0.62 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2459, 1380, 27, 2345, 24}
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.
\(\displaystyle \int \frac {e^4 \left (50 x^2-200 x\right )-25 e^2 x^2}{x^4-4 x^3+4 x^2+e^4 \left (16 x^2-64 x+64\right )+e^2 \left (8 x^3-32 x^2+32 x\right )} \, dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int \frac {-25 e^2 \left (1-2 e^2\right ) \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^2-50 e^2 \left (1+4 e^4\right ) \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )-25 e^2 \left (1-2 e^2\right ) \left (1+2 e^2\right )^2}{\left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^4-2 \left (1+2 e^2\right )^2 \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^2+\left (1+2 e^2\right )^4}d\left (x+\frac {1}{4} \left (8 e^2-4\right )\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int -\frac {25 \left (e^2 \left (1-2 e^2\right ) \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^2+2 e^2 \left (1+4 e^4\right ) \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )+e^2 \left (1-2 e^2\right ) \left (1+2 e^2\right )^2\right )}{\left (\left (1+2 e^2\right )^2-\left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^2\right )^2}d\left (x+\frac {1}{4} \left (8 e^2-4\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -25 \int \frac {e^2 \left (1-2 e^2\right ) \left (x+\frac {1}{4} \left (-4+8 e^2\right )\right )^2+2 e^2 \left (1+4 e^4\right ) \left (x+\frac {1}{4} \left (-4+8 e^2\right )\right )+e^2 \left (1-2 e^2\right ) \left (1+2 e^2\right )^2}{\left (\left (1+2 e^2\right )^2-\left (x+\frac {1}{4} \left (-4+8 e^2\right )\right )^2\right )^2}d\left (x+\frac {1}{4} \left (-4+8 e^2\right )\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -25 \left (\frac {e^2 \left (\left (1+2 e^2-4 e^4-8 e^6\right ) \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )+\left (1+2 e^2\right )^2 \left (1+4 e^4\right )\right )}{\left (1+2 e^2\right )^2 \left (\left (1+2 e^2\right )^2-\left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^2\right )}-\frac {\int 0d\left (x+\frac {1}{4} \left (-4+8 e^2\right )\right )}{2 \left (1+2 e^2\right )^2}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {25 e^2 \left (\left (1+2 e^2-4 e^4-8 e^6\right ) \left (x+\frac {1}{4} \left (8 e^2-4\right )\right )+\left (1+2 e^2\right )^2 \left (1+4 e^4\right )\right )}{\left (1+2 e^2\right )^2 \left (\left (1+2 e^2\right )^2-\left (x+\frac {1}{4} \left (8 e^2-4\right )\right )^2\right )}\) |
Input:
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]
Output:
(-25*E^2*((1 + 2*E^2)^2*(1 + 4*E^4) + (1 + 2*E^2 - 4*E^4 - 8*E^6)*((-4 + 8 *E^2)/4 + x)))/((1 + 2*E^2)^2*((1 + 2*E^2)^2 - ((-4 + 8*E^2)/4 + x)^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[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] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.26 (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 | \(-\frac {50 x \,{\mathrm e}^{4}-100 \,{\mathrm e}^{4}-25 \,{\mathrm e}^{2} x}{4 \,{\mathrm e}^{2} x +x^{2}-8 \,{\mathrm e}^{2}-2 x}\) | \(40\) |
Input:
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)
Output:
((-50*exp(2)^2+25*exp(2))*x+100*exp(2)^2)/(-2+x)/(x+4*exp(2))
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 \, {\left (2 \, {\left (x - 2\right )} e^{4} - x e^{2}\right )}}{x^{2} + 4 \, {\left (x - 2\right )} e^{2} - 2 \, x} \] Input:
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")
Output:
-25*(2*(x - 2)*e^4 - x*e^2)/(x^2 + 4*(x - 2)*e^2 - 2*x)
Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=- \frac {x \left (- 25 e^{2} + 50 e^{4}\right ) - 100 e^{4}}{x^{2} + x \left (-2 + 4 e^{2}\right ) - 8 e^{2}} \] Input:
integrate(((50*x**2-200*x)*exp(2)**2-25*x**2*exp(2))/((16*x**2-64*x+64)*ex p(2)**2+(8*x**3-32*x**2+32*x)*exp(2)+x**4-4*x**3+4*x**2),x)
Output:
-(x*(-25*exp(2) + 50*exp(4)) - 100*exp(4))/(x**2 + x*(-2 + 4*exp(2)) - 8*e xp(2))
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 \, {\left (x {\left (2 \, e^{4} - e^{2}\right )} - 4 \, e^{4}\right )}}{x^{2} + 2 \, x {\left (2 \, e^{2} - 1\right )} - 8 \, e^{2}} \] Input:
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")
Output:
-25*(x*(2*e^4 - e^2) - 4*e^4)/(x^2 + 2*x*(2*e^2 - 1) - 8*e^2)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=-\frac {25 \, {\left (2 \, x e^{4} - x e^{2} - 4 \, e^{4}\right )}}{x^{2} + 4 \, x e^{2} - 2 \, x - 8 \, e^{2}} \] Input:
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")
Output:
-25*(2*x*e^4 - x*e^2 - 4*e^4)/(x^2 + 4*x*e^2 - 2*x - 8*e^2)
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25\,{\mathrm {e}}^2\,\left (x+4\,{\mathrm {e}}^2-2\,x\,{\mathrm {e}}^2\right )}{\left (x+4\,{\mathrm {e}}^2\right )\,\left (x-2\right )} \] Input:
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)
Output:
(25*exp(2)*(x + 4*exp(2) - 2*x*exp(2)))/((x + 4*exp(2))*(x - 2))
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-25 e^2 x^2+e^4 \left (-200 x+50 x^2\right )}{4 x^2-4 x^3+x^4+e^4 \left (64-64 x+16 x^2\right )+e^2 \left (32 x-32 x^2+8 x^3\right )} \, dx=\frac {25 e^{2} x^{2}}{8 e^{2} x -16 e^{2}+2 x^{2}-4 x} \] Input:
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)
Output:
(25*e**2*x**2)/(2*(4*e**2*x - 8*e**2 + x**2 - 2*x))