\(\int \frac {-4+e^{24} (256-4 x^2)+e^{48} (-4096-6 x+128 x^2-x^4)}{4+e^{24} (-256+4 x^2)+e^{48} (4096-128 x^2+x^4)} \, dx\) [5513]

Optimal result

Integrand size = 62, antiderivative size = 21 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=-\frac {2}{5}-x+\frac {3}{-64+\frac {2}{e^{24}}+x^2} \]

[Out]

3/(x^2-64+2/exp(24))-2/5-x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {2019, 28, 1828, 21, 8} \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=\frac {3 e^{24}}{e^{24} x^2+2 \left (1-32 e^{24}\right )}-x \]

[In]

Int[(-4 + E^24*(256 - 4*x^2) + E^48*(-4096 - 6*x + 128*x^2 - x^4))/(4 + E^24*(-256 + 4*x^2) + E^48*(4096 - 128
*x^2 + x^4)),x]

[Out]

-x + (3*E^24)/(2*(1 - 32*E^24) + E^24*x^2)

Rule 8

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 2019

Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && PolyQ[Pq, x] && TrinomialQ
[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4 \left (1-32 e^{24}\right )^2+4 e^{24} \left (1-32 e^{24}\right ) x^2+e^{48} x^4} \, dx \\ & = e^{48} \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{\left (2 e^{24} \left (1-32 e^{24}\right )+e^{48} x^2\right )^2} \, dx \\ & = \frac {3 e^{24}}{2 \left (1-32 e^{24}\right )+e^{24} x^2}-\frac {e^{24} \int \frac {8 \left (1-32 e^{24}\right )^2+4 e^{24} \left (1-32 e^{24}\right ) x^2}{2 e^{24} \left (1-32 e^{24}\right )+e^{48} x^2} \, dx}{4 \left (1-32 e^{24}\right )} \\ & = \frac {3 e^{24}}{2 \left (1-32 e^{24}\right )+e^{24} x^2}-\int 1 \, dx \\ & = -x+\frac {3 e^{24}}{2 \left (1-32 e^{24}\right )+e^{24} x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=-x+\frac {3 e^{24}}{2+e^{24} \left (-64+x^2\right )} \]

[In]

Integrate[(-4 + E^24*(256 - 4*x^2) + E^48*(-4096 - 6*x + 128*x^2 - x^4))/(4 + E^24*(-256 + 4*x^2) + E^48*(4096
- 128*x^2 + x^4)),x]

[Out]

-x + (3*E^24)/(2 + E^24*(-64 + x^2))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10

method	result	size
risch	\(-x +\frac {3 \,{\mathrm e}^{24}}{x^{2} {\mathrm e}^{24}-64 \,{\mathrm e}^{24}+2}\)	\(23\)
gosper	\(-\frac {x^{3} {\mathrm e}^{24}-64 x \,{\mathrm e}^{24}-3 \,{\mathrm e}^{24}+2 x}{x^{2} {\mathrm e}^{24}-64 \,{\mathrm e}^{24}+2}\)	\(36\)
parallelrisch	\(-\frac {\left (x^{3} {\mathrm e}^{48}-64 x \,{\mathrm e}^{48}-3 \,{\mathrm e}^{48}+2 x \,{\mathrm e}^{24}\right ) {\mathrm e}^{-24}}{x^{2} {\mathrm e}^{24}-64 \,{\mathrm e}^{24}+2}\)	\(48\)

[In]

int(((-x^4+128*x^2-6*x-4096)*exp(24)^2+(-4*x^2+256)*exp(24)-4)/((x^4-128*x^2+4096)*exp(24)^2+(4*x^2-256)*exp(2
4)+4),x,method=_RETURNVERBOSE)

[Out]

-x+3*exp(24)/(x^2*exp(24)-64*exp(24)+2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=-\frac {{\left (x^{3} - 64 \, x - 3\right )} e^{24} + 2 \, x}{{\left (x^{2} - 64\right )} e^{24} + 2} \]

[In]

integrate(((-x^4+128*x^2-6*x-4096)*exp(24)^2+(-4*x^2+256)*exp(24)-4)/((x^4-128*x^2+4096)*exp(24)^2+(4*x^2-256)
*exp(24)+4),x, algorithm="fricas")

[Out]

-((x^3 - 64*x - 3)*e^24 + 2*x)/((x^2 - 64)*e^24 + 2)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=- x + \frac {3 e^{24}}{x^{2} e^{24} - 64 e^{24} + 2} \]

[In]

integrate(((-x**4+128*x**2-6*x-4096)*exp(24)**2+(-4*x**2+256)*exp(24)-4)/((x**4-128*x**2+4096)*exp(24)**2+(4*x
**2-256)*exp(24)+4),x)

[Out]

-x + 3*exp(24)/(x**2*exp(24) - 64*exp(24) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=-x + \frac {3 \, e^{24}}{x^{2} e^{24} - 64 \, e^{24} + 2} \]

[In]

integrate(((-x^4+128*x^2-6*x-4096)*exp(24)^2+(-4*x^2+256)*exp(24)-4)/((x^4-128*x^2+4096)*exp(24)^2+(4*x^2-256)
*exp(24)+4),x, algorithm="maxima")

[Out]

-x + 3*e^24/(x^2*e^24 - 64*e^24 + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=-x + \frac {3 \, e^{24}}{x^{2} e^{24} - 64 \, e^{24} + 2} \]

[In]

integrate(((-x^4+128*x^2-6*x-4096)*exp(24)^2+(-4*x^2+256)*exp(24)-4)/((x^4-128*x^2+4096)*exp(24)^2+(4*x^2-256)
*exp(24)+4),x, algorithm="giac")

[Out]

-x + 3*e^24/(x^2*e^24 - 64*e^24 + 2)

Mupad [B] (verification not implemented)

Time = 12.49 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-4+e^{24} \left (256-4 x^2\right )+e^{48} \left (-4096-6 x+128 x^2-x^4\right )}{4+e^{24} \left (-256+4 x^2\right )+e^{48} \left (4096-128 x^2+x^4\right )} \, dx=\frac {3\,{\mathrm {e}}^{24}}{{\mathrm {e}}^{24}\,x^2-64\,{\mathrm {e}}^{24}+2}-x \]

[In]

int(-(exp(48)*(6*x - 128*x^2 + x^4 + 4096) + exp(24)*(4*x^2 - 256) + 4)/(exp(48)*(x^4 - 128*x^2 + 4096) + exp(
24)*(4*x^2 - 256) + 4),x)

[Out]

(3*exp(24))/(x^2*exp(24) - 64*exp(24) + 2) - x