\(\int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} (390625+1250 x+x^2)+e^{10} (1250 x^2+2 x^3)} \, dx\) [9163]

Optimal result

Integrand size = 45, antiderivative size = 17 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4}{e^{10} \left (1+\frac {625}{x}\right )+x} \]

[Out]

4/((1+625/x)*exp(5)^2+x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1694, 12, 1828, 8} \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4 x}{x^2+e^{10} x+625 e^{10}} \]

[In]

Int[(2500*E^10 - 4*x^2)/(x^4 + E^20*(390625 + 1250*x + x^2) + E^10*(1250*x^2 + 2*x^3)),x]

[Out]

(4*x)/(625*E^10 + E^10*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 {16 \left (e^{10} \left (2500-e^{10}\right )+4 e^{10} x-4 x^2\right )}{\left (2500 e^{10}-e^{20}+4 x^2\right )^2} \, dx,x,\frac {e^{10}}{2}+x\right ) \\ & = 16 \text {Subst}\left (\int \frac {e^{10} \left (2500-e^{10}\right )+4 e^{10} x-4 x^2}{\left (2500 e^{10}-e^{20}+4 x^2\right )^2} \, dx,x,\frac {e^{10}}{2}+x\right ) \\ & = \frac {4 x}{625 e^{10}+e^{10} x+x^2}-\frac {8 \text {Subst}\left (\int 0 \, dx,x,\frac {e^{10}}{2}+x\right )}{e^{10} \left (2500-e^{10}\right )} \\ & = \frac {4 x}{625 e^{10}+e^{10} x+x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4 x}{x^2+e^{10} (625+x)} \]

[In]

Integrate[(2500*E^10 - 4*x^2)/(x^4 + E^20*(390625 + 1250*x + x^2) + E^10*(1250*x^2 + 2*x^3)),x]

[Out]

(4*x)/(x^2 + E^10*(625 + x))

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {4 x}{x \,{\mathrm e}^{10}+625 \,{\mathrm e}^{10}+x^{2}}\)	\(18\)
gosper	\(\frac {4 x}{x \,{\mathrm e}^{10}+625 \,{\mathrm e}^{10}+x^{2}}\)	\(22\)
norman	\(\frac {4 x}{x \,{\mathrm e}^{10}+625 \,{\mathrm e}^{10}+x^{2}}\)	\(22\)
parallelrisch	\(\frac {4 x}{x \,{\mathrm e}^{10}+625 \,{\mathrm e}^{10}+x^{2}}\)	\(22\)
default	\(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3} {\mathrm e}^{10}+\left ({\mathrm e}^{20}+1250 \,{\mathrm e}^{10}\right ) \textit {\_Z}^{2}+1250 \textit {\_Z} \,{\mathrm e}^{20}+390625 \,{\mathrm e}^{20}\right )}{\sum }\frac {\left (625 \,{\mathrm e}^{10}-\textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R} \,{\mathrm e}^{20}+625 \,{\mathrm e}^{20}+3 \textit {\_R}^{2} {\mathrm e}^{10}+1250 \textit {\_R} \,{\mathrm e}^{10}+2 \textit {\_R}^{3}}\right )\)	\(83\)

[In]

int((2500*exp(5)^2-4*x^2)/((x^2+1250*x+390625)*exp(5)^4+(2*x^3+1250*x^2)*exp(5)^2+x^4),x,method=_RETURNVERBOSE
)

[Out]

4*x/(x*exp(10)+625*exp(10)+x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4 \, x}{x^{2} + {\left (x + 625\right )} e^{10}} \]

[In]

integrate((2500*exp(5)^2-4*x^2)/((x^2+1250*x+390625)*exp(5)^4+(2*x^3+1250*x^2)*exp(5)^2+x^4),x, algorithm="fri
cas")

[Out]

4*x/(x^2 + (x + 625)*e^10)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4 x}{x^{2} + x e^{10} + 625 e^{10}} \]

[In]

integrate((2500*exp(5)**2-4*x**2)/((x**2+1250*x+390625)*exp(5)**4+(2*x**3+1250*x**2)*exp(5)**2+x**4),x)

[Out]

4*x/(x**2 + x*exp(10) + 625*exp(10))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4 \, x}{x^{2} + x e^{10} + 625 \, e^{10}} \]

[In]

integrate((2500*exp(5)^2-4*x^2)/((x^2+1250*x+390625)*exp(5)^4+(2*x^3+1250*x^2)*exp(5)^2+x^4),x, algorithm="max
ima")

[Out]

4*x/(x^2 + x*e^10 + 625*e^10)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4 \, x}{x^{2} + x e^{10} + 625 \, e^{10}} \]

[In]

integrate((2500*exp(5)^2-4*x^2)/((x^2+1250*x+390625)*exp(5)^4+(2*x^3+1250*x^2)*exp(5)^2+x^4),x, algorithm="gia
c")

[Out]

4*x/(x^2 + x*e^10 + 625*e^10)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2500 e^{10}-4 x^2}{x^4+e^{20} \left (390625+1250 x+x^2\right )+e^{10} \left (1250 x^2+2 x^3\right )} \, dx=\frac {4\,x}{x^2+{\mathrm {e}}^{10}\,x+625\,{\mathrm {e}}^{10}} \]

[In]

int((2500*exp(10) - 4*x^2)/(exp(10)*(1250*x^2 + 2*x^3) + exp(20)*(1250*x + x^2 + 390625) + x^4),x)

[Out]

(4*x)/(625*exp(10) + x*exp(10) + x^2)