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

Optimal. Leaf size=17 \[ \frac {4}{e^{10} \left (1+\frac {625}{x}\right )+x} \]

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Rubi [A]  time = 0.08, antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1680, 12, 1814, 8} \begin {gather*} \frac {4 x}{x^2+e^{10} x+625 e^{10}} \end {gather*}

Antiderivative was successfully verified.

[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 1680

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 1814

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 {gather*} \begin {aligned} \text {integral} &=\operatorname {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 \operatorname {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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.94 \begin {gather*} \frac {4 x}{x^2+e^{10} (625+x)} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.52, size = 15, normalized size = 0.88 \begin {gather*} \frac {4 \, x}{x^{2} + {\left (x + 625\right )} e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (x^{2} - 625 \, e^{10}\right )}}{x^{4} + {\left (x^{2} + 1250 \, x + 390625\right )} e^{20} + 2 \, {\left (x^{3} + 625 \, x^{2}\right )} e^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

integrate(-4*(x^2 - 625*e^10)/(x^4 + (x^2 + 1250*x + 390625)*e^20 + 2*(x^3 + 625*x^2)*e^10), x)

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maple [A]  time = 0.17, size = 18, normalized size = 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	\(\frac {4 x}{x \,{\mathrm e}^{10}+625 \,{\mathrm e}^{10}+x^{2}}\)	\(22\)
default	\(2 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3} {\mathrm e}^{10}+\left ({\mathrm e}^{20}+1250 \,{\mathrm e}^{10}\right ) \textit {\_Z}^{2}+1250 \textit {\_Z} \,{\mathrm e}^{20}+390625 \,{\mathrm e}^{20}\right )}{\sum }\frac {\left (625 \,{\mathrm e}^{10}-\textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R} \,{\mathrm e}^{20}+625 \,{\mathrm e}^{20}+3 \,{\mathrm e}^{10} \textit {\_R}^{2}+1250 \,{\mathrm e}^{10} \textit {\_R} +2 \textit {\_R}^{3}}\right )\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.35, size = 17, normalized size = 1.00 \begin {gather*} \frac {4 \, x}{x^{2} + x e^{10} + 625 \, e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.19, size = 17, normalized size = 1.00 \begin {gather*} \frac {4\,x}{x^2+{\mathrm {e}}^{10}\,x+625\,{\mathrm {e}}^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.29, size = 15, normalized size = 0.88 \begin {gather*} \frac {4 x}{x^{2} + x e^{10} + 625 e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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