3.31.6 \(\int \frac {18750-1875 e^3+2700 x-675 x^2}{40000+625 e^6+10000 x+4225 x^2+450 x^3+81 x^4+e^3 (-10000-1250 x-450 x^2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {3 (-2+x)}{8-e^3+x+\frac {9 x^2}{25}} \]

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Rubi [A]  time = 0.08, antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 4, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {1680, 12, 1814, 8} \begin {gather*} -\frac {2700 (2-x)}{324 \left (x+\frac {25}{18}\right )^2+25 \left (263-36 e^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18750 - 1875*E^3 + 2700*x - 675*x^2)/(40000 + 625*E^6 + 10000*x + 4225*x^2 + 450*x^3 + 81*x^4 + E^3*(-100
00 - 1250*x - 450*x^2)),x]

[Out]

(-2700*(2 - x))/(25*(263 - 36*E^3) + 324*(25/18 + 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 {2700 \left (25 \left (263-36 e^3\right )+2196 x-324 x^2\right )}{\left (6575-900 e^3+324 x^2\right )^2} \, dx,x,\frac {25}{18}+x\right )\\ &=2700 \operatorname {Subst}\left (\int \frac {25 \left (263-36 e^3\right )+2196 x-324 x^2}{\left (6575-900 e^3+324 x^2\right )^2} \, dx,x,\frac {25}{18}+x\right )\\ &=-\frac {2700 (2-x)}{25 \left (263-36 e^3\right )+(25+18 x)^2}-\frac {54 \operatorname {Subst}\left (\int 0 \, dx,x,\frac {25}{18}+x\right )}{263-36 e^3}\\ &=-\frac {2700 (2-x)}{25 \left (263-36 e^3\right )+(25+18 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.09 \begin {gather*} -\frac {75 (2-x)}{200-25 e^3+25 x+9 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18750 - 1875*E^3 + 2700*x - 675*x^2)/(40000 + 625*E^6 + 10000*x + 4225*x^2 + 450*x^3 + 81*x^4 + E^3
*(-10000 - 1250*x - 450*x^2)),x]

[Out]

(-75*(2 - x))/(200 - 25*E^3 + 25*x + 9*x^2)

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fricas [A]  time = 0.88, size = 21, normalized size = 0.95 \begin {gather*} \frac {75 \, {\left (x - 2\right )}}{9 \, x^{2} + 25 \, x - 25 \, e^{3} + 200} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250*x-10000)*exp(3)+81*x^4+450*x^3+4225
*x^2+10000*x+40000),x, algorithm="fricas")

[Out]

75*(x - 2)/(9*x^2 + 25*x - 25*e^3 + 200)

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giac [B]  time = 0.33, size = 209, normalized size = 9.50 \begin {gather*} \frac {3 \, {\left (5 \, {\left (\sqrt {36 \, e^{3} - 263} + 5\right )}^{2} + 72 \, \sqrt {36 \, e^{3} - 263} + 180 \, e^{3} - 1440\right )} \log \left (x + \frac {5}{18} \, \sqrt {36 \, e^{3} - 263} + \frac {25}{18}\right )}{{\left (\sqrt {36 \, e^{3} - 263} + 5\right )}^{3} - 15 \, {\left (\sqrt {36 \, e^{3} - 263} + 5\right )}^{2} - 36 \, {\left (\sqrt {36 \, e^{3} - 263} + 5\right )} e^{3} + 338 \, \sqrt {36 \, e^{3} - 263} + 180 \, e^{3} + 250} - \frac {3 \, {\left (5 \, {\left (\sqrt {36 \, e^{3} - 263} - 5\right )}^{2} - 72 \, \sqrt {36 \, e^{3} - 263} + 180 \, e^{3} - 1440\right )} \log \left (x - \frac {5}{18} \, \sqrt {36 \, e^{3} - 263} + \frac {25}{18}\right )}{{\left (\sqrt {36 \, e^{3} - 263} - 5\right )}^{3} + 15 \, {\left (\sqrt {36 \, e^{3} - 263} - 5\right )}^{2} - 36 \, {\left (\sqrt {36 \, e^{3} - 263} - 5\right )} e^{3} + 338 \, \sqrt {36 \, e^{3} - 263} - 180 \, e^{3} - 250} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250*x-10000)*exp(3)+81*x^4+450*x^3+4225
*x^2+10000*x+40000),x, algorithm="giac")

[Out]

3*(5*(sqrt(36*e^3 - 263) + 5)^2 + 72*sqrt(36*e^3 - 263) + 180*e^3 - 1440)*log(x + 5/18*sqrt(36*e^3 - 263) + 25
/18)/((sqrt(36*e^3 - 263) + 5)^3 - 15*(sqrt(36*e^3 - 263) + 5)^2 - 36*(sqrt(36*e^3 - 263) + 5)*e^3 + 338*sqrt(
36*e^3 - 263) + 180*e^3 + 250) - 3*(5*(sqrt(36*e^3 - 263) - 5)^2 - 72*sqrt(36*e^3 - 263) + 180*e^3 - 1440)*log
(x - 5/18*sqrt(36*e^3 - 263) + 25/18)/((sqrt(36*e^3 - 263) - 5)^3 + 15*(sqrt(36*e^3 - 263) - 5)^2 - 36*(sqrt(3
6*e^3 - 263) - 5)*e^3 + 338*sqrt(36*e^3 - 263) - 180*e^3 - 250)

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maple [A]  time = 0.08, size = 21, normalized size = 0.95




method result size



risch \(\frac {-3 x +6}{-\frac {9 x^{2}}{25}+{\mathrm e}^{3}-x -8}\) \(21\)
gosper \(-\frac {75 \left (x -2\right )}{-9 x^{2}+25 \,{\mathrm e}^{3}-25 x -200}\) \(22\)
norman \(\frac {-75 x +150}{-9 x^{2}+25 \,{\mathrm e}^{3}-25 x -200}\) \(23\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250*x-10000)*exp(3)+81*x^4+450*x^3+4225*x^2+1
0000*x+40000),x,method=_RETURNVERBOSE)

[Out]

(-3*x+6)/(-9/25*x^2+exp(3)-x-8)

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maxima [A]  time = 0.42, size = 21, normalized size = 0.95 \begin {gather*} \frac {75 \, {\left (x - 2\right )}}{9 \, x^{2} + 25 \, x - 25 \, e^{3} + 200} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1875*exp(3)-675*x^2+2700*x+18750)/(625*exp(3)^2+(-450*x^2-1250*x-10000)*exp(3)+81*x^4+450*x^3+4225
*x^2+10000*x+40000),x, algorithm="maxima")

[Out]

75*(x - 2)/(9*x^2 + 25*x - 25*e^3 + 200)

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mupad [B]  time = 0.11, size = 22, normalized size = 1.00 \begin {gather*} \frac {75\,x-150}{9\,x^2+25\,x-25\,{\mathrm {e}}^3+200} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2700*x - 1875*exp(3) - 675*x^2 + 18750)/(10000*x + 625*exp(6) - exp(3)*(1250*x + 450*x^2 + 10000) + 4225*
x^2 + 450*x^3 + 81*x^4 + 40000),x)

[Out]

(75*x - 150)/(25*x - 25*exp(3) + 9*x^2 + 200)

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sympy [A]  time = 0.38, size = 20, normalized size = 0.91 \begin {gather*} - \frac {150 - 75 x}{9 x^{2} + 25 x - 25 e^{3} + 200} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1875*exp(3)-675*x**2+2700*x+18750)/(625*exp(3)**2+(-450*x**2-1250*x-10000)*exp(3)+81*x**4+450*x**3
+4225*x**2+10000*x+40000),x)

[Out]

-(150 - 75*x)/(9*x**2 + 25*x - 25*exp(3) + 200)

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