3.9.17 \(\int \frac {-300+25 e^3-200 x}{144 x^2+e^6 x^2+96 x^3+16 x^4+e^3 (-24 x^2-8 x^3)} \, dx\)

Optimal. Leaf size=17 \[ \frac {25}{x \left (12-e^3+4 x\right )} \]

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Rubi [A]  time = 0.07, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6, 1680, 12, 261} \begin {gather*} \frac {25}{x \left (4 x-e^3+12\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-300 + 25*E^3 - 200*x)/(144*x^2 + E^6*x^2 + 96*x^3 + 16*x^4 + E^3*(-24*x^2 - 8*x^3)),x]

[Out]

25/(x*(12 - E^3 + 4*x))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-300+25 e^3-200 x}{\left (144+e^6\right ) x^2+96 x^3+16 x^4+e^3 \left (-24 x^2-8 x^3\right )} \, dx\\ &=\operatorname {Subst}\left (\int -\frac {51200 x}{\left (144-24 e^3+e^6-64 x^2\right )^2} \, dx,x,\frac {1}{64} \left (96-8 e^3\right )+x\right )\\ &=-\left (51200 \operatorname {Subst}\left (\int \frac {x}{\left (144-24 e^3+e^6-64 x^2\right )^2} \, dx,x,\frac {1}{64} \left (96-8 e^3\right )+x\right )\right )\\ &=\frac {25}{x \left (12-e^3+4 x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.12 \begin {gather*} \frac {25}{12 x-e^3 x+4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-300 + 25*E^3 - 200*x)/(144*x^2 + E^6*x^2 + 96*x^3 + 16*x^4 + E^3*(-24*x^2 - 8*x^3)),x]

[Out]

25/(12*x - E^3*x + 4*x^2)

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fricas [A]  time = 0.52, size = 18, normalized size = 1.06 \begin {gather*} \frac {25}{4 \, x^{2} - x e^{3} + 12 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(3)-200*x-300)/(x^2*exp(3)^2+(-8*x^3-24*x^2)*exp(3)+16*x^4+96*x^3+144*x^2),x, algorithm="fric
as")

[Out]

25/(4*x^2 - x*e^3 + 12*x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(3)-200*x-300)/(x^2*exp(3)^2+(-8*x^3-24*x^2)*exp(3)+16*x^4+96*x^3+144*x^2),x, algorithm="giac
")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -25*((-4*exp(6)+4*exp(3)^2)/(exp(6)^2-48
*exp(6)*exp(3)+288*exp(6)+576*exp(3)^2-6912*exp(3)+20736)*ln(16*sageVARx^2-8*sageVARx*exp(3)+96*sageVARx+exp(6
)-24*exp(3)+144)+(96*

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maple [A]  time = 14.43, size = 15, normalized size = 0.88

method	result	size
gosper	\(-\frac {25}{x \left (-12+{\mathrm e}^{3}-4 x \right )}\)	\(15\)
norman	\(-\frac {25}{x \left (-12+{\mathrm e}^{3}-4 x \right )}\)	\(15\)
risch	\(-\frac {25}{x \left (-12+{\mathrm e}^{3}-4 x \right )}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(3)-200*x-300)/(x^2*exp(3)^2+(-8*x^3-24*x^2)*exp(3)+16*x^4+96*x^3+144*x^2),x,method=_RETURNVERBOSE)

[Out]

-25/x/(-12+exp(3)-4*x)

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maxima [A]  time = 0.37, size = 17, normalized size = 1.00 \begin {gather*} \frac {25}{4 \, x^{2} - x {\left (e^{3} - 12\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(3)-200*x-300)/(x^2*exp(3)^2+(-8*x^3-24*x^2)*exp(3)+16*x^4+96*x^3+144*x^2),x, algorithm="maxi
ma")

[Out]

25/(4*x^2 - x*(e^3 - 12))

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mupad [B]  time = 0.74, size = 16, normalized size = 0.94 \begin {gather*} \frac {25}{x\,\left (4\,x-{\mathrm {e}}^3+12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(200*x - 25*exp(3) + 300)/(x^2*exp(6) - exp(3)*(24*x^2 + 8*x^3) + 144*x^2 + 96*x^3 + 16*x^4),x)

[Out]

25/(x*(4*x - exp(3) + 12))

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sympy [A]  time = 0.31, size = 12, normalized size = 0.71 \begin {gather*} \frac {25}{4 x^{2} + x \left (12 - e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(3)-200*x-300)/(x**2*exp(3)**2+(-8*x**3-24*x**2)*exp(3)+16*x**4+96*x**3+144*x**2),x)

[Out]

25/(4*x**2 + x*(12 - exp(3)))

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