3.46.90 \(\int \frac {-16-144 e^3+16 e^{e^9}+288 x}{x^2+81 e^6 x^2+e^{2 e^9} x^2-18 x^3+81 x^4+e^3 (18 x^2-162 x^3)+e^{e^9} (-2 x^2-18 e^3 x^2+18 x^3)} \, dx\) [4590]

Optimal. Leaf size=25 \[ \frac {16}{\left (1-e^{e^9}+9 \left (e^3-x\right )\right ) x} \]

[Out]

16/(-9*x+9*exp(3)-exp(exp(9))+1)/x

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Rubi [A]
time = 0.11, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 1694, 12, 267} \begin {gather*} \frac {16}{\left (-9 x-e^{e^9}+9 e^3+1\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 144*E^3 + 16*E^E^9 + 288*x)/(x^2 + 81*E^6*x^2 + E^(2*E^9)*x^2 - 18*x^3 + 81*x^4 + E^3*(18*x^2 - 162
*x^3) + E^E^9*(-2*x^2 - 18*E^3*x^2 + 18*x^3)),x]

[Out]

16/((1 + 9*E^3 - E^E^9 - 9*x)*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 267

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16-144 e^3+16 e^{e^9}+288 x}{e^{2 e^9} x^2+\left (1+81 e^6\right ) x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx\\ &=\int \frac {-16-144 e^3+16 e^{e^9}+288 x}{\left (1+81 e^6+e^{2 e^9}\right ) x^2-18 x^3+81 x^4+e^3 \left (18 x^2-162 x^3\right )+e^{e^9} \left (-2 x^2-18 e^3 x^2+18 x^3\right )} \, dx\\ &=\text {Subst}\left (\int \frac {373248 x}{\left (1+18 e^3+81 e^6-2 e^{e^9}+e^{2 e^9}-18 e^{3+e^9}-324 x^2\right )^2} \, dx,x,\frac {1}{324} \left (-18-162 e^3+18 e^{e^9}\right )+x\right )\\ &=373248 \text {Subst}\left (\int \frac {x}{\left (1+18 e^3+81 e^6-2 e^{e^9}+e^{2 e^9}-18 e^{3+e^9}-324 x^2\right )^2} \, dx,x,\frac {1}{324} \left (-18-162 e^3+18 e^{e^9}\right )+x\right )\\ &=\frac {16}{\left (1+9 e^3-e^{e^9}-9 x\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.02, size = 25, normalized size = 1.00 \begin {gather*} \frac {16}{x+9 e^3 x-e^{e^9} x-9 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 144*E^3 + 16*E^E^9 + 288*x)/(x^2 + 81*E^6*x^2 + E^(2*E^9)*x^2 - 18*x^3 + 81*x^4 + E^3*(18*x^2
- 162*x^3) + E^E^9*(-2*x^2 - 18*E^3*x^2 + 18*x^3)),x]

[Out]

16/(x + 9*E^3*x - E^E^9*x - 9*x^2)

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Maple [A]
time = 0.43, size = 22, normalized size = 0.88

method	result	size
gosper	\(\frac {16}{\left (-9 x +9 \,{\mathrm e}^{3}-{\mathrm e}^{{\mathrm e}^{9}}+1\right ) x}\)	\(22\)
norman	\(\frac {16}{\left (-9 x +9 \,{\mathrm e}^{3}-{\mathrm e}^{{\mathrm e}^{9}}+1\right ) x}\)	\(22\)
risch	\(\frac {16}{\left (-9 x +9 \,{\mathrm e}^{3}-{\mathrm e}^{{\mathrm e}^{9}}+1\right ) x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2*exp(3)+18*x^3-2*x^2)*exp(exp(9))+81*x^2*e
xp(3)^2+(-162*x^3+18*x^2)*exp(3)+81*x^4-18*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

16/(-9*x+9*exp(3)-exp(exp(9))+1)/x

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Maxima [A]
time = 0.29, size = 24, normalized size = 0.96 \begin {gather*} -\frac {16}{9 \, x^{2} - x {\left (9 \, e^{3} - e^{\left (e^{9}\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2*exp(3)+18*x^3-2*x^2)*exp(exp(9))+81
*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3)+81*x^4-18*x^3+x^2),x, algorithm="maxima")

[Out]

-16/(9*x^2 - x*(9*e^3 - e^(e^9) + 1))

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Fricas [A]
time = 0.34, size = 23, normalized size = 0.92 \begin {gather*} -\frac {16}{9 \, x^{2} - 9 \, x e^{3} + x e^{\left (e^{9}\right )} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2*exp(3)+18*x^3-2*x^2)*exp(exp(9))+81
*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3)+81*x^4-18*x^3+x^2),x, algorithm="fricas")

[Out]

-16/(9*x^2 - 9*x*e^3 + x*e^(e^9) - x)

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Sympy [A]
time = 0.43, size = 20, normalized size = 0.80 \begin {gather*} - \frac {16}{9 x^{2} + x \left (- 9 e^{3} - 1 + e^{e^{9}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x**2*exp(exp(9))**2+(-18*x**2*exp(3)+18*x**3-2*x**2)*exp(exp(9
))+81*x**2*exp(3)**2+(-162*x**3+18*x**2)*exp(3)+81*x**4-18*x**3+x**2),x)

[Out]

-16/(9*x**2 + x*(-9*exp(3) - 1 + exp(exp(9))))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(exp(9))-144*exp(3)+288*x-16)/(x^2*exp(exp(9))^2+(-18*x^2*exp(3)+18*x^3-2*x^2)*exp(exp(9))+81
*x^2*exp(3)^2+(-162*x^3+18*x^2)*exp(3)+81*x^4-18*x^3+x^2),x, algorithm="giac")

[Out]

undef

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((288*x - 144*exp(3) + 16*exp(exp(9)) - 16)/(x^2*exp(2*exp(9)) + exp(3)*(18*x^2 - 162*x^3) - exp(exp(9))*(1
8*x^2*exp(3) + 2*x^2 - 18*x^3) + 81*x^2*exp(6) + x^2 - 18*x^3 + 81*x^4),x)

[Out]

\text{Hanged}

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