3.84.47 \(\int \frac {-9 e^6-162 x+81 x^2}{e^{12} x^2+81 x^4-54 x^5+9 x^6+e^6 (18 x^3-6 x^4)} \, dx\) [8347]

Optimal. Leaf size=26 \[ \frac {9}{\left (-3+\frac {e^6}{(-3+x) x}\right ) (-3+x) x^2} \]

[Out]

9/x^2/(-3+exp(3)^2/(-3+x)/x)/(-3+x)

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(26)=52\).
time = 0.12, antiderivative size = 57, normalized size of antiderivative = 2.19, number of steps used = 7, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {2099, 652, 632, 212} \begin {gather*} \frac {9}{e^6 x}-\frac {27 \left (3 \left (27+4 e^6\right )-\left (27+4 e^6\right ) x\right )}{e^6 \left (27+4 e^6\right ) \left (-3 x^2+9 x+e^6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*E^6 - 162*x + 81*x^2)/(E^12*x^2 + 81*x^4 - 54*x^5 + 9*x^6 + E^6*(18*x^3 - 6*x^4)),x]

[Out]

9/(E^6*x) - (27*(3*(27 + 4*E^6) - (27 + 4*E^6)*x))/(E^6*(27 + 4*E^6)*(E^6 + 9*x - 3*x^2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {9}{e^6 x^2}+\frac {27 \left (27+2 e^6-9 x\right )}{e^6 \left (e^6+9 x-3 x^2\right )^2}-\frac {27}{e^6 \left (e^6+9 x-3 x^2\right )}\right ) \, dx\\ &=\frac {9}{e^6 x}+\frac {27 \int \frac {27+2 e^6-9 x}{\left (e^6+9 x-3 x^2\right )^2} \, dx}{e^6}-\frac {27 \int \frac {1}{e^6+9 x-3 x^2} \, dx}{e^6}\\ &=\frac {9}{e^6 x}-\frac {27 \left (3 \left (27+4 e^6\right )-\left (27+4 e^6\right ) x\right )}{e^6 \left (27+4 e^6\right ) \left (e^6+9 x-3 x^2\right )}+\frac {27 \int \frac {1}{e^6+9 x-3 x^2} \, dx}{e^6}+\frac {54 \text {Subst}\left (\int \frac {1}{3 \left (27+4 e^6\right )-x^2} \, dx,x,9-6 x\right )}{e^6}\\ &=\frac {9}{e^6 x}-\frac {27 \left (3 \left (27+4 e^6\right )-\left (27+4 e^6\right ) x\right )}{e^6 \left (27+4 e^6\right ) \left (e^6+9 x-3 x^2\right )}+\frac {18 \sqrt {\frac {3}{27+4 e^6}} \tanh ^{-1}\left (\sqrt {\frac {3}{27+4 e^6}} (3-2 x)\right )}{e^6}-\frac {54 \text {Subst}\left (\int \frac {1}{3 \left (27+4 e^6\right )-x^2} \, dx,x,9-6 x\right )}{e^6}\\ &=\frac {9}{e^6 x}-\frac {27 \left (3 \left (27+4 e^6\right )-\left (27+4 e^6\right ) x\right )}{e^6 \left (27+4 e^6\right ) \left (e^6+9 x-3 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.03, size = 19, normalized size = 0.73 \begin {gather*} -\frac {9}{x \left (-e^6+3 (-3+x) x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*E^6 - 162*x + 81*x^2)/(E^12*x^2 + 81*x^4 - 54*x^5 + 9*x^6 + E^6*(18*x^3 - 6*x^4)),x]

[Out]

-9/(x*(-E^6 + 3*(-3 + x)*x))

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Maple [A]
time = 0.51, size = 19, normalized size = 0.73

method result size
risch \(\frac {9}{x \left ({\mathrm e}^{6}-3 x^{2}+9 x \right )}\) \(19\)
gosper \(\frac {9}{x \left ({\mathrm e}^{6}-3 x^{2}+9 x \right )}\) \(21\)
norman \(\frac {9}{x \left ({\mathrm e}^{6}-3 x^{2}+9 x \right )}\) \(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*exp(3)^2+81*x^2-162*x)/(x^2*exp(3)^4+(-6*x^4+18*x^3)*exp(3)^2+9*x^6-54*x^5+81*x^4),x,method=_RETURNVER
BOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 20, normalized size = 0.77 \begin {gather*} -\frac {9}{3 \, x^{3} - 9 \, x^{2} - x e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(3)^2+81*x^2-162*x)/(x^2*exp(3)^4+(-6*x^4+18*x^3)*exp(3)^2+9*x^6-54*x^5+81*x^4),x, algorithm=
"maxima")

[Out]

-9/(3*x^3 - 9*x^2 - x*e^6)

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Fricas [A]
time = 0.38, size = 20, normalized size = 0.77 \begin {gather*} -\frac {9}{3 \, x^{3} - 9 \, x^{2} - x e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(3)^2+81*x^2-162*x)/(x^2*exp(3)^4+(-6*x^4+18*x^3)*exp(3)^2+9*x^6-54*x^5+81*x^4),x, algorithm=
"fricas")

[Out]

-9/(3*x^3 - 9*x^2 - x*e^6)

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Sympy [A]
time = 0.32, size = 17, normalized size = 0.65 \begin {gather*} - \frac {9}{3 x^{3} - 9 x^{2} - x e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(3)**2+81*x**2-162*x)/(x**2*exp(3)**4+(-6*x**4+18*x**3)*exp(3)**2+9*x**6-54*x**5+81*x**4),x)

[Out]

-9/(3*x**3 - 9*x**2 - x*exp(6))

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Giac [A]
time = 0.42, size = 20, normalized size = 0.77 \begin {gather*} -\frac {9}{3 \, x^{3} - 9 \, x^{2} - x e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(3)^2+81*x^2-162*x)/(x^2*exp(3)^4+(-6*x^4+18*x^3)*exp(3)^2+9*x^6-54*x^5+81*x^4),x, algorithm=
"giac")

[Out]

-9/(3*x^3 - 9*x^2 - x*e^6)

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Mupad [B]
time = 5.26, size = 18, normalized size = 0.69 \begin {gather*} \frac {9}{x\,\left (-3\,x^2+9\,x+{\mathrm {e}}^6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(162*x + 9*exp(6) - 81*x^2)/(exp(6)*(18*x^3 - 6*x^4) + x^2*exp(12) + 81*x^4 - 54*x^5 + 9*x^6),x)

[Out]

9/(x*(9*x + exp(6) - 3*x^2))

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