3.56.94 \(\int -\frac {5}{18-12 x+2 x^2+e^{16 e^{-8 e^9}} (9-6 x+x^2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {5}{\left (2+e^{16 e^{-8 e^9}}\right ) (-3+x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 1981, 27, 32} \begin {gather*} -\frac {5}{\left (2+e^{16 e^{-8 e^9}}\right ) (3-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-5/(18 - 12*x + 2*x^2 + E^(16/E^(8*E^9))*(9 - 6*x + x^2)),x]

[Out]

-5/((2 + E^(16/E^(8*E^9)))*(3 - 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (5 \int \frac {1}{18-12 x+2 x^2+e^{16 e^{-8 e^9}} \left (9-6 x+x^2\right )} \, dx\right )\\ &=-\left (5 \int \frac {1}{9 \left (2+e^{16 e^{-8 e^9}}\right )-6 \left (2+e^{16 e^{-8 e^9}}\right ) x+\left (2+e^{16 e^{-8 e^9}}\right ) x^2} \, dx\right )\\ &=-\left (5 \int \frac {1}{\left (2+e^{16 e^{-8 e^9}}\right ) (-3+x)^2} \, dx\right )\\ &=-\frac {5 \int \frac {1}{(-3+x)^2} \, dx}{2+e^{16 e^{-8 e^9}}}\\ &=-\frac {5}{\left (2+e^{16 e^{-8 e^9}}\right ) (3-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {5}{\left (2+e^{16 e^{-8 e^9}}\right ) (-3+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-5/(18 - 12*x + 2*x^2 + E^(16/E^(8*E^9))*(9 - 6*x + x^2)),x]

[Out]

5/((2 + E^(16/E^(8*E^9)))*(-3 + x))

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fricas [A]  time = 0.60, size = 21, normalized size = 0.95 \begin {gather*} \frac {5}{{\left (x - 3\right )} e^{\left (16 \, e^{\left (-8 \, e^{9}\right )}\right )} + 2 \, x - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/((x^2-6*x+9)*exp(16/exp(4*exp(9))^2)+2*x^2-12*x+18),x, algorithm="fricas")

[Out]

5/((x - 3)*e^(16*e^(-8*e^9)) + 2*x - 6)

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giac [A]  time = 0.11, size = 19, normalized size = 0.86 \begin {gather*} \frac {5}{{\left (x - 3\right )} {\left (e^{\left (16 \, e^{\left (-8 \, e^{9}\right )}\right )} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/((x^2-6*x+9)*exp(16/exp(4*exp(9))^2)+2*x^2-12*x+18),x, algorithm="giac")

[Out]

5/((x - 3)*(e^(16*e^(-8*e^9)) + 2))

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maple [A]  time = 0.67, size = 20, normalized size = 0.91

method	result	size
norman	\(\frac {5}{\left ({\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}}+2\right ) \left (x -3\right )}\)	\(20\)
default	\(\frac {5}{\left ({\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}}+2\right ) \left (x -3\right )}\)	\(22\)
meijerg	\(-\frac {5 x}{9 \left ({\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}}+2\right ) \left (1-\frac {x}{3}\right )}\)	\(23\)
risch	\(\frac {5}{{\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}} x -3 \,{\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}}+2 x -6}\)	\(30\)
gosper	\(\frac {5}{{\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}} x -3 \,{\mathrm e}^{16 \,{\mathrm e}^{-8 \,{\mathrm e}^{9}}}+2 x -6}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/((x^2-6*x+9)*exp(16/exp(4*exp(9))^2)+2*x^2-12*x+18),x,method=_RETURNVERBOSE)

[Out]

5/(exp(1/exp(exp(9))^8)^16+2)/(x-3)

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maxima [A]  time = 0.38, size = 28, normalized size = 1.27 \begin {gather*} \frac {5}{x {\left (e^{\left (16 \, e^{\left (-8 \, e^{9}\right )}\right )} + 2\right )} - 3 \, e^{\left (16 \, e^{\left (-8 \, e^{9}\right )}\right )} - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/((x^2-6*x+9)*exp(16/exp(4*exp(9))^2)+2*x^2-12*x+18),x, algorithm="maxima")

[Out]

5/(x*(e^(16*e^(-8*e^9)) + 2) - 3*e^(16*e^(-8*e^9)) - 6)

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mupad [B]  time = 0.12, size = 19, normalized size = 0.86 \begin {gather*} \frac {5}{\left ({\mathrm {e}}^{16\,{\mathrm {e}}^{-8\,{\mathrm {e}}^9}}+2\right )\,\left (x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(exp(16*exp(-8*exp(9)))*(x^2 - 6*x + 9) - 12*x + 2*x^2 + 18),x)

[Out]

5/((exp(16*exp(-8*exp(9))) + 2)*(x - 3))

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sympy [A]  time = 0.21, size = 27, normalized size = 1.23 \begin {gather*} \frac {5}{x \left (e^{\frac {16}{e^{8 e^{9}}}} + 2\right ) - 6 - 3 e^{\frac {16}{e^{8 e^{9}}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/((x**2-6*x+9)*exp(16/exp(4*exp(9))**2)+2*x**2-12*x+18),x)

[Out]

5/(x*(exp(16*exp(-8*exp(9))) + 2) - 6 - 3*exp(16*exp(-8*exp(9))))

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