3.51.10 \(\int \frac {16 x}{81-72 x^2+16 x^4} \, dx\)

Optimal. Leaf size=14 \[ -\frac {2}{\left (4-\frac {9}{x^2}\right ) x^2} \]

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Rubi [A]  time = 0.00, antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 28, 261} \begin {gather*} \frac {2}{9-4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x)/(81 - 72*x^2 + 16*x^4),x]

[Out]

2/(9 - 4*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=16 \int \frac {x}{81-72 x^2+16 x^4} \, dx\\ &=256 \int \frac {x}{\left (-36+16 x^2\right )^2} \, dx\\ &=\frac {2}{9-4 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.79 \begin {gather*} \frac {16}{72-32 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x)/(81 - 72*x^2 + 16*x^4),x]

[Out]

16/(72 - 32*x^2)

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fricas [A]  time = 0.98, size = 11, normalized size = 0.79 \begin {gather*} -\frac {2}{4 \, x^{2} - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*x/(16*x^4-72*x^2+81),x, algorithm="fricas")

[Out]

-2/(4*x^2 - 9)

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giac [A]  time = 0.15, size = 11, normalized size = 0.79 \begin {gather*} -\frac {2}{4 \, x^{2} - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*x/(16*x^4-72*x^2+81),x, algorithm="giac")

[Out]

-2/(4*x^2 - 9)

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maple [A]  time = 0.02, size = 10, normalized size = 0.71

method	result	size
risch	\(-\frac {1}{2 \left (x^{2}-\frac {9}{4}\right )}\)	\(10\)
gosper	\(-\frac {2}{4 x^{2}-9}\)	\(12\)
default	\(-\frac {2}{4 x^{2}-9}\)	\(12\)
norman	\(-\frac {2}{4 x^{2}-9}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16*x/(16*x^4-72*x^2+81),x,method=_RETURNVERBOSE)

[Out]

-1/2/(x^2-9/4)

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maxima [A]  time = 0.34, size = 11, normalized size = 0.79 \begin {gather*} -\frac {2}{4 \, x^{2} - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*x/(16*x^4-72*x^2+81),x, algorithm="maxima")

[Out]

-2/(4*x^2 - 9)

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mupad [B]  time = 0.04, size = 11, normalized size = 0.79 \begin {gather*} -\frac {1}{2\,\left (x^2-\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x)/(16*x^4 - 72*x^2 + 81),x)

[Out]

-1/(2*(x^2 - 9/4))

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sympy [A]  time = 0.07, size = 8, normalized size = 0.57 \begin {gather*} - \frac {16}{32 x^{2} - 72} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*x/(16*x**4-72*x**2+81),x)

[Out]

-16/(32*x**2 - 72)

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