3.963 \(\int \frac {x}{\sqrt {16-x^4}} \, dx\)

Optimal. Leaf size=12 \[ \frac {1}{2} \sin ^{-1}\left (\frac {x^2}{4}\right ) \]

[Out]

1/2*arcsin(1/4*x^2)

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {275, 216} \[ \frac {1}{2} \sin ^{-1}\left (\frac {x^2}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[16 - x^4],x]

[Out]

ArcSin[x^2/4]/2

Rule 216

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {16-x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {16-x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \sin ^{-1}\left (\frac {x^2}{4}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.00 \[ \frac {1}{2} \sin ^{-1}\left (\frac {x^2}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[16 - x^4],x]

[Out]

ArcSin[x^2/4]/2

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fricas [B]  time = 0.86, size = 18, normalized size = 1.50 \[ -\arctan \left (\frac {\sqrt {-x^{4} + 16} - 4}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x^4+16)^(1/2),x, algorithm="fricas")

[Out]

-arctan((sqrt(-x^4 + 16) - 4)/x^2)

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giac [A]  time = 0.17, size = 8, normalized size = 0.67 \[ \frac {1}{2} \, \arcsin \left (\frac {1}{4} \, x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x^4+16)^(1/2),x, algorithm="giac")

[Out]

1/2*arcsin(1/4*x^2)

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maple [A]  time = 0.01, size = 9, normalized size = 0.75 \[ \frac {\arcsin \left (\frac {x^{2}}{4}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-x^4+16)^(1/2),x)

[Out]

1/2*arcsin(1/4*x^2)

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maxima [A]  time = 2.96, size = 16, normalized size = 1.33 \[ -\frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{4} + 16}}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x^4+16)^(1/2),x, algorithm="maxima")

[Out]

-1/2*arctan(sqrt(-x^4 + 16)/x^2)

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mupad [B]  time = 0.17, size = 16, normalized size = 1.33 \[ \frac {\mathrm {atan}\left (\frac {x^2}{\sqrt {16-x^4}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(16 - x^4)^(1/2),x)

[Out]

atan(x^2/(16 - x^4)^(1/2))/2

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sympy [A]  time = 1.19, size = 24, normalized size = 2.00 \[ \begin {cases} - \frac {i \operatorname {acosh}{\left (\frac {x^{2}}{4} \right )}}{2} & \text {for}\: \frac {\left |{x^{4}}\right |}{16} > 1 \\\frac {\operatorname {asin}{\left (\frac {x^{2}}{4} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x**4+16)**(1/2),x)

[Out]

Piecewise((-I*acosh(x**2/4)/2, Abs(x**4)/16 > 1), (asin(x**2/4)/2, True))

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