3.10.63 \(\int \frac {x}{\sqrt {16-x^4}} \, dx\) [963]

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 = {281, 222} \begin {gather*} \frac {1}{2} \text {ArcSin}\left (\frac {x^2}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[16 - x^4],x]

[Out]

ArcSin[x^2/4]/2

Rule 222

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 281

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} \text {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 = 0.08, size = 20, normalized size = 1.67 \begin {gather*} -\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {16-x^4}}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[16 - x^4],x]

[Out]

-1/2*ArcTan[Sqrt[16 - x^4]/x^2]

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Maple [A]
time = 0.20, size = 9, normalized size = 0.75

method	result	size
default	\(\frac {\arcsin \left (\frac {x^{2}}{4}\right )}{2}\)	\(9\)
meijerg	\(\frac {\arcsin \left (\frac {x^{2}}{4}\right )}{2}\)	\(9\)
elliptic	\(\frac {\arcsin \left (\frac {x^{2}}{4}\right )}{2}\)	\(9\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}+16}+x^{2}\right )}{2}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 16, normalized size = 1.33 \begin {gather*} -\frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{4} + 16}}{x^{2}}\right ) \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (8) = 16\).
time = 0.36, size = 18, normalized size = 1.50 \begin {gather*} -\arctan \left (\frac {\sqrt {-x^{4} + 16} - 4}{x^{2}}\right ) \end {gather*}

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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Sympy [C] Result contains complex when optimal does not.
time = 0.41, size = 22, normalized size = 1.83 \begin {gather*} \begin {cases} - \frac {i \operatorname {acosh}{\left (\frac {x^{2}}{4} \right )}}{2} & \text {for}\: \left |{x^{4}}\right | > 16 \\\frac {\operatorname {asin}{\left (\frac {x^{2}}{4} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

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), (asin(x**2/4)/2, True))

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Giac [A]
time = 1.62, size = 8, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, \arcsin \left (\frac {1}{4} \, x^{2}\right ) \end {gather*}

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

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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