\(\int \frac {x}{\sqrt {1-x^4}} \, dx\) [880]

Optimal result

Integrand size = 13, antiderivative size = 8 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=\frac {\arcsin \left (x^2\right )}{2} \]

[Out]

1/2*arcsin(x^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {281, 222} \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=\frac {\arcsin \left (x^2\right )}{2} \]

[In]

Int[x/Sqrt[1 - x^4],x]

[Out]

ArcSin[x^2]/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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \sin ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=\frac {\arcsin \left (x^2\right )}{2} \]

[In]

Integrate[x/Sqrt[1 - x^4],x]

[Out]

ArcSin[x^2]/2

Maple [A] (verified)

Time = 4.06 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\arcsin \left (x^{2}\right )}{2}\)	\(7\)
meijerg	\(\frac {\arcsin \left (x^{2}\right )}{2}\)	\(7\)
elliptic	\(\frac {\arcsin \left (x^{2}\right )}{2}\)	\(7\)
pseudoelliptic	\(\frac {\arcsin \left (x^{2}\right )}{2}\)	\(7\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}+1}+x^{2}\right )}{2}\)	\(30\)

[In]

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

[Out]

1/2*arcsin(x^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (6) = 12\).

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.25 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=-\arctan \left (\frac {\sqrt {-x^{4} + 1} - 1}{x^{2}}\right ) \]

[In]

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

[Out]

-arctan((sqrt(-x^4 + 1) - 1)/x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 2.38 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=\begin {cases} - \frac {i \operatorname {acosh}{\left (x^{2} \right )}}{2} & \text {for}\: \left |{x^{4}}\right | > 1 \\\frac {\operatorname {asin}{\left (x^{2} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (6) = 12\).

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 2.00 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{4} + 1}}{x^{2}}\right ) \]

[In]

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

[Out]

-1/2*arctan(sqrt(-x^4 + 1)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=\frac {1}{2} \, \arcsin \left (x^{2}\right ) \]

[In]

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

[Out]

1/2*arcsin(x^2)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 2.00 \[ \int \frac {x}{\sqrt {1-x^4}} \, dx=\frac {\mathrm {atan}\left (\frac {x^2}{\sqrt {1-x^4}}\right )}{2} \]

[In]

int(x/(1 - x^4)^(1/2),x)

[Out]

atan(x^2/(1 - x^4)^(1/2))/2