\(\int \frac {x^5}{\sqrt {1-x^4}} \, dx\) [879]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=-\frac {1}{4} x^2 \sqrt {1-x^4}+\frac {\arcsin \left (x^2\right )}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {281, 327, 222} \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=\frac {\arcsin \left (x^2\right )}{4}-\frac {1}{4} x^2 \sqrt {1-x^4} \]

[In]

Int[x^5/Sqrt[1 - x^4],x]

[Out]

-1/4*(x^2*Sqrt[1 - x^4]) + ArcSin[x^2]/4

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]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2}} \, dx,x,x^2\right ) \\ & = -\frac {1}{4} x^2 \sqrt {1-x^4}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,x^2\right ) \\ & = -\frac {1}{4} x^2 \sqrt {1-x^4}+\frac {1}{4} \sin ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=-\frac {1}{4} x^2 \sqrt {1-x^4}+\frac {\arcsin \left (x^2\right )}{4} \]

[In]

Integrate[x^5/Sqrt[1 - x^4],x]

[Out]

-1/4*(x^2*Sqrt[1 - x^4]) + ArcSin[x^2]/4

Maple [A] (verified)

Time = 4.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\arcsin \left (x^{2}\right )}{4}-\frac {x^{2} \sqrt {-x^{4}+1}}{4}\)	\(22\)
elliptic	\(\frac {\arcsin \left (x^{2}\right )}{4}-\frac {x^{2} \sqrt {-x^{4}+1}}{4}\)	\(22\)
pseudoelliptic	\(\frac {\arcsin \left (x^{2}\right )}{4}-\frac {x^{2} \sqrt {-x^{4}+1}}{4}\)	\(22\)
risch	\(\frac {x^{2} \left (x^{4}-1\right )}{4 \sqrt {-x^{4}+1}}+\frac {\arcsin \left (x^{2}\right )}{4}\)	\(27\)
meijerg	\(\frac {i \left (i \sqrt {\pi }\, x^{2} \sqrt {-x^{4}+1}-i \sqrt {\pi }\, \arcsin \left (x^{2}\right )\right )}{4 \sqrt {\pi }}\)	\(36\)
trager	\(-\frac {x^{2} \sqrt {-x^{4}+1}}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}+1}+x^{2}\right )}{4}\)	\(45\)

[In]

int(x^5/(-x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=-\frac {1}{4} \, \sqrt {-x^{4} + 1} x^{2} - \frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{4} + 1} - 1}{x^{2}}\right ) \]

[In]

integrate(x^5/(-x^4+1)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(-x^4 + 1)*x^2 - 1/2*arctan((sqrt(-x^4 + 1) - 1)/x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.00 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=\begin {cases} - \frac {i x^{2} \sqrt {x^{4} - 1}}{4} - \frac {i \operatorname {acosh}{\left (x^{2} \right )}}{4} & \text {for}\: \left |{x^{4}}\right | > 1 \\\frac {x^{6}}{4 \sqrt {1 - x^{4}}} - \frac {x^{2}}{4 \sqrt {1 - x^{4}}} + \frac {\operatorname {asin}{\left (x^{2} \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5/(-x**4+1)**(1/2),x)

[Out]

Piecewise((-I*x**2*sqrt(x**4 - 1)/4 - I*acosh(x**2)/4, Abs(x**4) > 1), (x**6/(4*sqrt(1 - x**4)) - x**2/(4*sqrt
(1 - x**4)) + asin(x**2)/4, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (21) = 42\).

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=\frac {\sqrt {-x^{4} + 1}}{4 \, x^{2} {\left (\frac {x^{4} - 1}{x^{4}} - 1\right )}} - \frac {1}{4} \, \arctan \left (\frac {\sqrt {-x^{4} + 1}}{x^{2}}\right ) \]

[In]

integrate(x^5/(-x^4+1)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(-x^4 + 1)/(x^2*((x^4 - 1)/x^4 - 1)) - 1/4*arctan(sqrt(-x^4 + 1)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=-\frac {1}{4} \, \sqrt {-x^{4} + 1} x^{2} + \frac {1}{4} \, \arcsin \left (x^{2}\right ) \]

[In]

integrate(x^5/(-x^4+1)^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt(-x^4 + 1)*x^2 + 1/4*arcsin(x^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\sqrt {1-x^4}} \, dx=\int \frac {x^5}{\sqrt {1-x^4}} \,d x \]

[In]

int(x^5/(1 - x^4)^(1/2),x)

[Out]

int(x^5/(1 - x^4)^(1/2), x)