\(\int \frac {x^5}{\sqrt {16-x^4}} \, dx\) [961]

Optimal result

Integrand size = 15, antiderivative size = 29 \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=-\frac {1}{4} x^2 \sqrt {16-x^4}+4 \arcsin \left (\frac {x^2}{4}\right ) \]

[Out]

4*arcsin(1/4*x^2)-1/4*x^2*(-x^4+16)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, 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} \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=4 \arcsin \left (\frac {x^2}{4}\right )-\frac {1}{4} x^2 \sqrt {16-x^4} \]

[In]

Int[x^5/Sqrt[16 - x^4],x]

[Out]

-1/4*(x^2*Sqrt[16 - x^4]) + 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 {16-x^2}} \, dx,x,x^2\right ) \\ & = -\frac {1}{4} x^2 \sqrt {16-x^4}+4 \text {Subst}\left (\int \frac {1}{\sqrt {16-x^2}} \, dx,x,x^2\right ) \\ & = -\frac {1}{4} x^2 \sqrt {16-x^4}+4 \sin ^{-1}\left (\frac {x^2}{4}\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=-\frac {1}{4} x^2 \sqrt {16-x^4}-4 i \log \left (i x^2+\sqrt {16-x^4}\right ) \]

[In]

Integrate[x^5/Sqrt[16 - x^4],x]

[Out]

-1/4*(x^2*Sqrt[16 - x^4]) - (4*I)*Log[I*x^2 + Sqrt[16 - x^4]]

Maple [A] (verified)

Time = 4.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
default	\(4 \arcsin \left (\frac {x^{2}}{4}\right )-\frac {x^{2} \sqrt {-x^{4}+16}}{4}\)	\(24\)
elliptic	\(4 \arcsin \left (\frac {x^{2}}{4}\right )-\frac {x^{2} \sqrt {-x^{4}+16}}{4}\)	\(24\)
pseudoelliptic	\(4 \arcsin \left (\frac {x^{2}}{4}\right )-\frac {x^{2} \sqrt {-x^{4}+16}}{4}\)	\(24\)
risch	\(\frac {x^{2} \left (x^{4}-16\right )}{4 \sqrt {-x^{4}+16}}+4 \arcsin \left (\frac {x^{2}}{4}\right )\)	\(29\)
meijerg	\(\frac {4 i \left (\frac {i \sqrt {\pi }\, x^{2} \sqrt {1-\frac {x^{4}}{16}}}{4}-i \sqrt {\pi }\, \arcsin \left (\frac {x^{2}}{4}\right )\right )}{\sqrt {\pi }}\)	\(38\)
trager	\(-\frac {x^{2} \sqrt {-x^{4}+16}}{4}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}+16}+x^{2}\right )\)	\(46\)

[In]

int(x^5/(-x^4+16)^(1/2),x,method=_RETURNVERBOSE)

[Out]

4*arcsin(1/4*x^2)-1/4*x^2*(-x^4+16)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=-\frac {1}{4} \, \sqrt {-x^{4} + 16} x^{2} - 8 \, \arctan \left (\frac {\sqrt {-x^{4} + 16} - 4}{x^{2}}\right ) \]

[In]

integrate(x^5/(-x^4+16)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(-x^4 + 16)*x^2 - 8*arctan((sqrt(-x^4 + 16) - 4)/x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.69 \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=\begin {cases} - \frac {i x^{6}}{4 \sqrt {x^{4} - 16}} + \frac {4 i x^{2}}{\sqrt {x^{4} - 16}} - 4 i \operatorname {acosh}{\left (\frac {x^{2}}{4} \right )} & \text {for}\: \left |{x^{4}}\right | > 16 \\\frac {x^{6}}{4 \sqrt {16 - x^{4}}} - \frac {4 x^{2}}{\sqrt {16 - x^{4}}} + 4 \operatorname {asin}{\left (\frac {x^{2}}{4} \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5/(-x**4+16)**(1/2),x)

[Out]

Piecewise((-I*x**6/(4*sqrt(x**4 - 16)) + 4*I*x**2/sqrt(x**4 - 16) - 4*I*acosh(x**2/4), Abs(x**4) > 16), (x**6/
(4*sqrt(16 - x**4)) - 4*x**2/sqrt(16 - x**4) + 4*asin(x**2/4), True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=\frac {4 \, \sqrt {-x^{4} + 16}}{x^{2} {\left (\frac {x^{4} - 16}{x^{4}} - 1\right )}} - 4 \, \arctan \left (\frac {\sqrt {-x^{4} + 16}}{x^{2}}\right ) \]

[In]

integrate(x^5/(-x^4+16)^(1/2),x, algorithm="maxima")

[Out]

4*sqrt(-x^4 + 16)/(x^2*((x^4 - 16)/x^4 - 1)) - 4*arctan(sqrt(-x^4 + 16)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=-\frac {1}{4} \, \sqrt {-x^{4} + 16} x^{2} + 4 \, \arcsin \left (\frac {1}{4} \, x^{2}\right ) \]

[In]

integrate(x^5/(-x^4+16)^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt(-x^4 + 16)*x^2 + 4*arcsin(1/4*x^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\sqrt {16-x^4}} \, dx=\int \frac {x^5}{\sqrt {16-x^4}} \,d x \]

[In]

int(x^5/(16 - x^4)^(1/2),x)

[Out]

int(x^5/(16 - x^4)^(1/2), x)