3.10.16 \(\int \frac {1}{(1-x^4)^{5/2}} \, dx\) [916]

Optimal. Leaf size=41 \[ \frac {x}{6 \left (1-x^4\right )^{3/2}}+\frac {5 x}{12 \sqrt {1-x^4}}+\frac {5}{12} F\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

1/6*x/(-x^4+1)^(3/2)+5/12*EllipticF(x,I)+5/12*x/(-x^4+1)^(1/2)

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Rubi [A]
time = 0.00, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {205, 227} \begin {gather*} \frac {5}{12} F(\text {ArcSin}(x)|-1)+\frac {5 x}{12 \sqrt {1-x^4}}+\frac {x}{6 \left (1-x^4\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^4)^(-5/2),x]

[Out]

x/(6*(1 - x^4)^(3/2)) + (5*x)/(12*Sqrt[1 - x^4]) + (5*EllipticF[ArcSin[x], -1])/12

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (1-x^4\right )^{5/2}} \, dx &=\frac {x}{6 \left (1-x^4\right )^{3/2}}+\frac {5}{6} \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx\\ &=\frac {x}{6 \left (1-x^4\right )^{3/2}}+\frac {5 x}{12 \sqrt {1-x^4}}+\frac {5}{12} \int \frac {1}{\sqrt {1-x^4}} \, dx\\ &=\frac {x}{6 \left (1-x^4\right )^{3/2}}+\frac {5 x}{12 \sqrt {1-x^4}}+\frac {5}{12} F\left (\left .\sin ^{-1}(x)\right |-1\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.28, size = 51, normalized size = 1.24 \begin {gather*} \frac {x}{6 \left (1-x^4\right )^{3/2}}+\frac {5 x}{12 \sqrt {1-x^4}}+\frac {5}{12} x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^4)^(-5/2),x]

[Out]

x/(6*(1 - x^4)^(3/2)) + (5*x)/(12*Sqrt[1 - x^4]) + (5*x*Hypergeometric2F1[1/4, 1/2, 5/4, x^4])/12

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (31 ) = 62\).
time = 0.17, size = 64, normalized size = 1.56

method	result	size
meijerg	\(x \hypergeom \left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {5}{4}\right ], x^{4}\right )\)	\(12\)
risch	\(\frac {x \left (5 x^{4}-7\right )}{12 \left (x^{4}-1\right ) \sqrt {-x^{4}+1}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{12 \sqrt {-x^{4}+1}}\)	\(59\)
default	\(\frac {x \sqrt {-x^{4}+1}}{6 \left (x^{4}-1\right )^{2}}+\frac {5 x}{12 \sqrt {-x^{4}+1}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{12 \sqrt {-x^{4}+1}}\)	\(64\)
elliptic	\(\frac {x \sqrt {-x^{4}+1}}{6 \left (x^{4}-1\right )^{2}}+\frac {5 x}{12 \sqrt {-x^{4}+1}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \EllipticF \left (x , i\right )}{12 \sqrt {-x^{4}+1}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(-x^4+1)^(1/2)/(x^4-1)^2+5/12*x/(-x^4+1)^(1/2)+5/12*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(1/2)*Elliptic
F(x,I)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-x^4 + 1)^(-5/2), x)

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Fricas [A]
time = 0.07, size = 51, normalized size = 1.24 \begin {gather*} \frac {5 \, {\left (x^{8} - 2 \, x^{4} + 1\right )} F(\arcsin \left (x\right )\,|\,-1) - {\left (5 \, x^{5} - 7 \, x\right )} \sqrt {-x^{4} + 1}}{12 \, {\left (x^{8} - 2 \, x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(5*(x^8 - 2*x^4 + 1)*elliptic_f(arcsin(x), -1) - (5*x^5 - 7*x)*sqrt(-x^4 + 1))/(x^8 - 2*x^4 + 1)

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Sympy [A]
time = 0.38, size = 29, normalized size = 0.71 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), x**4*exp_polar(2*I*pi))/(4*gamma(5/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-x^4 + 1)^(-5/2), x)

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Mupad [B]
time = 1.07, size = 10, normalized size = 0.24 \begin {gather*} x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {5}{2};\ \frac {5}{4};\ x^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - x^4)^(5/2),x)

[Out]

x*hypergeom([1/4, 5/2], 5/4, x^4)

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