\(\int \frac {1}{(1-x^4)^{3/2}} \, dx\) [907]

Optimal result

Integrand size = 11, antiderivative size = 25 \[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\frac {x}{2 \sqrt {1-x^4}}+\frac {1}{2} \operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

1/2*EllipticF(x,I)+1/2*x/(-x^4+1)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {205, 227} \[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\frac {1}{2} \operatorname {EllipticF}(\arcsin (x),-1)+\frac {x}{2 \sqrt {1-x^4}} \]

[In]

Int[(1 - x^4)^(-3/2),x]

[Out]

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

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*} \text {integral}& = \frac {x}{2 \sqrt {1-x^4}}+\frac {1}{2} \int \frac {1}{\sqrt {1-x^4}} \, dx \\ & = \frac {x}{2 \sqrt {1-x^4}}+\frac {1}{2} F\left (\left .\sin ^{-1}(x)\right |-1\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.42 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\frac {1}{2} x \left (\frac {1}{\sqrt {1-x^4}}+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},x^4\right )\right ) \]

[In]

Integrate[(1 - x^4)^(-3/2),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48

method	result	size
meijerg	\(x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {3}{2};\frac {5}{4};x^{4}\right )\)	\(12\)
default	\(\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{2 \sqrt {-x^{4}+1}}\)	\(45\)
risch	\(\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{2 \sqrt {-x^{4}+1}}\)	\(45\)
elliptic	\(\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{2 \sqrt {-x^{4}+1}}\)	\(45\)

[In]

int(1/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

x*hypergeom([1/4,3/2],[5/4],x^4)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\frac {{\left (x^{4} - 1\right )} F(\arcsin \left (x\right )\,|\,-1) - \sqrt {-x^{4} + 1} x}{2 \, {\left (x^{4} - 1\right )}} \]

[In]

integrate(1/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(1/(-x**4+1)**(3/2),x)

[Out]

x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), x**4*exp_polar(2*I*pi))/(4*gamma(5/4))

Maxima [F]

\[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((-x^4 + 1)^(-3/2), x)

Giac [F]

\[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate((-x^4 + 1)^(-3/2), x)

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\left (1-x^4\right )^{3/2}} \, dx=x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{2};\ \frac {5}{4};\ x^4\right ) \]

[In]

int(1/(1 - x^4)^(3/2),x)

[Out]

x*hypergeom([1/4, 3/2], 5/4, x^4)