\(\int \frac {1}{x^2 (1-x^4)^{3/2}} \, dx\) [914]

Optimal result

Integrand size = 15, antiderivative size = 53 \[ \int \frac {1}{x^2 \left (1-x^4\right )^{3/2}} \, dx=\frac {1}{2 x \sqrt {1-x^4}}-\frac {3 \sqrt {1-x^4}}{2 x}-\frac {3}{2} E(\arcsin (x)|-1)+\frac {3}{2} \operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

-3/2*EllipticE(x,I)+3/2*EllipticF(x,I)+1/2/x/(-x^4+1)^(1/2)-3/2*(-x^4+1)^(1/2)/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {296, 331, 313, 227, 1195, 435} \[ \int \frac {1}{x^2 \left (1-x^4\right )^{3/2}} \, dx=\frac {3}{2} \operatorname {EllipticF}(\arcsin (x),-1)-\frac {3}{2} E(\arcsin (x)|-1)-\frac {3 \sqrt {1-x^4}}{2 x}+\frac {1}{2 x \sqrt {1-x^4}} \]

[In]

Int[1/(x^2*(1 - x^4)^(3/2)),x]

[Out]

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

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]

Rule 296

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

Rule 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 331

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

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 x \sqrt {1-x^4}}+\frac {3}{2} \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x \sqrt {1-x^4}}-\frac {3 \sqrt {1-x^4}}{2 x}-\frac {3}{2} \int \frac {x^2}{\sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x \sqrt {1-x^4}}-\frac {3 \sqrt {1-x^4}}{2 x}+\frac {3}{2} \int \frac {1}{\sqrt {1-x^4}} \, dx-\frac {3}{2} \int \frac {1+x^2}{\sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x \sqrt {1-x^4}}-\frac {3 \sqrt {1-x^4}}{2 x}+\frac {3}{2} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {3}{2} \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx \\ & = \frac {1}{2 x \sqrt {1-x^4}}-\frac {3 \sqrt {1-x^4}}{2 x}-\frac {3}{2} E\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {3}{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 = 10.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^2 \left (1-x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},x^4\right )}{x} \]

[In]

Integrate[1/(x^2*(1 - x^4)^(3/2)),x]

[Out]

-(Hypergeometric2F1[-1/4, 3/2, 3/4, x^4]/x)

Maple [C] (verified)

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

Time = 4.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.28

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{4},\frac {3}{2};\frac {3}{4};x^{4}\right )}{x}\)	\(15\)
risch	\(\frac {3 x^{4}-2}{2 x \sqrt {-x^{4}+1}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(61\)
default	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}-\frac {\sqrt {-x^{4}+1}}{x}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(68\)
elliptic	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}-\frac {\sqrt {-x^{4}+1}}{x}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^2 \left (1-x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (x^{5} - x\right )} E(\arcsin \left (x\right )\,|\,-1) - 3 \, {\left (x^{5} - x\right )} F(\arcsin \left (x\right )\,|\,-1) + {\left (3 \, x^{4} - 2\right )} \sqrt {-x^{4} + 1}}{2 \, {\left (x^{5} - x\right )}} \]

[In]

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

[Out]

-1/2*(3*(x^5 - x)*elliptic_e(arcsin(x), -1) - 3*(x^5 - x)*elliptic_f(arcsin(x), -1) + (3*x^4 - 2)*sqrt(-x^4 +
1))/(x^5 - x)

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^2 \left (1-x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.45 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^2 \left (1-x^4\right )^{3/2}} \, dx=-\frac {{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{2};\ \frac {3}{4};\ x^4\right )}{x} \]

[In]

int(1/(x^2*(1 - x^4)^(3/2)),x)

[Out]

-hypergeom([-1/4, 3/2], 3/4, x^4)/x