\(\int \frac {1}{x^6 (1-x^4)^{3/2}} \, dx\) [915]

Optimal result

Integrand size = 15, antiderivative size = 71 \[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=\frac {1}{2 x^5 \sqrt {1-x^4}}-\frac {7 \sqrt {1-x^4}}{10 x^5}-\frac {21 \sqrt {1-x^4}}{10 x}-\frac {21}{10} E(\arcsin (x)|-1)+\frac {21}{10} \operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

-21/10*EllipticE(x,I)+21/10*EllipticF(x,I)+1/2/x^5/(-x^4+1)^(1/2)-7/10*(-x^4+1)^(1/2)/x^5-21/10*(-x^4+1)^(1/2)
/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {296, 331, 313, 227, 1195, 435} \[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=\frac {21}{10} \operatorname {EllipticF}(\arcsin (x),-1)-\frac {21}{10} E(\arcsin (x)|-1)-\frac {21 \sqrt {1-x^4}}{10 x}-\frac {7 \sqrt {1-x^4}}{10 x^5}+\frac {1}{2 x^5 \sqrt {1-x^4}} \]

[In]

Int[1/(x^6*(1 - x^4)^(3/2)),x]

[Out]

1/(2*x^5*Sqrt[1 - x^4]) - (7*Sqrt[1 - x^4])/(10*x^5) - (21*Sqrt[1 - x^4])/(10*x) - (21*EllipticE[ArcSin[x], -1
])/10 + (21*EllipticF[ArcSin[x], -1])/10

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^5 \sqrt {1-x^4}}+\frac {7}{2} \int \frac {1}{x^6 \sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1-x^4}}-\frac {7 \sqrt {1-x^4}}{10 x^5}+\frac {21}{10} \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1-x^4}}-\frac {7 \sqrt {1-x^4}}{10 x^5}-\frac {21 \sqrt {1-x^4}}{10 x}-\frac {21}{10} \int \frac {x^2}{\sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1-x^4}}-\frac {7 \sqrt {1-x^4}}{10 x^5}-\frac {21 \sqrt {1-x^4}}{10 x}+\frac {21}{10} \int \frac {1}{\sqrt {1-x^4}} \, dx-\frac {21}{10} \int \frac {1+x^2}{\sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1-x^4}}-\frac {7 \sqrt {1-x^4}}{10 x^5}-\frac {21 \sqrt {1-x^4}}{10 x}+\frac {21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {21}{10} \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx \\ & = \frac {1}{2 x^5 \sqrt {1-x^4}}-\frac {7 \sqrt {1-x^4}}{10 x^5}-\frac {21 \sqrt {1-x^4}}{10 x}-\frac {21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {21}{10} 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 = 20, normalized size of antiderivative = 0.28 \[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},x^4\right )}{5 x^5} \]

[In]

Integrate[1/(x^6*(1 - x^4)^(3/2)),x]

[Out]

-1/5*Hypergeometric2F1[-5/4, 3/2, -1/4, x^4]/x^5

Maple [C] (verified)

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

Time = 4.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.21

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {5}{4},\frac {3}{2};-\frac {1}{4};x^{4}\right )}{5 x^{5}}\)	\(15\)
risch	\(\frac {21 x^{8}-14 x^{4}-2}{10 x^{5} \sqrt {-x^{4}+1}}+\frac {21 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{10 \sqrt {-x^{4}+1}}\)	\(66\)
default	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}-\frac {\sqrt {-x^{4}+1}}{5 x^{5}}-\frac {8 \sqrt {-x^{4}+1}}{5 x}+\frac {21 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{10 \sqrt {-x^{4}+1}}\)	\(82\)
elliptic	\(\frac {x^{3}}{2 \sqrt {-x^{4}+1}}-\frac {\sqrt {-x^{4}+1}}{5 x^{5}}-\frac {8 \sqrt {-x^{4}+1}}{5 x}+\frac {21 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{10 \sqrt {-x^{4}+1}}\)	\(82\)

[In]

int(1/x^6/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=-\frac {21 \, {\left (x^{9} - x^{5}\right )} E(\arcsin \left (x\right )\,|\,-1) - 21 \, {\left (x^{9} - x^{5}\right )} F(\arcsin \left (x\right )\,|\,-1) + {\left (21 \, x^{8} - 14 \, x^{4} - 2\right )} \sqrt {-x^{4} + 1}}{10 \, {\left (x^{9} - x^{5}\right )}} \]

[In]

integrate(1/x^6/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

-1/10*(21*(x^9 - x^5)*elliptic_e(arcsin(x), -1) - 21*(x^9 - x^5)*elliptic_f(arcsin(x), -1) + (21*x^8 - 14*x^4
- 2)*sqrt(-x^4 + 1))/(x^9 - x^5)

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.52 \[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \]

[In]

integrate(1/x**6/(-x**4+1)**(3/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-x^4 + 1)^(3/2)*x^6), x)

Giac [F]

\[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((-x^4 + 1)^(3/2)*x^6), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \left (1-x^4\right )^{3/2}} \, dx=\int \frac {1}{x^6\,{\left (1-x^4\right )}^{3/2}} \,d x \]

[In]

int(1/(x^6*(1 - x^4)^(3/2)),x)

[Out]

int(1/(x^6*(1 - x^4)^(3/2)), x)