\(\int \frac {1}{x^8 (1-x^4)^{3/2}} \, dx\) [909]

Optimal result

Integrand size = 15, antiderivative size = 63 \[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\frac {1}{2 x^7 \sqrt {1-x^4}}-\frac {9 \sqrt {1-x^4}}{14 x^7}-\frac {15 \sqrt {1-x^4}}{14 x^3}+\frac {15}{14} \operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

15/14*EllipticF(x,I)+1/2/x^7/(-x^4+1)^(1/2)-9/14*(-x^4+1)^(1/2)/x^7-15/14*(-x^4+1)^(1/2)/x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {296, 331, 227} \[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\frac {15}{14} \operatorname {EllipticF}(\arcsin (x),-1)-\frac {9 \sqrt {1-x^4}}{14 x^7}+\frac {1}{2 x^7 \sqrt {1-x^4}}-\frac {15 \sqrt {1-x^4}}{14 x^3} \]

[In]

Int[1/(x^8*(1 - x^4)^(3/2)),x]

[Out]

1/(2*x^7*Sqrt[1 - x^4]) - (9*Sqrt[1 - x^4])/(14*x^7) - (15*Sqrt[1 - x^4])/(14*x^3) + (15*EllipticF[ArcSin[x],
-1])/14

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 x^7 \sqrt {1-x^4}}+\frac {9}{2} \int \frac {1}{x^8 \sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^7 \sqrt {1-x^4}}-\frac {9 \sqrt {1-x^4}}{14 x^7}+\frac {45}{14} \int \frac {1}{x^4 \sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^7 \sqrt {1-x^4}}-\frac {9 \sqrt {1-x^4}}{14 x^7}-\frac {15 \sqrt {1-x^4}}{14 x^3}+\frac {15}{14} \int \frac {1}{\sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^7 \sqrt {1-x^4}}-\frac {9 \sqrt {1-x^4}}{14 x^7}-\frac {15 \sqrt {1-x^4}}{14 x^3}+\frac {15}{14} 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 = 0.32 \[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{2},-\frac {3}{4},x^4\right )}{7 x^7} \]

[In]

Integrate[1/(x^8*(1 - x^4)^(3/2)),x]

[Out]

-1/7*Hypergeometric2F1[-7/4, 3/2, -3/4, x^4]/x^7

Maple [C] (verified)

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

Time = 4.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.24

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {7}{4},\frac {3}{2};-\frac {3}{4};x^{4}\right )}{7 x^{7}}\)	\(15\)
risch	\(\frac {15 x^{8}-6 x^{4}-2}{14 x^{7} \sqrt {-x^{4}+1}}+\frac {15 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{14 \sqrt {-x^{4}+1}}\)	\(59\)
default	\(-\frac {\sqrt {-x^{4}+1}}{7 x^{7}}-\frac {4 \sqrt {-x^{4}+1}}{7 x^{3}}+\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {15 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{14 \sqrt {-x^{4}+1}}\)	\(73\)
elliptic	\(-\frac {\sqrt {-x^{4}+1}}{7 x^{7}}-\frac {4 \sqrt {-x^{4}+1}}{7 x^{3}}+\frac {x}{2 \sqrt {-x^{4}+1}}+\frac {15 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{14 \sqrt {-x^{4}+1}}\)	\(73\)

[In]

int(1/x^8/(-x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/7/x^7*hypergeom([-7/4,3/2],[-3/4],x^4)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\frac {15 \, {\left (x^{11} - x^{7}\right )} F(\arcsin \left (x\right )\,|\,-1) - {\left (15 \, x^{8} - 6 \, x^{4} - 2\right )} \sqrt {-x^{4} + 1}}{14 \, {\left (x^{11} - x^{7}\right )}} \]

[In]

integrate(1/x^8/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

1/14*(15*(x^11 - x^7)*elliptic_f(arcsin(x), -1) - (15*x^8 - 6*x^4 - 2)*sqrt(-x^4 + 1))/(x^11 - x^7)

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

[In]

integrate(1/x**8/(-x**4+1)**(3/2),x)

[Out]

gamma(-7/4)*hyper((-7/4, 3/2), (-3/4,), x**4*exp_polar(2*I*pi))/(4*x**7*gamma(-3/4))

Maxima [F]

\[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-x^4 + 1)^(3/2)*x^8), x)

Giac [F]

\[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + 1\right )}^{\frac {3}{2}} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((-x^4 + 1)^(3/2)*x^8), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^8 \left (1-x^4\right )^{3/2}} \, dx=\int \frac {1}{x^8\,{\left (1-x^4\right )}^{3/2}} \,d x \]

[In]

int(1/(x^8*(1 - x^4)^(3/2)),x)

[Out]

int(1/(x^8*(1 - x^4)^(3/2)), x)