\(\int \frac {1}{x^4 (1-x^4)^{3/2}} \, dx\) [908]

Optimal result

Integrand size = 15, antiderivative size = 45 \[ \int \frac {1}{x^4 \left (1-x^4\right )^{3/2}} \, dx=\frac {1}{2 x^3 \sqrt {1-x^4}}-\frac {5 \sqrt {1-x^4}}{6 x^3}+\frac {5}{6} \operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {296, 331, 227} \[ \int \frac {1}{x^4 \left (1-x^4\right )^{3/2}} \, dx=\frac {5}{6} \operatorname {EllipticF}(\arcsin (x),-1)-\frac {5 \sqrt {1-x^4}}{6 x^3}+\frac {1}{2 x^3 \sqrt {1-x^4}} \]

[In]

Int[1/(x^4*(1 - x^4)^(3/2)),x]

[Out]

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

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^3 \sqrt {1-x^4}}+\frac {5}{2} \int \frac {1}{x^4 \sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^3 \sqrt {1-x^4}}-\frac {5 \sqrt {1-x^4}}{6 x^3}+\frac {5}{6} \int \frac {1}{\sqrt {1-x^4}} \, dx \\ & = \frac {1}{2 x^3 \sqrt {1-x^4}}-\frac {5 \sqrt {1-x^4}}{6 x^3}+\frac {5}{6} 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 = 0.44 \[ \int \frac {1}{x^4 \left (1-x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},x^4\right )}{3 x^3} \]

[In]

Integrate[1/(x^4*(1 - x^4)^(3/2)),x]

[Out]

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

Maple [C] (verified)

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

Time = 4.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.33

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};x^{4}\right )}{3 x^{3}}\)	\(15\)
risch	\(\frac {5 x^{4}-2}{6 x^{3} \sqrt {-x^{4}+1}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{6 \sqrt {-x^{4}+1}}\)	\(54\)
default	\(\frac {x}{2 \sqrt {-x^{4}+1}}-\frac {\sqrt {-x^{4}+1}}{3 x^{3}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{6 \sqrt {-x^{4}+1}}\)	\(59\)
elliptic	\(\frac {x}{2 \sqrt {-x^{4}+1}}-\frac {\sqrt {-x^{4}+1}}{3 x^{3}}+\frac {5 \sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, F\left (x , i\right )}{6 \sqrt {-x^{4}+1}}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^4 \left (1-x^4\right )^{3/2}} \, dx=\frac {5 \, {\left (x^{7} - x^{3}\right )} F(\arcsin \left (x\right )\,|\,-1) - {\left (5 \, x^{4} - 2\right )} \sqrt {-x^{4} + 1}}{6 \, {\left (x^{7} - x^{3}\right )}} \]

[In]

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

[Out]

1/6*(5*(x^7 - x^3)*elliptic_f(arcsin(x), -1) - (5*x^4 - 2)*sqrt(-x^4 + 1))/(x^7 - x^3)

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^4 \left (1-x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(x^4*(1 - x^4)^(3/2)),x)

[Out]

int(1/(x^4*(1 - x^4)^(3/2)), x)