\(\int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx\) [979]

Optimal result

Integrand size = 13, antiderivative size = 74 \[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=\frac {\sqrt {-1+x^4}}{3 x^3}+\frac {\sqrt {-1+x^2} \sqrt {1+x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right ),\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {-1+x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {331, 228} \[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=\frac {\sqrt {x^2-1} \sqrt {x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right ),\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {x^4-1}}+\frac {\sqrt {x^4-1}}{3 x^3} \]

[In]

Int[1/(x^4*Sqrt[-1 + x^4]),x]

[Out]

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

Rule 228

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*b, 2]}, Simp[Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2
)/q]/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]))*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

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 {\sqrt {-1+x^4}}{3 x^3}+\frac {1}{3} \int \frac {1}{\sqrt {-1+x^4}} \, dx \\ & = \frac {\sqrt {-1+x^4}}{3 x^3}+\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {-1+x^4}} \\ \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 = 40, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=-\frac {\sqrt {1-x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},x^4\right )}{3 x^3 \sqrt {-1+x^4}} \]

[In]

Integrate[1/(x^4*Sqrt[-1 + x^4]),x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.42 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.45

method	result	size
meijerg	\(-\frac {\sqrt {-\operatorname {signum}\left (x^{4}-1\right )}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};x^{4}\right )}{3 \sqrt {\operatorname {signum}\left (x^{4}-1\right )}\, x^{3}}\)	\(33\)
default	\(\frac {\sqrt {x^{4}-1}}{3 x^{3}}-\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, F\left (i x , i\right )}{3 \sqrt {x^{4}-1}}\)	\(47\)
risch	\(\frac {\sqrt {x^{4}-1}}{3 x^{3}}-\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, F\left (i x , i\right )}{3 \sqrt {x^{4}-1}}\)	\(47\)
elliptic	\(\frac {\sqrt {x^{4}-1}}{3 x^{3}}-\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, F\left (i x , i\right )}{3 \sqrt {x^{4}-1}}\)	\(47\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.30 \[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=\frac {-i \, x^{3} F(\arcsin \left (x\right )\,|\,-1) + \sqrt {x^{4} - 1}}{3 \, x^{3}} \]

[In]

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

[Out]

1/3*(-I*x^3*elliptic_f(arcsin(x), -1) + sqrt(x^4 - 1))/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=- \frac {i \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} - 1} x^{4}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(x^4 - 1)*x^4), x)

Giac [F]

\[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} - 1} x^{4}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(x^4 - 1)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \sqrt {-1+x^4}} \, dx=\int \frac {1}{x^4\,\sqrt {x^4-1}} \,d x \]

[In]

int(1/(x^4*(x^4 - 1)^(1/2)),x)

[Out]

int(1/(x^4*(x^4 - 1)^(1/2)), x)