\(\int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx\) [982]

Optimal result

Integrand size = 13, antiderivative size = 140 \[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=-\frac {x \left (1+x^2\right )}{\sqrt {-1+x^4}}+\frac {\sqrt {-1+x^4}}{x}+\frac {\sqrt {2} \sqrt {-1+x^2} \sqrt {1+x^2} E\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {-1+x^4}}-\frac {\sqrt {-1+x^2} \sqrt {1+x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {331, 312, 228, 1199} \[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=-\frac {\sqrt {x^2-1} \sqrt {x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4-1}}+\frac {\sqrt {2} \sqrt {x^2-1} \sqrt {x^2+1} E\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right )|\frac {1}{2}\right )}{\sqrt {x^4-1}}+\frac {\sqrt {x^4-1}}{x}-\frac {x \left (x^2+1\right )}{\sqrt {x^4-1}} \]

[In]

Int[1/(x^2*Sqrt[-1 + x^4]),x]

[Out]

-((x*(1 + x^2))/Sqrt[-1 + x^4]) + Sqrt[-1 + x^4]/x + (Sqrt[2]*Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticE[ArcSin[(S
qrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/Sqrt[-1 + x^4] - (Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticF[ArcSin[(Sqrt[2]*x)/S
qrt[-1 + x^2]], 1/2])/(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 312

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

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[e*x*((q + c*x
^2)/(c*Sqrt[a + c*x^4])), x] - Simp[Sqrt[2]*e*q*Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2)/q]/(Sqrt[-a]*c*Sqrt[a + c*x
^4]))*EllipticE[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; EqQ[c*d + e*q, 0] && IntegerQ[q]] /; FreeQ[{a,
c, d, e}, x] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-1+x^4}}{x}-\int \frac {x^2}{\sqrt {-1+x^4}} \, dx \\ & = \frac {\sqrt {-1+x^4}}{x}-\int \frac {1}{\sqrt {-1+x^4}} \, dx+\int \frac {1-x^2}{\sqrt {-1+x^4}} \, dx \\ & = -\frac {x \left (1+x^2\right )}{\sqrt {-1+x^4}}+\frac {\sqrt {-1+x^4}}{x}+\frac {\sqrt {2} \sqrt {-1+x^2} \sqrt {1+x^2} E\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {-1+x^4}}-\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 )}{\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 = 38, normalized size of antiderivative = 0.27 \[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=-\frac {\sqrt {1-x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},x^4\right )}{x \sqrt {-1+x^4}} \]

[In]

Integrate[1/(x^2*Sqrt[-1 + x^4]),x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 4.37 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.24

method	result	size
meijerg	\(-\frac {\sqrt {-\operatorname {signum}\left (x^{4}-1\right )}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};x^{4}\right )}{\sqrt {\operatorname {signum}\left (x^{4}-1\right )}\, x}\)	\(33\)
default	\(\frac {\sqrt {x^{4}-1}}{x}+\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (F\left (i x , i\right )-E\left (i x , i\right )\right )}{\sqrt {x^{4}-1}}\)	\(56\)
risch	\(\frac {\sqrt {x^{4}-1}}{x}+\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (F\left (i x , i\right )-E\left (i x , i\right )\right )}{\sqrt {x^{4}-1}}\)	\(56\)
elliptic	\(\frac {\sqrt {x^{4}-1}}{x}+\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (F\left (i x , i\right )-E\left (i x , i\right )\right )}{\sqrt {x^{4}-1}}\)	\(56\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.19 \[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=\frac {i \, x E(\arcsin \left (x\right )\,|\,-1) - i \, x F(\arcsin \left (x\right )\,|\,-1) + \sqrt {x^{4} - 1}}{x} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.21 \[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=- \frac {i \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} - 1} x^{2}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(x^4 - 1)*x^2), x)

Giac [F]

\[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} - 1} x^{2}} \,d x } \]

[In]

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

[Out]

integrate(1/(sqrt(x^4 - 1)*x^2), x)

Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.13 \[ \int \frac {1}{x^2 \sqrt {-1+x^4}} \, dx=-\frac {\sqrt {\frac {1}{x^4}}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ \frac {1}{x^4}\right )}{3\,x} \]

[In]

int(1/(x^2*(x^4 - 1)^(1/2)),x)

[Out]

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