\(\int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx\) [406]

Optimal result

Integrand size = 13, antiderivative size = 33 \[ \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx=\arctan \left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right ) \]

[Out]

arctan(x/(x^4-x^2)^(1/4))+arctanh(x/(x^4-x^2)^(1/4))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(33)=66\).

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.70, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2036, 335, 246, 218, 212, 209} \[ \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx=\frac {\sqrt {x} \sqrt [4]{x^2-1} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}+\frac {\sqrt {x} \sqrt [4]{x^2-1} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}} \]

[In]

Int[(-x^2 + x^4)^(-1/4),x]

[Out]

(Sqrt[x]*(-1 + x^2)^(1/4)*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4) + (Sqrt[x]*(-1 + x^2)^(1/4)*Arc
Tanh[Sqrt[x]/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{-1+x^2}} \, dx}{\sqrt [4]{-x^2+x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}} \\ & = \frac {\sqrt {x} \sqrt [4]{-1+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \left (\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )\right )}{\sqrt [4]{x^2 \left (-1+x^2\right )}} \]

[In]

Integrate[(-x^2 + x^4)^(-1/4),x]

[Out]

(Sqrt[x]*(-1 + x^2)^(1/4)*(ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] + ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)]))/(x^2*(-1 + x
^2))^(1/4)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00

method	result	size
meijerg	\(\frac {2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}} \sqrt {x}\, \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], x^{2}\right )}{\operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}}}\)	\(33\)
pseudoelliptic	\(-\frac {\ln \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}}-x}{x}\right )}{2}+\frac {\ln \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}}+x}{x}\right )}{2}-\arctan \left (\frac {\left (x^{4}-x^{2}\right )^{\frac {1}{4}}}{x}\right )\)	\(62\)
trager	\(-\frac {\ln \left (\frac {2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-x^{2}}\, x +2 x^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}-2 x^{3}+x}{x}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-x^{2}}\, x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x}{x}\right )}{2}\)	\(141\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (29) = 58\).

Time = 1.69 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.88 \[ \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {2 \, x^{3} + 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} - x^{2}} x - x + 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \]

[In]

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

[Out]

-1/2*arctan(2*((x^4 - x^2)^(1/4)*x^2 + (x^4 - x^2)^(3/4))/x) + 1/2*log((2*x^3 + 2*(x^4 - x^2)^(1/4)*x^2 + 2*sq
rt(x^4 - x^2)*x - x + 2*(x^4 - x^2)^(3/4))/x)

Sympy [F]

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

[In]

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

[Out]

Integral((x**4 - x**2)**(-1/4), x)

Maxima [F]

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

[In]

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

[Out]

integrate((x^4 - x^2)^(-1/4), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx=-\arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{2} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \]

[In]

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

[Out]

-arctan((-1/x^2 + 1)^(1/4)) + 1/2*log((-1/x^2 + 1)^(1/4) + 1) - 1/2*log(-(-1/x^2 + 1)^(1/4) + 1)

Mupad [B] (verification not implemented)

Time = 5.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx=\frac {2\,x\,{\left (1-x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ x^2\right )}{{\left (x^4-x^2\right )}^{1/4}} \]

[In]

int(1/(x^4 - x^2)^(1/4),x)

[Out]

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