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

Optimal. Leaf size=33 \[ \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^2}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^2}}\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 89, normalized size of antiderivative = 2.70, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2011, 329, 240, 212, 206, 203} \begin {gather*} \frac {\sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}+\frac {\sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}} \end {gather*}

Antiderivative was successfully verified.

[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 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 240

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 329

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 2011

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*} \int \frac {1}{\sqrt [4]{-x^2+x^4}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 61, normalized size = 1.85 \begin {gather*} \frac {\sqrt {x} \sqrt [4]{x^2-1} \left (\tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )+\tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )}{\sqrt [4]{x^2 \left (x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[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)

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IntegrateAlgebraic [A]  time = 0.11, size = 33, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-x^2 + x^4)^(-1/4),x]

[Out]

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

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fricas [B]  time = 3.03, size = 95, normalized size = 2.88 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.35, size = 41, normalized size = 1.24 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 1.30, size = 33, normalized size = 1.00

method	result	size
meijerg	\(\frac {2 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{4}} \hypergeom \left (\left [\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], x^{2}\right ) \sqrt {x}}{\mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{4}}}\)	\(33\)
trager	\(\frac {\ln \left (\frac {2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{2}}\, x +2 x^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}+2 x^{3}-x}{x}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-x^{2}}\, x +2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{2}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x}{x}\right )}{2}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 - x^2)^(-1/4), x)

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mupad [B]  time = 0.40, size = 31, normalized size = 0.94 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{x^{4} - x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**4 - x**2)**(-1/4), x)

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