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

Optimal. Leaf size=47 \[ -\frac {\sqrt [4]{x^4-1}}{x}-\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {277, 331, 298, 203, 206} \begin {gather*} -\frac {\sqrt [4]{x^4-1}}{x}-\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^4)^(1/4)/x^2,x]

[Out]

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

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx &=-\frac {\sqrt [4]{-1+x^4}}{x}+\int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{-1+x^4}}{x}+\operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {\sqrt [4]{-1+x^4}}{x}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {\sqrt [4]{-1+x^4}}{x}-\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.81 \begin {gather*} -\frac {\sqrt [4]{x^4-1} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};x^4\right )}{x \sqrt [4]{1-x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^4)^(1/4)/x^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 47, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{-1+x^4}}{x}-\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 2.33, size = 85, normalized size = 1.81 \begin {gather*} \frac {x \arctan \left (2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right ) + x \log \left (2 \, x^{4} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} - 1} x^{2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1\right ) - 4 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 59, normalized size = 1.26 \begin {gather*} \frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {1}{4} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.40, size = 33, normalized size = 0.70

method	result	size
meijerg	\(-\frac {\mathrm {signum}\left (x^{4}-1\right )^{\frac {1}{4}} \hypergeom \left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], x^{4}\right )}{\left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {1}{4}} x}\)	\(33\)
risch	\(-\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}+\frac {\left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right ) x^{3}}{3 \mathrm {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\)	\(46\)
trager	\(-\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}+\frac {\ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+2 x^{4}-1\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{4}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 59, normalized size = 1.26 \begin {gather*} -\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{4} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.68, size = 29, normalized size = 0.62 \begin {gather*} -\frac {{\left (x^4-1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ x^4\right )}{x\,{\left (1-x^4\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - 1)^(1/4)/x^2,x)

[Out]

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

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sympy [C]  time = 0.83, size = 34, normalized size = 0.72 \begin {gather*} \frac {e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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