\(\int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx\) [594]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=-\frac {\sqrt [4]{-1+x^4}}{x}-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {283, 338, 304, 209, 212} \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\sqrt [4]{x^4-1}}{x} \]

[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 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 283

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 304

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 338

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=-\frac {\sqrt [4]{-1+x^4}}{x}-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \]

[In]

Integrate[(-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

Maple [C] (warning: unable to verify)

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

Time = 2.87 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.70

method	result	size
meijerg	\(-\frac {\operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], x^{4}\right )}{{\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x}\)	\(33\)
risch	\(-\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}+\frac {{\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {3}{4}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{3 \operatorname {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\)	\(46\)
pseudoelliptic	\(\frac {-\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right ) x +\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}+x}{x}\right ) x +2 \arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right ) x -4 \left (x^{4}-1\right )^{\frac {1}{4}}}{4 x}\)	\(66\)
trager	\(-\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}+\frac {\ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+2 x^{4}-1\right )}{4}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{4}\)	\(127\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (37) = 74\).

Time = 1.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.81 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=\frac {x \arctan \left (2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right ) + x \log \left (2 \, x^{4} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} - 1} x^{2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1\right ) - 4 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{4 \, x} \]

[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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=\frac {e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{4} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{4} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 5.68 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx=-\frac {{\left (x^4-1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ x^4\right )}{x\,{\left (1-x^4\right )}^{1/4}} \]

[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))