\(\int \frac {\sqrt [4]{-1+x^4}}{x} \, dx\) [1161]

Optimal result

Integrand size = 13, antiderivative size = 86 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=\sqrt [4]{-1+x^4}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{-1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.60, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {272, 52, 65, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{2 \sqrt {2}}-\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{2 \sqrt {2}}+\sqrt [4]{x^4-1}+\frac {\log \left (\sqrt {x^4-1}-\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{4 \sqrt {2}}-\frac {\log \left (\sqrt {x^4-1}+\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{4 \sqrt {2}} \]

[In]

Int[(-1 + x^4)^(1/4)/x,x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

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

Rule 217

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

Rule 272

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

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

Rule 1179

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {\sqrt [4]{-1+x}}{x} \, dx,x,x^4\right ) \\ & = \sqrt [4]{-1+x^4}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{(-1+x)^{3/4} x} \, dx,x,x^4\right ) \\ & = \sqrt [4]{-1+x^4}-\text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right ) \\ & = \sqrt [4]{-1+x^4}-\frac {1}{2} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right ) \\ & = \sqrt [4]{-1+x^4}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )+\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )}{4 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )}{4 \sqrt {2}} \\ & = \sqrt [4]{-1+x^4}+\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}} \\ & = \sqrt [4]{-1+x^4}+\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} \sqrt [4]{-1+x^4}\right )}{2 \sqrt {2}}+\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{4 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=\frac {1}{4} \left (4 \sqrt [4]{-1+x^4}-\sqrt {2} \arctan \left (\frac {-1+\sqrt {-1+x^4}}{\sqrt {2} \sqrt [4]{-1+x^4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )\right ) \]

[In]

Integrate[(-1 + x^4)^(1/4)/x,x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 6.38 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74

method	result	size
meijerg	\(-\frac {\operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} \left (\Gamma \left (\frac {3}{4}\right ) x^{4} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 1, 1\right ], \left [2, 2\right ], x^{4}\right )-4 \left (4-3 \ln \left (2\right )+\frac {\pi }{2}+4 \ln \left (x \right )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{16 \Gamma \left (\frac {3}{4}\right ) {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\)	\(64\)
pseudoelliptic	\(\left (x^{4}-1\right )^{\frac {1}{4}}-\frac {\ln \left (\frac {-\left (x^{4}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {x^{4}-1}-1}{\left (x^{4}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {x^{4}-1}-1}\right ) \sqrt {2}}{8}-\frac {\arctan \left (\left (x^{4}-1\right )^{\frac {1}{4}} \sqrt {2}+1\right ) \sqrt {2}}{4}-\frac {\arctan \left (\left (x^{4}-1\right )^{\frac {1}{4}} \sqrt {2}-1\right ) \sqrt {2}}{4}\)	\(101\)
trager	\(\left (x^{4}-1\right )^{\frac {1}{4}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}}{x^{4}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {-2 \sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{4}}\right )}{4}\)	\(163\)

[In]

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

[Out]

-1/16/GAMMA(3/4)*signum(x^4-1)^(1/4)/(-signum(x^4-1))^(1/4)*(GAMMA(3/4)*x^4*hypergeom([3/4,1,1],[2,2],x^4)-4*(
4-3*ln(2)+1/2*Pi+4*ln(x)+I*Pi)*GAMMA(3/4))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=-\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) + {\left (x^{4} - 1\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

-(1/8*I + 1/8)*sqrt(2)*log((I + 1)*sqrt(2) + 2*(x^4 - 1)^(1/4)) + (1/8*I - 1/8)*sqrt(2)*log(-(I - 1)*sqrt(2) +
2*(x^4 - 1)^(1/4)) - (1/8*I - 1/8)*sqrt(2)*log((I - 1)*sqrt(2) + 2*(x^4 - 1)^(1/4)) + (1/8*I + 1/8)*sqrt(2)*l
og(-(I + 1)*sqrt(2) + 2*(x^4 - 1)^(1/4)) + (x^4 - 1)^(1/4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=- \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + {\left (x^{4} - 1\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^4 - 1)^(1/4))) - 1/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*
(x^4 - 1)^(1/4))) - 1/8*sqrt(2)*log(sqrt(2)*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) + 1/8*sqrt(2)*log(-sqrt(2)*(x
^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) + (x^4 - 1)^(1/4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + {\left (x^{4} - 1\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^4 - 1)^(1/4))) - 1/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*
(x^4 - 1)^(1/4))) - 1/8*sqrt(2)*log(sqrt(2)*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) + 1/8*sqrt(2)*log(-sqrt(2)*(x
^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) + (x^4 - 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx={\left (x^4-1\right )}^{1/4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^4-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^4-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int((x^4 - 1)^(1/4)/x,x)

[Out]

(x^4 - 1)^(1/4) - 2^(1/2)*atan(2^(1/2)*(x^4 - 1)^(1/4)*(1/2 + 1i/2))*(1/4 - 1i/4) - 2^(1/2)*atan(2^(1/2)*(x^4
- 1)^(1/4)*(1/2 - 1i/2))*(1/4 + 1i/4)