Integrand size = 13, antiderivative size = 86 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=\sqrt [4]{-1+x^4}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{-1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )}{2 \sqrt {2}} \]
(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)
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=\frac {1}{4} \left (4 \sqrt [4]{-1+x^4}-\sqrt {2} \arctan \left (\frac {-1+\sqrt {-1+x^4}}{\sqrt {2} \sqrt [4]{-1+x^4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )\right ) \]
(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
Time = 0.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.63, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {798, 60, 73, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt [4]{x^4-1}}{x} \, dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt [4]{x^4-1}}{x^4}dx^4\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-\int \frac {1}{x^4 \left (x^4-1\right )^{3/4}}dx^4\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \int \frac {1}{x^{16}+1}d\sqrt [4]{x^4-1}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \int \frac {1-x^8}{x^{16}+1}d\sqrt [4]{x^4-1}+\frac {1}{2} \int \frac {x^8+1}{x^{16}+1}d\sqrt [4]{x^4-1}\right )\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^8-\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}+\frac {1}{2} \int \frac {1}{x^8+\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}\right )+\frac {1}{2} \int \frac {1-x^8}{x^{16}+1}d\sqrt [4]{x^4-1}\right )\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-x^8-1}d\left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x^8-1}d\left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-x^8}{x^{16}+1}d\sqrt [4]{x^4-1}\right )\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \int \frac {1-x^8}{x^{16}+1}d\sqrt [4]{x^4-1}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{\sqrt {2}}\right )\right )\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt [4]{x^4-1}}{x^8-\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{x^8+\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{\sqrt {2}}\right )\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt [4]{x^4-1}}{x^8-\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{x^8+\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{\sqrt {2}}\right )\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt [4]{x^4-1}}{x^8-\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt [4]{x^4-1}+1}{x^8+\sqrt {2} \sqrt [4]{x^4-1}+1}d\sqrt [4]{x^4-1}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{\sqrt {2}}\right )\right )\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{4} \left (4 \sqrt [4]{x^4-1}-4 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^8+\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^8-\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{2 \sqrt {2}}\right )\right )\right )\) |
(4*(-1 + x^4)^(1/4) - 4*((-(ArcTan[1 - Sqrt[2]*(-1 + x^4)^(1/4)]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*(-1 + x^4)^(1/4)]/Sqrt[2])/2 + (-1/2*Log[1 + x^8 - Sq rt[2]*(-1 + x^4)^(1/4)]/Sqrt[2] + Log[1 + x^8 + Sqrt[2]*(-1 + x^4)^(1/4)]/ (2*Sqrt[2]))/2))/4
3.12.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[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]]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
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 | \(\left (x^{4}-1\right )^{\frac {1}{4}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}}{x^{4}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {-2 \sqrt {x^{4}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{4}}\right )}{4}\) | \(163\) |
-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))
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=-\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right ) + {\left (x^{4} - 1\right )}^{\frac {1}{4}} \]
-(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)*sq rt(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)) + (x^4 - 1)^(1/4)
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=- \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + {\left (x^{4} - 1\right )}^{\frac {1}{4}} \]
-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(sq rt(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)
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) + {\left (x^{4} - 1\right )}^{\frac {1}{4}} \]
-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(sq rt(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)
Time = 5.87 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt [4]{-1+x^4}}{x} \, dx={\left (x^4-1\right )}^{1/4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^4-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^4-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]