3.13.84 \(\int \frac {\sqrt [4]{-1+x^4}}{x^5} \, dx\)

Optimal. Leaf size=93 \[ -\frac {\sqrt [4]{x^4-1}}{4 x^4}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^4-1}}{\sqrt {x^4-1}-1}\right )}{8 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^4-1}}{\sqrt {x^4-1}+1}\right )}{8 \sqrt {2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 145, normalized size of antiderivative = 1.56, number of steps used = 12, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {266, 47, 63, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {\sqrt [4]{x^4-1}}{4 x^4}-\frac {\log \left (\sqrt {x^4-1}-\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{16 \sqrt {2}}+\frac {\log \left (\sqrt {x^4-1}+\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{16 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{x^4-1}\right )}{8 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt [4]{x^4-1}+1\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^4)^(1/4)/x^5,x]

[Out]

-1/4*(-1 + x^4)^(1/4)/x^4 - ArcTan[1 - Sqrt[2]*(-1 + x^4)^(1/4)]/(8*Sqrt[2]) + ArcTan[1 + Sqrt[2]*(-1 + x^4)^(
1/4)]/(8*Sqrt[2]) - Log[1 - Sqrt[2]*(-1 + x^4)^(1/4) + Sqrt[-1 + x^4]]/(16*Sqrt[2]) + Log[1 + Sqrt[2]*(-1 + x^
4)^(1/4) + Sqrt[-1 + x^4]]/(16*Sqrt[2])

Rule 47

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

Rule 63

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 204

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

Rule 211

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 266

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 617

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 628

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 1162

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 1165

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*} \int \frac {\sqrt [4]{-1+x^4}}{x^5} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt [4]{-1+x}}{x^2} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{-1+x^4}}{4 x^4}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{3/4} x} \, dx,x,x^4\right )\\ &=-\frac {\sqrt [4]{-1+x^4}}{4 x^4}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right )\\ &=-\frac {\sqrt [4]{-1+x^4}}{4 x^4}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^4}\right )\\ &=-\frac {\sqrt [4]{-1+x^4}}{4 x^4}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )+\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )}{16 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^4}\right )}{16 \sqrt {2}}\\ &=-\frac {\sqrt [4]{-1+x^4}}{4 x^4}-\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{16 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{16 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+x^4}\right )}{8 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+x^4}\right )}{8 \sqrt {2}}\\ &=-\frac {\sqrt [4]{-1+x^4}}{4 x^4}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{-1+x^4}\right )}{8 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt [4]{-1+x^4}\right )}{8 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{16 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+x^4}+\sqrt {-1+x^4}\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 28, normalized size = 0.30 \begin {gather*} \frac {1}{5} \left (x^4-1\right )^{5/4} \, _2F_1\left (\frac {5}{4},2;\frac {9}{4};1-x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^4)^(1/4)/x^5,x]

[Out]

((-1 + x^4)^(5/4)*Hypergeometric2F1[5/4, 2, 9/4, 1 - x^4])/5

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IntegrateAlgebraic [A]  time = 0.12, size = 98, normalized size = 1.05 \begin {gather*} -\frac {\sqrt [4]{-1+x^4}}{4 x^4}+\frac {\tan ^{-1}\left (\frac {-\frac {1}{\sqrt {2}}+\frac {\sqrt {-1+x^4}}{\sqrt {2}}}{\sqrt [4]{-1+x^4}}\right )}{8 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^4}}{1+\sqrt {-1+x^4}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^4)^(1/4)/x^5,x]

[Out]

-1/4*(-1 + x^4)^(1/4)/x^4 + ArcTan[(-(1/Sqrt[2]) + Sqrt[-1 + x^4]/Sqrt[2])/(-1 + x^4)^(1/4)]/(8*Sqrt[2]) + Arc
Tanh[(Sqrt[2]*(-1 + x^4)^(1/4))/(1 + Sqrt[-1 + x^4])]/(8*Sqrt[2])

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fricas [B]  time = 0.56, size = 180, normalized size = 1.94 \begin {gather*} -\frac {4 \, \sqrt {2} x^{4} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1} - \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} - 1\right ) + 4 \, \sqrt {2} x^{4} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{4} - 1} + 4} - \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + 1\right ) - \sqrt {2} x^{4} \log \left (4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{4} - 1} + 4\right ) + \sqrt {2} x^{4} \log \left (-4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{4} - 1} + 4\right ) + 8 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{32 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(1/4)/x^5,x, algorithm="fricas")

[Out]

-1/32*(4*sqrt(2)*x^4*arctan(sqrt(2)*sqrt(sqrt(2)*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) - sqrt(2)*(x^4 - 1)^(1/4
) - 1) + 4*sqrt(2)*x^4*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*(x^4 - 1)^(1/4) + 4*sqrt(x^4 - 1) + 4) - sqrt(2)*(x^
4 - 1)^(1/4) + 1) - sqrt(2)*x^4*log(4*sqrt(2)*(x^4 - 1)^(1/4) + 4*sqrt(x^4 - 1) + 4) + sqrt(2)*x^4*log(-4*sqrt
(2)*(x^4 - 1)^(1/4) + 4*sqrt(x^4 - 1) + 4) + 8*(x^4 - 1)^(1/4))/x^4

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giac [A]  time = 0.17, size = 114, normalized size = 1.23 \begin {gather*} \frac {1}{16} \, \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}{16} \, \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}{32} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) - \frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(1/4)/x^5,x, algorithm="giac")

[Out]

1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^4 - 1)^(1/4))) + 1/16*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2
*(x^4 - 1)^(1/4))) + 1/32*sqrt(2)*log(sqrt(2)*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) - 1/32*sqrt(2)*log(-sqrt(2)
*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) - 1/4*(x^4 - 1)^(1/4)/x^4

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maple [C]  time = 7.89, size = 72, normalized size = 0.77

method	result	size
meijerg	\(\frac {\mathrm {signum}\left (x^{4}-1\right )^{\frac {1}{4}} \left (-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{4} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 3\right ], x^{4}\right )}{8}-\left (-3 \ln \relax (2)+\frac {\pi }{2}-1+4 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )-\frac {4 \Gamma \left (\frac {3}{4}\right )}{x^{4}}\right )}{16 \Gamma \left (\frac {3}{4}\right ) \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {1}{4}}}\)	\(72\)
risch	\(-\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{4 x^{4}}+\frac {\left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{4} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{4}\right )}{4}+\left (-3 \ln \relax (2)+\frac {\pi }{2}+4 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{16 \Gamma \left (\frac {3}{4}\right ) \mathrm {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\)	\(76\)
trager	\(-\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{4 x^{4}}+\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-\RootOf \left (\textit {\_Z}^{4}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )}{x^{4}}\right )}{16}-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}}{x^{4}}\right )}{16}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)^(1/4)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/16/GAMMA(3/4)*signum(x^4-1)^(1/4)/(-signum(x^4-1))^(1/4)*(-3/8*GAMMA(3/4)*x^4*hypergeom([1,1,7/4],[2,3],x^4)
-(-3*ln(2)+1/2*Pi-1+4*ln(x)+I*Pi)*GAMMA(3/4)-4*GAMMA(3/4)/x^4)

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maxima [A]  time = 0.66, size = 114, normalized size = 1.23 \begin {gather*} \frac {1}{16} \, \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}{16} \, \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}{32} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{4} - 1} + 1\right ) - \frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(1/4)/x^5,x, algorithm="maxima")

[Out]

1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^4 - 1)^(1/4))) + 1/16*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2
*(x^4 - 1)^(1/4))) + 1/32*sqrt(2)*log(sqrt(2)*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) - 1/32*sqrt(2)*log(-sqrt(2)
*(x^4 - 1)^(1/4) + sqrt(x^4 - 1) + 1) - 1/4*(x^4 - 1)^(1/4)/x^4

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mupad [B]  time = 0.97, size = 57, normalized size = 0.61 \begin {gather*} -\frac {{\left (x^4-1\right )}^{1/4}}{4\,x^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}{16}+\frac {1}{16}{}\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}{16}-\frac {1}{16}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - 1)^(1/4)/x^5,x)

[Out]

2^(1/2)*atan(2^(1/2)*(x^4 - 1)^(1/4)*(1/2 - 1i/2))*(1/16 + 1i/16) + 2^(1/2)*atan(2^(1/2)*(x^4 - 1)^(1/4)*(1/2
+ 1i/2))*(1/16 - 1i/16) - (x^4 - 1)^(1/4)/(4*x^4)

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sympy [C]  time = 0.98, size = 34, normalized size = 0.37 \begin {gather*} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-1)**(1/4)/x**5,x)

[Out]

-gamma(3/4)*hyper((-1/4, 3/4), (7/4,), exp_polar(2*I*pi)/x**4)/(4*x**3*gamma(7/4))

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