3.13.79 \(\int \frac {1}{x^3 (-1+x^2)^{3/4}} \, dx\)

Optimal. Leaf size=93 \[ \frac {\sqrt [4]{x^2-1}}{2 x^2}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^2-1}}{\sqrt {x^2-1}-1}\right )}{4 \sqrt {2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^2-1}}{\sqrt {x^2-1}+1}\right )}{4 \sqrt {2}} \]

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Rubi [A]  time = 0.11, 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, 51, 63, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {\sqrt [4]{x^2-1}}{2 x^2}-\frac {3 \log \left (\sqrt {x^2-1}-\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{8 \sqrt {2}}+\frac {3 \log \left (\sqrt {x^2-1}+\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{8 \sqrt {2}}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{x^2-1}\right )}{4 \sqrt {2}}+\frac {3 \tan ^{-1}\left (\sqrt {2} \sqrt [4]{x^2-1}+1\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-1 + x^2)^(3/4)),x]

[Out]

(-1 + x^2)^(1/4)/(2*x^2) - (3*ArcTan[1 - Sqrt[2]*(-1 + x^2)^(1/4)])/(4*Sqrt[2]) + (3*ArcTan[1 + Sqrt[2]*(-1 +
x^2)^(1/4)])/(4*Sqrt[2]) - (3*Log[1 - Sqrt[2]*(-1 + x^2)^(1/4) + Sqrt[-1 + x^2]])/(8*Sqrt[2]) + (3*Log[1 + Sqr
t[2]*(-1 + x^2)^(1/4) + Sqrt[-1 + x^2]])/(8*Sqrt[2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{x^3 \left (-1+x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{3/4} x^2} \, dx,x,x^2\right )\\ &=\frac {\sqrt [4]{-1+x^2}}{2 x^2}+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{3/4} x} \, dx,x,x^2\right )\\ &=\frac {\sqrt [4]{-1+x^2}}{2 x^2}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt [4]{-1+x^2}\right )\\ &=\frac {\sqrt [4]{-1+x^2}}{2 x^2}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^2}\right )+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt [4]{-1+x^2}\right )\\ &=\frac {\sqrt [4]{-1+x^2}}{2 x^2}+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )-\frac {3 \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )}{8 \sqrt {2}}-\frac {3 \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+x^2}\right )}{8 \sqrt {2}}\\ &=\frac {\sqrt [4]{-1+x^2}}{2 x^2}-\frac {3 \log \left (1-\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{8 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{8 \sqrt {2}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+x^2}\right )}{4 \sqrt {2}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+x^2}\right )}{4 \sqrt {2}}\\ &=\frac {\sqrt [4]{-1+x^2}}{2 x^2}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{-1+x^2}\right )}{4 \sqrt {2}}+\frac {3 \tan ^{-1}\left (1+\sqrt {2} \sqrt [4]{-1+x^2}\right )}{4 \sqrt {2}}-\frac {3 \log \left (1-\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{8 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [4]{-1+x^2}+\sqrt {-1+x^2}\right )}{8 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 26, normalized size = 0.28 \begin {gather*} 2 \sqrt [4]{x^2-1} \, _2F_1\left (\frac {1}{4},2;\frac {5}{4};1-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(-1 + x^2)^(3/4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 98, normalized size = 1.05 \begin {gather*} \frac {\sqrt [4]{-1+x^2}}{2 x^2}+\frac {3 \tan ^{-1}\left (\frac {-\frac {1}{\sqrt {2}}+\frac {\sqrt {-1+x^2}}{\sqrt {2}}}{\sqrt [4]{-1+x^2}}\right )}{4 \sqrt {2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{1+\sqrt {-1+x^2}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^3*(-1 + x^2)^(3/4)),x]

[Out]

(-1 + x^2)^(1/4)/(2*x^2) + (3*ArcTan[(-(1/Sqrt[2]) + Sqrt[-1 + x^2]/Sqrt[2])/(-1 + x^2)^(1/4)])/(4*Sqrt[2]) +
(3*ArcTanh[(Sqrt[2]*(-1 + x^2)^(1/4))/(1 + Sqrt[-1 + x^2])])/(4*Sqrt[2])

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fricas [B]  time = 0.46, size = 181, normalized size = 1.95 \begin {gather*} -\frac {12 \, \sqrt {2} x^{2} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1} - \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) + 12 \, \sqrt {2} x^{2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1} + 4} - \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) - 3 \, \sqrt {2} x^{2} \log \left (4 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1} + 4\right ) + 3 \, \sqrt {2} x^{2} \log \left (-4 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {x^{2} - 1} + 4\right ) - 8 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^2-1)^(3/4),x, algorithm="fricas")

[Out]

-1/16*(12*sqrt(2)*x^2*arctan(sqrt(2)*sqrt(sqrt(2)*(x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) - sqrt(2)*(x^2 - 1)^(1/
4) - 1) + 12*sqrt(2)*x^2*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*(x^2 - 1)^(1/4) + 4*sqrt(x^2 - 1) + 4) - sqrt(2)*(
x^2 - 1)^(1/4) + 1) - 3*sqrt(2)*x^2*log(4*sqrt(2)*(x^2 - 1)^(1/4) + 4*sqrt(x^2 - 1) + 4) + 3*sqrt(2)*x^2*log(-
4*sqrt(2)*(x^2 - 1)^(1/4) + 4*sqrt(x^2 - 1) + 4) - 8*(x^2 - 1)^(1/4))/x^2

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giac [A]  time = 0.28, size = 114, normalized size = 1.23 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) + \frac {{\left (x^{2} - 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^2-1)^(3/4),x, algorithm="giac")

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^2 - 1)^(1/4))) + 3/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(
x^2 - 1)^(1/4))) + 3/16*sqrt(2)*log(sqrt(2)*(x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) - 3/16*sqrt(2)*log(-sqrt(2)*(
x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) + 1/2*(x^2 - 1)^(1/4)/x^2

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maple [C]  time = 2.61, size = 71, normalized size = 0.76

method	result	size
meijerg	\(-\frac {\left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} \left (-\frac {21 \Gamma \left (\frac {3}{4}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {11}{4}\right ], \left [2, 3\right ], x^{2}\right )}{32}-\frac {3 \left (\frac {1}{3}-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )}{4}+\frac {\Gamma \left (\frac {3}{4}\right )}{x^{2}}\right )}{2 \Gamma \left (\frac {3}{4}\right ) \mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\)	\(71\)
risch	\(\frac {\left (x^{2}-1\right )^{\frac {1}{4}}}{2 x^{2}}+\frac {3 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {3}{4}} \left (\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], x^{2}\right )}{4}+\left (-3 \ln \relax (2)+\frac {\pi }{2}+2 \ln \relax (x )+i \pi \right ) \Gamma \left (\frac {3}{4}\right )\right )}{8 \Gamma \left (\frac {3}{4}\right ) \mathrm {signum}\left (x^{2}-1\right )^{\frac {3}{4}}}\)	\(76\)
trager	\(\frac {\left (x^{2}-1\right )^{\frac {1}{4}}}{2 x^{2}}+\frac {3 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {2 \sqrt {x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \left (x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-\RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{8}-\frac {3 \RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+2 \left (x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )+2 \left (x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}}{x^{2}}\right )}{8}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^2-1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-1/2/GAMMA(3/4)/signum(x^2-1)^(3/4)*(-signum(x^2-1))^(3/4)*(-21/32*GAMMA(3/4)*x^2*hypergeom([1,1,11/4],[2,3],x
^2)-3/4*(1/3-3*ln(2)+1/2*Pi+2*ln(x)+I*Pi)*GAMMA(3/4)+GAMMA(3/4)/x^2)

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maxima [A]  time = 0.76, size = 114, normalized size = 1.23 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {x^{2} - 1} + 1\right ) + \frac {{\left (x^{2} - 1\right )}^{\frac {1}{4}}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^2-1)^(3/4),x, algorithm="maxima")

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^2 - 1)^(1/4))) + 3/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(
x^2 - 1)^(1/4))) + 3/16*sqrt(2)*log(sqrt(2)*(x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) - 3/16*sqrt(2)*log(-sqrt(2)*(
x^2 - 1)^(1/4) + sqrt(x^2 - 1) + 1) + 1/2*(x^2 - 1)^(1/4)/x^2

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mupad [B]  time = 0.95, size = 57, normalized size = 0.61 \begin {gather*} \frac {{\left (x^2-1\right )}^{1/4}}{2\,x^2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^2-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{8}+\frac {3}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (x^2-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{8}-\frac {3}{8}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(x^2 - 1)^(3/4)),x)

[Out]

2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 - 1i/2))*(3/8 + 3i/8) + 2^(1/2)*atan(2^(1/2)*(x^2 - 1)^(1/4)*(1/2 +
1i/2))*(3/8 - 3i/8) + (x^2 - 1)^(1/4)/(2*x^2)

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sympy [C]  time = 0.95, size = 34, normalized size = 0.37 \begin {gather*} - \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(x**2-1)**(3/4),x)

[Out]

-gamma(7/4)*hyper((3/4, 7/4), (11/4,), exp_polar(2*I*pi)/x**2)/(2*x**(7/2)*gamma(11/4))

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