3.675 \(\int \frac {\sqrt {-1+\frac {1}{x^2}} (-1+x^2)^3}{x} \, dx\)

Optimal. Leaf size=76 \[ -\frac {35}{48} \left (\frac {1}{x^2}-1\right )^{3/2} x^2+\frac {35}{16} \sqrt {\frac {1}{x^2}-1}-\frac {35}{16} \tan ^{-1}\left (\sqrt {\frac {1}{x^2}-1}\right )-\frac {1}{6} \left (\frac {1}{x^2}-1\right )^{7/2} x^6-\frac {7}{24} \left (\frac {1}{x^2}-1\right )^{5/2} x^4 \]

[Out]

-35/48*(-1+1/x^2)^(3/2)*x^2-7/24*(-1+1/x^2)^(5/2)*x^4-1/6*(-1+1/x^2)^(7/2)*x^6-35/16*arctan((-1+1/x^2)^(1/2))+
35/16*(-1+1/x^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {25, 266, 47, 50, 63, 203} \[ -\frac {1}{6} \left (\frac {1}{x^2}-1\right )^{7/2} x^6-\frac {7}{24} \left (\frac {1}{x^2}-1\right )^{5/2} x^4-\frac {35}{48} \left (\frac {1}{x^2}-1\right )^{3/2} x^2+\frac {35}{16} \sqrt {\frac {1}{x^2}-1}-\frac {35}{16} \tan ^{-1}\left (\sqrt {\frac {1}{x^2}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x^(-2)]*(-1 + x^2)^3)/x,x]

[Out]

(35*Sqrt[-1 + x^(-2)])/16 - (35*(-1 + x^(-2))^(3/2)*x^2)/48 - (7*(-1 + x^(-2))^(5/2)*x^4)/24 - ((-1 + x^(-2))^
(7/2)*x^6)/6 - (35*ArcTan[Sqrt[-1 + x^(-2)]])/16

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

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 50

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 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 203

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

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]]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^3}{x} \, dx &=-\int \left (-1+\frac {1}{x^2}\right )^{7/2} x^5 \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(-1+x)^{7/2}}{x^4} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {1}{6} \left (-1+\frac {1}{x^2}\right )^{7/2} x^6+\frac {7}{12} \operatorname {Subst}\left (\int \frac {(-1+x)^{5/2}}{x^3} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {7}{24} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {1}{6} \left (-1+\frac {1}{x^2}\right )^{7/2} x^6+\frac {35}{48} \operatorname {Subst}\left (\int \frac {(-1+x)^{3/2}}{x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {35}{48} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2-\frac {7}{24} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {1}{6} \left (-1+\frac {1}{x^2}\right )^{7/2} x^6+\frac {35}{32} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {35}{16} \sqrt {-1+\frac {1}{x^2}}-\frac {35}{48} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2-\frac {7}{24} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {1}{6} \left (-1+\frac {1}{x^2}\right )^{7/2} x^6-\frac {35}{32} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {35}{16} \sqrt {-1+\frac {1}{x^2}}-\frac {35}{48} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2-\frac {7}{24} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {1}{6} \left (-1+\frac {1}{x^2}\right )^{7/2} x^6-\frac {35}{16} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+\frac {1}{x^2}}\right )\\ &=\frac {35}{16} \sqrt {-1+\frac {1}{x^2}}-\frac {35}{48} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2-\frac {7}{24} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {1}{6} \left (-1+\frac {1}{x^2}\right )^{7/2} x^6-\frac {35}{16} \tan ^{-1}\left (\sqrt {-1+\frac {1}{x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.45 \[ \frac {\sqrt {\frac {1}{x^2}-1} \, _2F_1\left (-\frac {7}{2},-\frac {1}{2};\frac {1}{2};x^2\right )}{\sqrt {1-x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[-1 + x^(-2)]*(-1 + x^2)^3)/x,x]

[Out]

(Sqrt[-1 + x^(-2)]*Hypergeometric2F1[-7/2, -1/2, 1/2, x^2])/Sqrt[1 - x^2]

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fricas [A]  time = 0.48, size = 55, normalized size = 0.72 \[ \frac {1}{48} \, {\left (8 \, x^{6} - 38 \, x^{4} + 87 \, x^{2} + 48\right )} \sqrt {-\frac {x^{2} - 1}{x^{2}}} - \frac {35}{8} \, \arctan \left (\frac {x \sqrt {-\frac {x^{2} - 1}{x^{2}}} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*x^6 - 38*x^4 + 87*x^2 + 48)*sqrt(-(x^2 - 1)/x^2) - 35/8*arctan((x*sqrt(-(x^2 - 1)/x^2) - 1)/x)

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giac [A]  time = 0.39, size = 77, normalized size = 1.01 \[ \frac {1}{48} \, {\left (2 \, {\left (4 \, x^{2} \mathrm {sgn}\relax (x) - 19 \, \mathrm {sgn}\relax (x)\right )} x^{2} + 87 \, \mathrm {sgn}\relax (x)\right )} \sqrt {-x^{2} + 1} x + \frac {35}{16} \, \arcsin \relax (x) \mathrm {sgn}\relax (x) - \frac {x \mathrm {sgn}\relax (x)}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )} \mathrm {sgn}\relax (x)}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*x^2*sgn(x) - 19*sgn(x))*x^2 + 87*sgn(x))*sqrt(-x^2 + 1)*x + 35/16*arcsin(x)*sgn(x) - 1/2*x*sgn(x)/(
sqrt(-x^2 + 1) - 1) + 1/2*(sqrt(-x^2 + 1) - 1)*sgn(x)/x

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maple [A]  time = 0.02, size = 83, normalized size = 1.09 \[ \frac {\sqrt {-\frac {x^{2}-1}{x^{2}}}\, \left (-8 \left (-x^{2}+1\right )^{\frac {3}{2}} x^{4}+30 \left (-x^{2}+1\right )^{\frac {3}{2}} x^{2}+105 \sqrt {-x^{2}+1}\, x^{2}+105 x \arcsin \relax (x )+48 \left (-x^{2}+1\right )^{\frac {3}{2}}\right )}{48 \sqrt {-x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)^3*(-1+1/x^2)^(1/2)/x,x)

[Out]

1/48*(-(x^2-1)/x^2)^(1/2)*(-8*x^4*(-x^2+1)^(3/2)+30*x^2*(-x^2+1)^(3/2)+48*(-x^2+1)^(3/2)+105*x^2*(-x^2+1)^(1/2
)+105*arcsin(x)*x)/(-x^2+1)^(1/2)

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maxima [B]  time = 2.07, size = 120, normalized size = 1.58 \[ \frac {3}{2} \, x^{2} \sqrt {\frac {1}{x^{2}} - 1} + \sqrt {\frac {1}{x^{2}} - 1} - \frac {3 \, {\left (\frac {1}{x^{2}} - 1\right )}^{\frac {5}{2}} + 8 \, {\left (\frac {1}{x^{2}} - 1\right )}^{\frac {3}{2}} - 3 \, \sqrt {\frac {1}{x^{2}} - 1}}{48 \, {\left ({\left (\frac {1}{x^{2}} - 1\right )}^{3} + 3 \, {\left (\frac {1}{x^{2}} - 1\right )}^{2} + \frac {3}{x^{2}} - 2\right )}} + \frac {3 \, {\left ({\left (\frac {1}{x^{2}} - 1\right )}^{\frac {3}{2}} - \sqrt {\frac {1}{x^{2}} - 1}\right )}}{8 \, {\left ({\left (\frac {1}{x^{2}} - 1\right )}^{2} + \frac {2}{x^{2}} - 1\right )}} - \frac {35}{16} \, \arctan \left (\sqrt {\frac {1}{x^{2}} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*x^2*sqrt(1/x^2 - 1) + sqrt(1/x^2 - 1) - 1/48*(3*(1/x^2 - 1)^(5/2) + 8*(1/x^2 - 1)^(3/2) - 3*sqrt(1/x^2 - 1
))/((1/x^2 - 1)^3 + 3*(1/x^2 - 1)^2 + 3/x^2 - 2) + 3/8*((1/x^2 - 1)^(3/2) - sqrt(1/x^2 - 1))/((1/x^2 - 1)^2 +
2/x^2 - 1) - 35/16*arctan(sqrt(1/x^2 - 1))

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mupad [B]  time = 3.50, size = 54, normalized size = 0.71 \[ \sqrt {\frac {1}{x^2}-1}-\frac {35\,\mathrm {atan}\left (\sqrt {\frac {1}{x^2}-1}\right )}{16}+\frac {19\,x^6\,\sqrt {\frac {1}{x^2}-1}}{16}+\frac {17\,x^6\,{\left (\frac {1}{x^2}-1\right )}^{3/2}}{6}+\frac {29\,x^6\,{\left (\frac {1}{x^2}-1\right )}^{5/2}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/x^2 - 1)^(1/2)*(x^2 - 1)^3)/x,x)

[Out]

(1/x^2 - 1)^(1/2) - (35*atan((1/x^2 - 1)^(1/2)))/16 + (19*x^6*(1/x^2 - 1)^(1/2))/16 + (17*x^6*(1/x^2 - 1)^(3/2
))/6 + (29*x^6*(1/x^2 - 1)^(5/2))/16

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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