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

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

[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

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Rubi [A]  time = 0.026315, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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 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])

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*}

Mathematica [C]  time = 0.0104486, 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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Maple [A]  time = 0.011, size = 83, normalized size = 1.1 \begin{align*}{\frac{1}{48}\sqrt{-{\frac{{x}^{2}-1}{{x}^{2}}}} \left ( -8\,{x}^{4} \left ( -{x}^{2}+1 \right ) ^{3/2}+30\,{x}^{2} \left ( -{x}^{2}+1 \right ) ^{3/2}+48\, \left ( -{x}^{2}+1 \right ) ^{3/2}+105\,{x}^{2}\sqrt{-{x}^{2}+1}+105\,\arcsin \left ( x \right ) x \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

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 = 2.13629, size = 162, normalized size = 2.13 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 1.48702, size = 140, normalized size = 1.84 \begin{align*} \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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.1925, size = 104, normalized size = 1.37 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, x^{2} \mathrm{sgn}\left (x\right ) - 19 \, \mathrm{sgn}\left (x\right )\right )} x^{2} + 87 \, \mathrm{sgn}\left (x\right )\right )} \sqrt{-x^{2} + 1} x + \frac{35}{16} \, \arcsin \left (x\right ) \mathrm{sgn}\left (x\right ) - \frac{x \mathrm{sgn}\left (x\right )}{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}} + \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )} \mathrm{sgn}\left (x\right )}{2 \, x} \end{align*}

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