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

Optimal. Leaf size=44 \[ -\frac{1}{2} \left (\frac{1}{x^2}-1\right )^{3/2} x^2+\frac{3}{2} \sqrt{\frac{1}{x^2}-1}-\frac{3}{2} \tan ^{-1}\left (\sqrt{\frac{1}{x^2}-1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0138863, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {25, 266, 47, 50, 63, 203} \[ -\frac{1}{2} \left (\frac{1}{x^2}-1\right )^{3/2} x^2+\frac{3}{2} \sqrt{\frac{1}{x^2}-1}-\frac{3}{2} \tan ^{-1}\left (\sqrt{\frac{1}{x^2}-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0073557, size = 34, normalized size = 0.77 \[ \frac{\sqrt{\frac{1}{x^2}-1} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};x^2\right )}{\sqrt{1-x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-1 + x^(-2)]*Hypergeometric2F1[-3/2, -1/2, 1/2, x^2])/Sqrt[1 - x^2]

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Maple [A]  time = 0.006, size = 55, normalized size = 1.3 \begin{align*}{\frac{1}{2}\sqrt{-{\frac{{x}^{2}-1}{{x}^{2}}}} \left ( 2\, \left ( -{x}^{2}+1 \right ) ^{3/2}+3\,{x}^{2}\sqrt{-{x}^{2}+1}+3\,\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)*(-1+1/x^2)^(1/2)/x,x)

[Out]

1/2*(-(x^2-1)/x^2)^(1/2)*(2*(-x^2+1)^(3/2)+3*x^2*(-x^2+1)^(1/2)+3*arcsin(x)*x)/(-x^2+1)^(1/2)

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Maxima [A]  time = 2.05112, size = 41, normalized size = 0.93 \begin{align*} \frac{1}{2} \, x^{2} \sqrt{\frac{1}{x^{2}} - 1} + \sqrt{\frac{1}{x^{2}} - 1} - \frac{3}{2} \, \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)*(-1+1/x^2)^(1/2)/x,x, algorithm="maxima")

[Out]

1/2*x^2*sqrt(1/x^2 - 1) + sqrt(1/x^2 - 1) - 3/2*arctan(sqrt(1/x^2 - 1))

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Fricas [A]  time = 1.44859, size = 107, normalized size = 2.43 \begin{align*} \frac{1}{2} \,{\left (x^{2} + 2\right )} \sqrt{-\frac{x^{2} - 1}{x^{2}}} - 3 \, \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)*(-1+1/x^2)^(1/2)/x,x, algorithm="fricas")

[Out]

1/2*(x^2 + 2)*sqrt(-(x^2 - 1)/x^2) - 3*arctan((x*sqrt(-(x^2 - 1)/x^2) - 1)/x)

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Sympy [A]  time = 27.1288, size = 39, normalized size = 0.89 \begin{align*} \frac{x^{2} \sqrt{-1 + \frac{1}{x^{2}}}}{2} + \sqrt{-1 + \frac{1}{x^{2}}} - \frac{3 \operatorname{atan}{\left (\sqrt{-1 + \frac{1}{x^{2}}} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*sqrt(-1 + x**(-2))/2 + sqrt(-1 + x**(-2)) - 3*atan(sqrt(-1 + x**(-2)))/2

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Giac [A]  time = 1.16428, size = 77, normalized size = 1.75 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} + 1} x \mathrm{sgn}\left (x\right ) + \frac{3}{2} \, \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)*(-1+1/x^2)^(1/2)/x,x, algorithm="giac")

[Out]

1/2*sqrt(-x^2 + 1)*x*sgn(x) + 3/2*arcsin(x)*sgn(x) - 1/2*x*sgn(x)/(sqrt(-x^2 + 1) - 1) + 1/2*(sqrt(-x^2 + 1) -
1)*sgn(x)/x