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]

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

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, 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 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 )}{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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.45, size = 43, normalized size = 0.98 \[ \frac {1}{2} \, {\left (x^{2} + 2\right )} \sqrt {-\frac {x^{2} - 1}{x^{2}}} - 3 \, \arctan \left (\frac {x \sqrt {-\frac {x^{2} - 1}{x^{2}}} - 1}{x}\right ) \]

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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giac [A]  time = 0.38, size = 57, normalized size = 1.30 \[ \frac {1}{2} \, \sqrt {-x^{2} + 1} x \mathrm {sgn}\relax (x) + \frac {3}{2} \, \arcsin \relax (x) \mathrm {sgn}\relax (x) - \frac {x \mathrm {sgn}\relax (x)}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )} \mathrm {sgn}\relax (x)}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)*(-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

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maple [A]  time = 0.01, size = 55, normalized size = 1.25 \[ \frac {\sqrt {-\frac {x^{2}-1}{x^{2}}}\, \left (3 \sqrt {-x^{2}+1}\, x^{2}+3 x \arcsin \relax (x )+2 \left (-x^{2}+1\right )^{\frac {3}{2}}\right )}{2 \sqrt {-x^{2}+1}} \]

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

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maxima [A]  time = 1.99, size = 30, normalized size = 0.68 \[ \frac {1}{2} \, x^{2} \sqrt {\frac {1}{x^{2}} - 1} + \sqrt {\frac {1}{x^{2}} - 1} - \frac {3}{2} \, \arctan \left (\sqrt {\frac {1}{x^{2}} - 1}\right ) \]

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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mupad [B]  time = 3.57, size = 30, normalized size = 0.68 \[ \sqrt {\frac {1}{x^2}-1}-\frac {3\,\mathrm {atan}\left (\sqrt {\frac {1}{x^2}-1}\right )}{2}+\frac {x^2\,\sqrt {\frac {1}{x^2}-1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 43.65, size = 39, normalized size = 0.89 \[ \frac {x^{2} \sqrt {-1 + \frac {1}{x^{2}}}}{2} + \sqrt {-1 + \frac {1}{x^{2}}} - \frac {3 \operatorname {atan}{\left (\sqrt {-1 + \frac {1}{x^{2}}} \right )}}{2} \]

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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