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

Optimal. Leaf size=60 \[ \frac {5}{8} \left (\frac {1}{x^2}-1\right )^{3/2} x^2-\frac {15}{8} \sqrt {\frac {1}{x^2}-1}+\frac {15}{8} \tan ^{-1}\left (\sqrt {\frac {1}{x^2}-1}\right )+\frac {1}{4} \left (\frac {1}{x^2}-1\right )^{5/2} x^4 \]

[Out]

5/8*(-1+1/x^2)^(3/2)*x^2+1/4*(-1+1/x^2)^(5/2)*x^4+15/8*arctan((-1+1/x^2)^(1/2))-15/8*(-1+1/x^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {1}{4} \left (\frac {1}{x^2}-1\right )^{5/2} x^4+\frac {5}{8} \left (\frac {1}{x^2}-1\right )^{3/2} x^2-\frac {15}{8} \sqrt {\frac {1}{x^2}-1}+\frac {15}{8} \tan ^{-1}\left (\sqrt {\frac {1}{x^2}-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[-1 + x^(-2)])/8 + (5*(-1 + x^(-2))^(3/2)*x^2)/8 + ((-1 + x^(-2))^(5/2)*x^4)/4 + (15*ArcTan[Sqrt[-1 +
x^(-2)]])/8

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 )^2}{x} \, dx &=\int \left (-1+\frac {1}{x^2}\right )^{5/2} x^3 \, dx\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {(-1+x)^{5/2}}{x^3} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=\frac {1}{4} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {5}{8} \operatorname {Subst}\left (\int \frac {(-1+x)^{3/2}}{x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {5}{8} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4-\frac {15}{16} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {15}{8} \sqrt {-1+\frac {1}{x^2}}+\frac {5}{8} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4+\frac {15}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {15}{8} \sqrt {-1+\frac {1}{x^2}}+\frac {5}{8} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4+\frac {15}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+\frac {1}{x^2}}\right )\\ &=-\frac {15}{8} \sqrt {-1+\frac {1}{x^2}}+\frac {5}{8} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4+\frac {15}{8} \tan ^{-1}\left (\sqrt {-1+\frac {1}{x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.58 \[ -\frac {\sqrt {\frac {1}{x^2}-1} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};x^2\right )}{\sqrt {1-x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 50, normalized size = 0.83 \[ \frac {1}{8} \, {\left (2 \, x^{4} - 9 \, x^{2} - 8\right )} \sqrt {-\frac {x^{2} - 1}{x^{2}}} + \frac {15}{4} \, \arctan \left (\frac {x \sqrt {-\frac {x^{2} - 1}{x^{2}}} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x^4 - 9*x^2 - 8)*sqrt(-(x^2 - 1)/x^2) + 15/4*arctan((x*sqrt(-(x^2 - 1)/x^2) - 1)/x)

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giac [A]  time = 0.43, size = 67, normalized size = 1.12 \[ \frac {1}{8} \, {\left (2 \, x^{2} \mathrm {sgn}\relax (x) - 9 \, \mathrm {sgn}\relax (x)\right )} \sqrt {-x^{2} + 1} x - \frac {15}{8} \, \arcsin \relax (x) \mathrm {sgn}\relax (x) + \frac {x \mathrm {sgn}\relax (x)}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )} \mathrm {sgn}\relax (x)}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x^2*sgn(x) - 9*sgn(x))*sqrt(-x^2 + 1)*x - 15/8*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 = 69, normalized size = 1.15 \[ -\frac {\sqrt {-\frac {x^{2}-1}{x^{2}}}\, \left (2 \left (-x^{2}+1\right )^{\frac {3}{2}} x^{2}+15 \sqrt {-x^{2}+1}\, x^{2}+15 x \arcsin \relax (x )+8 \left (-x^{2}+1\right )^{\frac {3}{2}}\right )}{8 \sqrt {-x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-(x^2-1)/x^2)^(1/2)*(2*(-x^2+1)^(3/2)*x^2+8*(-x^2+1)^(3/2)+15*(-x^2+1)^(1/2)*x^2+15*x*arcsin(x))/(-x^2+1
)^(1/2)

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maxima [A]  time = 2.00, size = 67, normalized size = 1.12 \[ -x^{2} \sqrt {\frac {1}{x^{2}} - 1} - \sqrt {\frac {1}{x^{2}} - 1} - \frac {{\left (\frac {1}{x^{2}} - 1\right )}^{\frac {3}{2}} - \sqrt {\frac {1}{x^{2}} - 1}}{8 \, {\left ({\left (\frac {1}{x^{2}} - 1\right )}^{2} + \frac {2}{x^{2}} - 1\right )}} + \frac {15}{8} \, \arctan \left (\sqrt {\frac {1}{x^{2}} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*sqrt(1/x^2 - 1) - sqrt(1/x^2 - 1) - 1/8*((1/x^2 - 1)^(3/2) - sqrt(1/x^2 - 1))/((1/x^2 - 1)^2 + 2/x^2 - 1)
+ 15/8*arctan(sqrt(1/x^2 - 1))

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mupad [B]  time = 3.36, size = 44, normalized size = 0.73 \[ \frac {15\,\mathrm {atan}\left (\sqrt {\frac {1}{x^2}-1}\right )}{8}-\sqrt {\frac {1}{x^2}-1}-\frac {7\,x^4\,\sqrt {\frac {1}{x^2}-1}}{8}-\frac {9\,x^4\,{\left (\frac {1}{x^2}-1\right )}^{3/2}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*atan((1/x^2 - 1)^(1/2)))/8 - (1/x^2 - 1)^(1/2) - (7*x^4*(1/x^2 - 1)^(1/2))/8 - (9*x^4*(1/x^2 - 1)^(3/2))/8

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sympy [A]  time = 118.56, size = 60, normalized size = 1.00 \[ \frac {x^{4} \sqrt {-1 + \frac {1}{x^{2}}} \left (2 - \frac {1}{x^{2}}\right )}{8} - x^{2} \sqrt {-1 + \frac {1}{x^{2}}} - \sqrt {-1 + \frac {1}{x^{2}}} + \frac {15 \operatorname {atan}{\left (\sqrt {-1 + \frac {1}{x^{2}}} \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*sqrt(-1 + x**(-2))*(2 - 1/x**2)/8 - x**2*sqrt(-1 + x**(-2)) - sqrt(-1 + x**(-2)) + 15*atan(sqrt(-1 + x**(
-2)))/8

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