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

Optimal. Leaf size=22 \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\frac{1}{x}+1}}{\sqrt{2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0321021, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {1446, 1469, 627, 63, 207} \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\frac{1}{x}+1}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^(-1)]/(1 - x^2),x]

[Out]

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

Rule 1446

Int[((a_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[((d + e*x^n)^q*(c + a
*x^(2*n))^p)/x^(2*n*p), x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[mn2, -2*n] && IntegerQ[p]

Rule 1469

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Sim
plify[m - n + 1], 0]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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 207

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

Rubi steps

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

Mathematica [A]  time = 0.0212112, size = 22, normalized size = 1. \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\frac{1}{x}+1}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 41, normalized size = 1.9 \begin{align*}{\frac{x\sqrt{2}}{2}\sqrt{{\frac{1+x}{x}}}{\it Artanh} \left ({\frac{ \left ( 1+3\,x \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{{x}^{2}+x}}}} \right ){\frac{1}{\sqrt{x \left ( 1+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((1+x)/x)^(1/2)*x/(x*(1+x))^(1/2)*2^(1/2)*arctanh(1/4*(1+3*x)*2^(1/2)/(x^2+x)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sqrt{\frac{1}{x} + 1}}{x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(sqrt(1/x + 1)/(x^2 - 1), x)

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Fricas [A]  time = 1.50179, size = 90, normalized size = 4.09 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} x \sqrt{\frac{x + 1}{x}} + 3 \, x + 1}{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sqrt{1 + \frac{1}{x}}}{x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(sqrt(1 + 1/x)/(x**2 - 1), x)

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Giac [B]  time = 1.23347, size = 99, normalized size = 4.5 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right ) \mathrm{sgn}\left (x\right ) - \frac{1}{2} \, \sqrt{2} \log \left (\frac{{\left | -2 \, x - 2 \, \sqrt{2} + 2 \, \sqrt{x^{2} + x} + 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt{2} + 2 \, \sqrt{x^{2} + x} + 2 \right |}}\right ) \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log((sqrt(2) - 1)/(sqrt(2) + 1))*sgn(x) - 1/2*sqrt(2)*log(abs(-2*x - 2*sqrt(2) + 2*sqrt(x^2 + x) +
2)/abs(-2*x + 2*sqrt(2) + 2*sqrt(x^2 + x) + 2))*sgn(x)