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

Optimal. Leaf size=52 \[ \frac{\sqrt{-\frac{x^2}{1-x^2}} \sqrt{x^2-1} \tan ^{-1}\left (\frac{\sqrt{x^2-1}}{\sqrt{2}}\right )}{\sqrt{2} x} \]

[Out]

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

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Rubi [A]  time = 0.0989364, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6719, 444, 63, 203} \[ \frac{\sqrt{-\frac{x^2}{1-x^2}} \sqrt{x^2-1} \tan ^{-1}\left (\frac{\sqrt{x^2-1}}{\sqrt{2}}\right )}{\sqrt{2} x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^2/(-1 + x^2)]/(1 + x^2),x]

[Out]

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

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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

Mathematica [A]  time = 0.0160615, size = 49, normalized size = 0.94 \[ \frac{\sqrt{\frac{x^2}{x^2-1}} \sqrt{x^2-1} \tan ^{-1}\left (\frac{\sqrt{x^2-1}}{\sqrt{2}}\right )}{\sqrt{2} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 42, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2}}{2\,x}\sqrt{{\frac{{x}^{2}}{{x}^{2}-1}}}\sqrt{{x}^{2}-1}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{{x}^{2}-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.48377, size = 88, normalized size = 1.69 \begin{align*} \frac{1}{2} \, \sqrt{2} \arctan \left (\frac{\sqrt{2}{\left (x^{2} - 1\right )} \sqrt{\frac{x^{2}}{x^{2} - 1}}}{2 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16685, size = 55, normalized size = 1.06 \begin{align*} \frac{1}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{x^{2} - 1}\right ) \mathrm{sgn}\left (x^{2} - 1\right ) \mathrm{sgn}\left (x\right ) + \frac{1}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} i\right ) \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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