3.853 \(\int \sqrt{\frac{1+x^2}{-1+x^4}} \, dx\)

Optimal. Leaf size=25 \[ \sqrt{1-x^2} \sqrt{\frac{1}{x^2-1}} \sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0216021, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {6688, 6720, 217, 206} \[ \sqrt{\frac{1}{x^2-1}} \sqrt{x^2-1} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 + x^2)/(-1 + x^4)],x]

[Out]

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

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

Mathematica [B]  time = 0.0033965, size = 56, normalized size = 2.24 \[ \frac{1}{2} \sqrt{\frac{1}{x^2-1}} \sqrt{x^2-1} \left (\log \left (\frac{x}{\sqrt{x^2-1}}+1\right )-\log \left (1-\frac{x}{\sqrt{x^2-1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 + x^2)/(-1 + x^4)],x]

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 1.1 \begin{align*} \sqrt{ \left ({x}^{2}-1 \right ) ^{-1}}\sqrt{{x}^{2}-1}\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+1)/(x^4-1))^(1/2),x)

[Out]

(1/(x^2-1))^(1/2)*(x^2-1)^(1/2)*ln(x+(x^2-1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{x^{2} + 1}{x^{4} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+1)/(x^4-1))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt((x^2 + 1)/(x^4 - 1)), x)

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Fricas [A]  time = 1.4318, size = 35, normalized size = 1.4 \begin{align*} -\log \left (-x + \sqrt{x^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+1)/(x^4-1))^(1/2),x, algorithm="fricas")

[Out]

-log(-x + sqrt(x^2 - 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{x^{2} + 1}{x^{4} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+1)/(x**4-1))**(1/2),x)

[Out]

Integral(sqrt((x**2 + 1)/(x**4 - 1)), x)

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Giac [A]  time = 1.16616, size = 28, normalized size = 1.12 \begin{align*} -\log \left ({\left | -x + \sqrt{x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+1)/(x^4-1))^(1/2),x, algorithm="giac")

[Out]

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