\(\int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx\) [853]

Optimal result

Integrand size = 17, antiderivative size = 25 \[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=\sqrt {1-x^2} \sqrt {\frac {1}{-1+x^2}} \arcsin (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6820, 1973, 222} \[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=\sqrt {1-x^2} \sqrt {\frac {1}{x^2-1}} \arcsin (x) \]

[In]

Int[Sqrt[(1 + x^2)/(-1 + x^4)],x]

[Out]

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

Rule 222

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

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {\frac {1}{-1+x^2}} \, dx \\ & = \left (\sqrt {1-x^2} \sqrt {\frac {1}{-1+x^2}}\right ) \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = \sqrt {1-x^2} \sqrt {\frac {1}{-1+x^2}} \sin ^{-1}(x) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(25)=50\).

Time = 0.00 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24 \[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=\frac {1}{2} \sqrt {\frac {1}{-1+x^2}} \sqrt {-1+x^2} \left (-\log \left (1-\frac {x}{\sqrt {-1+x^2}}\right )+\log \left (1+\frac {x}{\sqrt {-1+x^2}}\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12

method	result	size
default	\(\sqrt {\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )\)	\(28\)
trager	\(\ln \left (\sqrt {\frac {1}{x^{2}-1}}\, x^{2}-\sqrt {\frac {1}{x^{2}-1}}+x \right )\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=-\log \left (-x + \sqrt {x^{2} - 1}\right ) \]

[In]

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

[Out]

-log(-x + sqrt(x^2 - 1))

Sympy [F]

\[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=\int \sqrt {\frac {x^{2} + 1}{x^{4} - 1}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=\int { \sqrt {\frac {x^{2} + 1}{x^{4} - 1}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=-\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (x^{2} - 1\right ) \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\frac {1+x^2}{-1+x^4}} \, dx=\int \sqrt {\frac {x^2+1}{x^4-1}} \,d x \]

[In]

int(((x^2 + 1)/(x^4 - 1))^(1/2),x)

[Out]

int(((x^2 + 1)/(x^4 - 1))^(1/2), x)