\(\int \frac {\sqrt {-1+x^2}}{\sqrt {-1+x^4}} \, dx\) [211]

Optimal result

Integrand size = 19, antiderivative size = 30 \[ \int \frac {\sqrt {-1+x^2}}{\sqrt {-1+x^4}} \, dx=\frac {\sqrt {-1+x^2} \sqrt {1+x^2} \text {arcsinh}(x)}{\sqrt {-1+x^4}} \]

[Out]

arcsinh(x)*(x^2-1)^(1/2)*(x^2+1)^(1/2)/(x^4-1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1166, 221} \[ \int \frac {\sqrt {-1+x^2}}{\sqrt {-1+x^4}} \, dx=\frac {\sqrt {x^2-1} \sqrt {x^2+1} \text {arcsinh}(x)}{\sqrt {x^4-1}} \]

[In]

Int[Sqrt[-1 + x^2]/Sqrt[-1 + x^4],x]

[Out]

(Sqrt[-1 + x^2]*Sqrt[1 + x^2]*ArcSinh[x])/Sqrt[-1 + x^4]

Rule 221

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

Rule 1166

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^FracPart[p]/((d + e*x
^2)^FracPart[p]*(a/d + c*(x^2/e))^FracPart[p]), Int[(d + e*x^2)^(p + q)*(a/d + (c/e)*x^2)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {1}{\sqrt {1+x^2}} \, dx}{\sqrt {-1+x^4}} \\ & = \frac {\sqrt {-1+x^2} \sqrt {1+x^2} \sinh ^{-1}(x)}{\sqrt {-1+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {-1+x^2}}{\sqrt {-1+x^4}} \, dx=-\log \left (1-x^2\right )+\log \left (-x+x^3+\sqrt {-1+x^2} \sqrt {-1+x^4}\right ) \]

[In]

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

[Out]

-Log[1 - x^2] + Log[-x + x^3 + Sqrt[-1 + x^2]*Sqrt[-1 + x^4]]

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\sqrt {x^{4}-1}\, \operatorname {arcsinh}\left (x \right )}{\sqrt {x^{2}-1}\, \sqrt {x^{2}+1}}\)	\(25\)

[In]

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

[Out]

1/(x^2-1)^(1/2)*(x^4-1)^(1/2)/(x^2+1)^(1/2)*arcsinh(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (24) = 48\).

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {-1+x^2}}{\sqrt {-1+x^4}} \, dx=\frac {1}{2} \, \log \left (\frac {x^{3} + \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - x}{x^{3} - x}\right ) - \frac {1}{2} \, \log \left (-\frac {x^{3} - \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - x}{x^{3} - x}\right ) \]

[In]

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

[Out]

1/2*log((x^3 + sqrt(x^4 - 1)*sqrt(x^2 - 1) - x)/(x^3 - x)) - 1/2*log(-(x^3 - sqrt(x^4 - 1)*sqrt(x^2 - 1) - x)/
(x^3 - x))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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