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

Optimal. Leaf size=73 \[ \frac{\sqrt{x^2-1} \sqrt{x^4-1} \sinh ^{-1}(x)}{\left (1-x^2\right ) \sqrt{x^2+1}}-\frac{\sqrt{x^4-1} \sin ^{-1}(x)}{\sqrt{1-x^2} \sqrt{x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.117558, antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {6742, 1152, 215, 217, 206} \[ \frac{\sqrt{x^2-1} \sqrt{x^2+1} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right )}{\sqrt{x^4-1}}-\frac{\sqrt{x^2-1} \sqrt{x^2+1} \sinh ^{-1}(x)}{\sqrt{x^4-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

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

Rule 1152

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*x^2)/e)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rule 215

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

Mathematica [A]  time = 0.0573332, size = 71, normalized size = 0.97 \[ \log \left (1-x^2\right )-\log \left (x^2+1\right )-\log \left (x^3+\sqrt{x^2-1} \sqrt{x^4-1}-x\right )+\log \left (x^3+\sqrt{x^2+1} \sqrt{x^4-1}+x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 59, normalized size = 0.8 \begin{align*} -{{\it Arcsinh} \left ( x \right ) \sqrt{{x}^{4}-1}{\frac{1}{\sqrt{{x}^{2}-1}}}{\frac{1}{\sqrt{{x}^{2}+1}}}}+{\sqrt{{x}^{4}-1}\ln \left ( x+\sqrt{{x}^{2}-1} \right ){\frac{1}{\sqrt{{x}^{2}-1}}}{\frac{1}{\sqrt{{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(x^2-1)^(1/2)*(x^4-1)^(1/2)/(x^2+1)^(1/2)*arcsinh(x)+1/(x^2+1)^(1/2)*(x^4-1)^(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 \frac{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}}{\sqrt{x^{4} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.9218, size = 331, normalized size = 4.53 \begin{align*} \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 ) - \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(x^2-1)^(1/2)+(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)) - 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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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