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

Optimal. Leaf size=22 \[ \cosh ^{-1}(x)-\frac{\sqrt{x-1} \sqrt{x+1}}{x} \]

[Out]

-((Sqrt[-1 + x]*Sqrt[1 + x])/x) + ArcCosh[x]

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Rubi [A]  time = 0.0041882, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {97, 52} \[ \cosh ^{-1}(x)-\frac{\sqrt{x-1} \sqrt{x+1}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x]*Sqrt[1 + x])/x^2,x]

[Out]

-((Sqrt[-1 + x]*Sqrt[1 + x])/x) + ArcCosh[x]

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{-1+x} \sqrt{1+x}}{x^2} \, dx &=-\frac{\sqrt{-1+x} \sqrt{1+x}}{x}+\int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx\\ &=-\frac{\sqrt{-1+x} \sqrt{1+x}}{x}+\cosh ^{-1}(x)\\ \end{align*}

Mathematica [B]  time = 0.0173428, size = 52, normalized size = 2.36 \[ \frac{-\sqrt{x+1} (x-1)-2 \sqrt{1-x} x \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right )}{\sqrt{x-1} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 44, normalized size = 2. \begin{align*}{\frac{1}{x}\sqrt{-1+x}\sqrt{1+x} \left ( \ln \left ( x+\sqrt{{x}^{2}-1} \right ) x-\sqrt{{x}^{2}-1} \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.51311, size = 36, normalized size = 1.64 \begin{align*} -\frac{\sqrt{x^{2} - 1}}{x} + \log \left (2 \, x + 2 \, \sqrt{x^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.49486, size = 96, normalized size = 4.36 \begin{align*} -\frac{x \log \left (\sqrt{x + 1} \sqrt{x - 1} - x\right ) + \sqrt{x + 1} \sqrt{x - 1} + x}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*log(sqrt(x + 1)*sqrt(x - 1) - x) + sqrt(x + 1)*sqrt(x - 1) + x)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.34247, size = 54, normalized size = 2.45 \begin{align*} -\frac{8}{{\left (\sqrt{x + 1} - \sqrt{x - 1}\right )}^{4} + 4} - \frac{1}{2} \, \log \left ({\left (\sqrt{x + 1} - \sqrt{x - 1}\right )}^{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/((sqrt(x + 1) - sqrt(x - 1))^4 + 4) - 1/2*log((sqrt(x + 1) - sqrt(x - 1))^4)