3.9.19 \(\int \frac {\sqrt {-1+x} \sqrt {1+x}}{x} \, dx\)

Optimal. Leaf size=34 \[ \sqrt {x-1} \sqrt {x+1}-\tan ^{-1}\left (\sqrt {x-1} \sqrt {x+1}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {101, 92, 203} \begin {gather*} \sqrt {x-1} \sqrt {x+1}-\tan ^{-1}\left (\sqrt {x-1} \sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x]*Sqrt[1 + x])/x,x]

[Out]

Sqrt[-1 + x]*Sqrt[1 + x] - ArcTan[Sqrt[-1 + x]*Sqrt[1 + x]]

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 101

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

Rule 203

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

Rubi steps

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

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Mathematica [B]  time = 0.13, size = 101, normalized size = 2.97 \begin {gather*} \frac {\sqrt {1-x} \left (x^2-\sqrt {x^2-1} \tan ^{-1}\left (\sqrt {x^2-1}\right )-1\right )+2 (x-1) \sqrt {x+1} \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{\sqrt {-(x-1)^2} \sqrt {x+1}}-2 \tanh ^{-1}\left (\sqrt {\frac {x-1}{x+1}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 48, normalized size = 1.41 \begin {gather*} -\frac {2 \sqrt {x-1}}{\sqrt {x+1} \left (\frac {x-1}{x+1}-1\right )}-2 \tan ^{-1}\left (\frac {\sqrt {x-1}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-1 + x]*Sqrt[1 + x])/x,x]

[Out]

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

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fricas [A]  time = 1.43, size = 30, normalized size = 0.88 \begin {gather*} \sqrt {x + 1} \sqrt {x - 1} - 2 \, \arctan \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x + 1)*sqrt(x - 1) - 2*arctan(sqrt(x + 1)*sqrt(x - 1) - x)

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giac [A]  time = 1.26, size = 32, normalized size = 0.94 \begin {gather*} \sqrt {x + 1} \sqrt {x - 1} + 2 \, \arctan \left (\frac {1}{2} \, {\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 35, normalized size = 1.03 \begin {gather*} \frac {\sqrt {x -1}\, \sqrt {x +1}\, \left (\arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right )+\sqrt {x^{2}-1}\right )}{\sqrt {x^{2}-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 13, normalized size = 0.38 \begin {gather*} \sqrt {x^{2} - 1} + \arcsin \left (\frac {1}{{\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 - 1) + arcsin(1/abs(x))

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mupad [B]  time = 2.09, size = 116, normalized size = 3.41 \begin {gather*} \ln \left (\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}-\ln \left (\frac {\sqrt {x-1}-\mathrm {i}}{\sqrt {x+1}-1}\right )\,1{}\mathrm {i}-\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^2\,8{}\mathrm {i}}{{\left (\sqrt {x+1}-1\right )}^2\,\left (1+\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {x+1}-1\right )}^4}-\frac {2\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x - 1} \sqrt {x + 1}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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