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

Optimal. Leaf size=16 \[ \tan ^{-1}\left (\sqrt{x-1} \sqrt{x+1}\right ) \]

[Out]

ArcTan[Sqrt[-1 + x]*Sqrt[1 + x]]

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Rubi [A]  time = 0.0020829, antiderivative size = 16, 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 = {92, 203} \[ \tan ^{-1}\left (\sqrt{x-1} \sqrt{x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-1 + x]*x*Sqrt[1 + x]),x]

[Out]

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

Mathematica [B]  time = 0.0068491, size = 34, normalized size = 2.12 \[ \frac{\sqrt{x^2-1} \tan ^{-1}\left (\sqrt{x^2-1}\right )}{\sqrt{x-1} \sqrt{x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 28, normalized size = 1.8 \begin{align*} -{\sqrt{-1+x}\sqrt{1+x}\arctan \left ({\frac{1}{\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+x)^(1/2)/(1+x)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.7236, size = 9, normalized size = 0.56 \begin{align*} -\arcsin \left (\frac{1}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(1/abs(x))

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Fricas [A]  time = 1.55455, size = 53, normalized size = 3.31 \begin{align*} 2 \, \arctan \left (\sqrt{x + 1} \sqrt{x - 1} - x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x + 1)*sqrt(x - 1) - x)

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Sympy [C]  time = 5.83691, size = 56, normalized size = 3.5 \begin{align*} - \frac{{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4}, 1 & 1, 1, \frac{3}{2} \\\frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2} & 0 \end{matrix} \middle |{\frac{1}{x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} + \frac{i{G_{6, 6}^{2, 6}\left (\begin{matrix} 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1 & \\\frac{1}{4}, \frac{3}{4} & 0, \frac{1}{2}, \frac{1}{2}, 0 \end{matrix} \middle |{\frac{e^{2 i \pi }}{x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-meijerg(((3/4, 5/4, 1), (1, 1, 3/2)), ((1/2, 3/4, 1, 5/4, 3/2), (0,)), x**(-2))/(4*pi**(3/2)) + I*meijerg(((0
, 1/4, 1/2, 3/4, 1, 1), ()), ((1/4, 3/4), (0, 1/2, 1/2, 0)), exp_polar(2*I*pi)/x**2)/(4*pi**(3/2))

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Giac [A]  time = 1.29603, size = 27, normalized size = 1.69 \begin{align*} -2 \, \arctan \left (\frac{1}{2} \,{\left (\sqrt{x + 1} - \sqrt{x - 1}\right )}^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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