3.9.23 \(\int \frac {1}{(-1+x)^{3/2} x (1+x)^{3/2}} \, dx\)

Optimal. Leaf size=35 \[ -\frac {1}{\sqrt {x-1} \sqrt {x+1}}-\tan ^{-1}\left (\sqrt {x-1} \sqrt {x+1}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 35, 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 = {104, 92, 203} \begin {gather*} -\frac {1}{\sqrt {x-1} \sqrt {x+1}}-\tan ^{-1}\left (\sqrt {x-1} \sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-1 + x)^(3/2)*x*(1 + x)^(3/2)),x]

[Out]

-(1/(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 104

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

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}{(-1+x)^{3/2} x (1+x)^{3/2}} \, dx &=-\frac {1}{\sqrt {-1+x} \sqrt {1+x}}-\int \frac {1}{\sqrt {-1+x} x \sqrt {1+x}} \, dx\\ &=-\frac {1}{\sqrt {-1+x} \sqrt {1+x}}-\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x} \sqrt {1+x}\right )\\ &=-\frac {1}{\sqrt {-1+x} \sqrt {1+x}}-\tan ^{-1}\left (\sqrt {-1+x} \sqrt {1+x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 31, normalized size = 0.89 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1-x^2\right )}{\sqrt {x-1} \sqrt {x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((-1 + x)^(3/2)*x*(1 + x)^(3/2)),x]

[Out]

-(Hypergeometric2F1[-1/2, 1, 1/2, 1 - x^2]/(Sqrt[-1 + x]*Sqrt[1 + x]))

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IntegrateAlgebraic [A]  time = 0.05, size = 48, normalized size = 1.37 \begin {gather*} \frac {\sqrt {x+1} \left (\frac {x-1}{x+1}-1\right )}{2 \sqrt {x-1}}-2 \tan ^{-1}\left (\frac {\sqrt {x-1}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((-1 + x)^(3/2)*x*(1 + x)^(3/2)),x]

[Out]

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

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fricas [A]  time = 1.33, size = 44, normalized size = 1.26 \begin {gather*} -\frac {2 \, {\left (x^{2} - 1\right )} \arctan \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) + \sqrt {x + 1} \sqrt {x - 1}}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1+x)^(3/2)/x/(1+x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.18, size = 54, normalized size = 1.54 \begin {gather*} -\frac {\sqrt {x + 1}}{2 \, \sqrt {x - 1}} + \frac {2}{{\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{2} + 2} + 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/(-1+x)^(3/2)/x/(1+x)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 51, normalized size = 1.46 \begin {gather*} \frac {x^{2} \arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right )-\arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right )-\sqrt {x^{2}-1}}{\sqrt {x^{2}-1}\, \sqrt {x +1}\, \sqrt {x -1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x-1)^(3/2)/x/(x+1)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.26, size = 15, normalized size = 0.43 \begin {gather*} -\frac {1}{\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/(-1+x)^(3/2)/x/(1+x)^(3/2),x, algorithm="maxima")

[Out]

-1/sqrt(x^2 - 1) + arcsin(1/abs(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{x\,{\left (x-1\right )}^{3/2}\,{\left (x+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x - 1)^(3/2)*(x + 1)^(3/2)),x)

[Out]

int(1/(x*(x - 1)^(3/2)*(x + 1)^(3/2)), x)

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sympy [C]  time = 7.08, size = 58, normalized size = 1.66 \begin {gather*} - \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & 1, 2, \frac {5}{2} \\\frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2, \frac {5}{2} & 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}}} - \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, 1 & \\\frac {3}{4}, \frac {5}{4} & 0, \frac {1}{2}, \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1+x)**(3/2)/x/(1+x)**(3/2),x)

[Out]

-meijerg(((5/4, 7/4, 1), (1, 2, 5/2)), ((5/4, 3/2, 7/4, 2, 5/2), (0,)), x**(-2))/(2*pi**(3/2)) - I*meijerg(((0
, 1/2, 3/4, 1, 5/4, 1), ()), ((3/4, 5/4), (0, 1/2, 3/2, 0)), exp_polar(2*I*pi)/x**2)/(2*pi**(3/2))

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