3.8.34 \(\int \frac {\sqrt {\frac {-1+x}{1+x}}}{x^2} \, dx\) [734]

Optimal. Leaf size=36 \[ -\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+\tan ^{-1}\left (\sqrt {-1+x} \sqrt {1+x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1978, 96, 94, 209} \begin {gather*} \text {ArcTan}\left (\sqrt {x-1} \sqrt {x+1}\right )-\frac {\sqrt {x-1} \sqrt {x+1}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[(-1 + x)/(1 + x)]/x^2,x]

[Out]

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

Rule 94

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 96

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 + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 209

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

Rule 1978

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*((a*e + b*e*
x^n)^p/(c + d*x^n)^p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - a*(d/b), 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 66, normalized size = 1.83 \begin {gather*} -\frac {\sqrt {\frac {-1+x}{1+x}} \left (\sqrt {-1+x} (1+x)+2 x \sqrt {1+x} \tan ^{-1}\left (x-\sqrt {-1+x} \sqrt {1+x}\right )\right )}{\sqrt {-1+x} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(58\) vs. \(2(28)=56\).
time = 0.12, size = 59, normalized size = 1.64

method	result	size
risch	\(-\frac {\left (1+x \right ) \sqrt {\frac {-1+x}{1+x}}}{x}-\frac {\arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right ) \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}}{-1+x}\)	\(56\)
default	\(\frac {\sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right ) \left (\left (x^{2}-1\right )^{\frac {3}{2}}-x^{2} \sqrt {x^{2}-1}-\arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right ) x \right )}{\sqrt {\left (1+x \right ) \left (-1+x \right )}\, x}\)	\(59\)
trager	\(-\frac {\left (1+x \right ) \sqrt {-\frac {1-x}{1+x}}}{x}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {1-x}{1+x}}\, x +\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {1-x}{1+x}}-1}{x}\right )\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-1+x)/(1+x))^(1/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

((-1+x)/(1+x))^(1/2)*(1+x)*((x^2-1)^(3/2)-x^2*(x^2-1)^(1/2)-arctan(1/(x^2-1)^(1/2))*x)/((1+x)*(-1+x))^(1/2)/x

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Maxima [A]
time = 0.50, size = 41, normalized size = 1.14 \begin {gather*} -\frac {2 \, \sqrt {\frac {x - 1}{x + 1}}}{\frac {x - 1}{x + 1} + 1} + 2 \, \arctan \left (\sqrt {\frac {x - 1}{x + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 36, normalized size = 1.00 \begin {gather*} \frac {2 \, x \arctan \left (\sqrt {\frac {x - 1}{x + 1}}\right ) - {\left (x + 1\right )} \sqrt {\frac {x - 1}{x + 1}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\frac {x - 1}{x + 1}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.50, size = 51, normalized size = 1.42 \begin {gather*} -\frac {1}{2} \, {\left (\pi - 2\right )} \mathrm {sgn}\left (x + 1\right ) + 2 \, \arctan \left (-x + \sqrt {x^{2} - 1}\right ) \mathrm {sgn}\left (x + 1\right ) - \frac {2 \, \mathrm {sgn}\left (x + 1\right )}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 41, normalized size = 1.14 \begin {gather*} 2\,\mathrm {atan}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {2\,\sqrt {\frac {x-1}{x+1}}}{\frac {x-1}{x+1}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x - 1)/(x + 1))^(1/2)/x^2,x)

[Out]

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

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