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

Optimal. Leaf size=24 \[ \sqrt {\frac {1}{x}-1} x-\tan ^{-1}\left (\sqrt {\frac {1}{x}-1}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1972, 242, 47, 63, 203} \[ \sqrt {\frac {1}{x}-1} x-\tan ^{-1}\left (\sqrt {\frac {1}{x}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 - x)/x],x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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])

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 1972

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && BinomialQ[u, x] &&  !BinomialMatchQ[
u, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ \sqrt {\frac {1}{x}-1} x-\tan ^{-1}\left (\sqrt {\frac {1}{x}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 - x)/x],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 26, normalized size = 1.08 \[ x \sqrt {-\frac {x - 1}{x}} - \arctan \left (\sqrt {-\frac {x - 1}{x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.40, size = 28, normalized size = 1.17 \[ \frac {1}{4} \, \pi \mathrm {sgn}\relax (x) + \frac {1}{2} \, \arcsin \left (2 \, x - 1\right ) \mathrm {sgn}\relax (x) + \sqrt {-x^{2} + x} \mathrm {sgn}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*pi*sgn(x) + 1/2*arcsin(2*x - 1)*sgn(x) + sqrt(-x^2 + x)*sgn(x)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 40, normalized size = 1.67 \[ \frac {\sqrt {-\frac {x -1}{x}}\, \left (\arcsin \left (2 x -1\right )+2 \sqrt {-x^{2}+x}\right ) x}{2 \sqrt {-\left (x -1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.97, size = 37, normalized size = 1.54 \[ -\frac {\sqrt {-\frac {x - 1}{x}}}{\frac {x - 1}{x} - 1} - \arctan \left (\sqrt {-\frac {x - 1}{x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-(x - 1)/x)/((x - 1)/x - 1) - arctan(sqrt(-(x - 1)/x))

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 20, normalized size = 0.83 \[ x\,\sqrt {\frac {1}{x}-1}-\mathrm {atan}\left (\sqrt {\frac {1}{x}-1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-(x - 1)/x)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((1 - x)/x), x)

________________________________________________________________________________________