3.10.75 \(\int \sqrt {\frac {1-x}{x}} \, dx\) [975]

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

[Out]

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

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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 = {1997, 248, 43, 65, 209} \begin {gather*} \sqrt {\frac {1}{x}-1} x-\text {ArcTan}\left (\sqrt {\frac {1}{x}-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 - x)/x],x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 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 248

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 1997

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

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Mathematica [A]
time = 0.02, size = 24, normalized size = 1.00 \begin {gather*} \sqrt {-1+\frac {1}{x}} x-\tan ^{-1}\left (\sqrt {-1+\frac {1}{x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 - x)/x],x]

[Out]

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

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Maple [A]
time = 0.10, size = 40, normalized size = 1.67

method	result	size
default	\(\frac {\sqrt {-\frac {-1+x}{x}}\, x \left (2 \sqrt {-x^{2}+x}+\arcsin \left (2 x -1\right )\right )}{2 \sqrt {-x \left (-1+x \right )}}\)	\(40\)
risch	\(\sqrt {-\frac {-1+x}{x}}\, x -\frac {\arcsin \left (2 x -1\right ) \sqrt {-\frac {-1+x}{x}}\, \sqrt {-x \left (-1+x \right )}}{2 \left (-1+x \right )}\)	\(45\)
trager	\(\sqrt {-\frac {-1+x}{x}}\, x -\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 \sqrt {-\frac {-1+x}{x}}\, x +2 x \RootOf \left (\textit {\_Z}^{2}+1\right )-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{2}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 37, normalized size = 1.54 \begin {gather*} -\frac {\sqrt {-\frac {x - 1}{x}}}{\frac {x - 1}{x} - 1} - \arctan \left (\sqrt {-\frac {x - 1}{x}}\right ) \end {gather*}

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))

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Fricas [A]
time = 0.35, size = 26, normalized size = 1.08 \begin {gather*} x \sqrt {-\frac {x - 1}{x}} - \arctan \left (\sqrt {-\frac {x - 1}{x}}\right ) \end {gather*}

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))

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

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)

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Giac [A]
time = 4.68, size = 28, normalized size = 1.17 \begin {gather*} \frac {1}{4} \, \pi \mathrm {sgn}\left (x\right ) + \frac {1}{2} \, \arcsin \left (2 \, x - 1\right ) \mathrm {sgn}\left (x\right ) + \sqrt {-x^{2} + x} \mathrm {sgn}\left (x\right ) \end {gather*}

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)

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

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))

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