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

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {1972, 242, 47, 63, 206} \[ \sqrt {\frac {x-1}{x}} x-\tanh ^{-1}\left (\sqrt {\frac {x-1}{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(-1 + x)/x],x]

[Out]

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 {\frac {-1+x}{x}} x+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\frac {1}{x}\right )\\ &=\sqrt {\frac {-1+x}{x}} x-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\frac {-1+x}{x}}\right )\\ &=\sqrt {\frac {-1+x}{x}} x-\tanh ^{-1}\left (\sqrt {\frac {-1+x}{x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.58 \[ \frac {\sqrt {x} (x-1)+\sqrt {1-x} \sin ^{-1}\left (\sqrt {1-x}\right )}{\sqrt {x-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(-1 + x)/x],x]

[Out]

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

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fricas [B]  time = 0.67, size = 40, normalized size = 1.67 \[ x \sqrt {\frac {x - 1}{x}} - \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x}} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x}} - 1\right ) \]

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

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giac [A]  time = 0.38, size = 35, normalized size = 1.46 \[ \frac {1}{2} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \mathrm {sgn}\relax (x) + \sqrt {x^{2} - x} \mathrm {sgn}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(abs(-2*x + 2*sqrt(x^2 - x) + 1))*sgn(x) + sqrt(x^2 - x)*sgn(x)

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maple [B]  time = 0.01, size = 45, normalized size = 1.88 \[ -\frac {\sqrt {\frac {x -1}{x}}\, \left (\ln \left (x -\frac {1}{2}+\sqrt {x^{2}-x}\right )-2 \sqrt {x^{2}-x}\right ) x}{2 \sqrt {\left (x -1\right ) x}} \]

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

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maxima [B]  time = 0.43, size = 51, normalized size = 2.12 \[ -\frac {\sqrt {\frac {x - 1}{x}}}{\frac {x - 1}{x} - 1} - \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x}} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x}} - 1\right ) \]

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

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mupad [B]  time = 0.03, size = 24, normalized size = 1.00 \[ x\,\sqrt {1-\frac {1}{x}}-\mathrm {atanh}\left (\sqrt {1-\frac {1}{x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x - 1)/x)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((x - 1)/x), x)

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