3.2391 \(\int \sqrt {3-\frac {1}{\sqrt {x}}} \, dx\)

Optimal. Leaf size=67 \[ \sqrt {3-\frac {1}{\sqrt {x}}} x-\frac {1}{6} \sqrt {3-\frac {1}{\sqrt {x}}} \sqrt {x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3-\frac {1}{\sqrt {x}}}}{\sqrt {3}}\right )}{6 \sqrt {3}} \]

[Out]

-1/18*arctanh(1/3*(3-1/x^(1/2))^(1/2)*3^(1/2))*3^(1/2)+x*(3-1/x^(1/2))^(1/2)-1/6*x^(1/2)*(3-1/x^(1/2))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 67, 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 = {190, 47, 51, 63, 206} \[ \sqrt {3-\frac {1}{\sqrt {x}}} x-\frac {1}{6} \sqrt {3-\frac {1}{\sqrt {x}}} \sqrt {x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3-\frac {1}{\sqrt {x}}}}{\sqrt {3}}\right )}{6 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[3 - 1/Sqrt[x]],x]

[Out]

-(Sqrt[3 - 1/Sqrt[x]]*Sqrt[x])/6 + Sqrt[3 - 1/Sqrt[x]]*x - ArcTanh[Sqrt[3 - 1/Sqrt[x]]/Sqrt[3]]/(6*Sqrt[3])

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 190

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

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])

Rubi steps

\begin {align*} \int \sqrt {3-\frac {1}{\sqrt {x}}} \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {\sqrt {3-x}}{x^3} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=\sqrt {3-\frac {1}{\sqrt {x}}} x+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-x} x^2} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {1}{6} \sqrt {3-\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {3-\frac {1}{\sqrt {x}}} x+\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-x} x} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {1}{6} \sqrt {3-\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {3-\frac {1}{\sqrt {x}}} x-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{3-x^2} \, dx,x,\sqrt {3-\frac {1}{\sqrt {x}}}\right )\\ &=-\frac {1}{6} \sqrt {3-\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {3-\frac {1}{\sqrt {x}}} x-\frac {\tanh ^{-1}\left (\frac {\sqrt {3-\frac {1}{\sqrt {x}}}}{\sqrt {3}}\right )}{6 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 36, normalized size = 0.54 \[ \frac {4}{81} \left (3-\frac {1}{\sqrt {x}}\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};1-\frac {1}{3 \sqrt {x}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[3 - 1/Sqrt[x]],x]

[Out]

(4*(3 - 1/Sqrt[x])^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 - 1/(3*Sqrt[x])])/81

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fricas [A]  time = 0.56, size = 63, normalized size = 0.94 \[ \frac {1}{6} \, {\left (6 \, x - \sqrt {x}\right )} \sqrt {\frac {3 \, x - \sqrt {x}}{x}} + \frac {1}{36} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {x} \sqrt {\frac {3 \, x - \sqrt {x}}{x}} - 6 \, \sqrt {x} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*x - sqrt(x))*sqrt((3*x - sqrt(x))/x) + 1/36*sqrt(3)*log(2*sqrt(3)*sqrt(x)*sqrt((3*x - sqrt(x))/x) - 6*s
qrt(x) + 1)

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giac [A]  time = 0.84, size = 59, normalized size = 0.88 \[ \frac {1}{36} \, {\left (6 \, \sqrt {3 \, x - \sqrt {x}} {\left (6 \, \sqrt {x} - 1\right )} + \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} \sqrt {x} - \sqrt {3 \, x - \sqrt {x}}\right )} + 1 \right |}\right )\right )} \mathrm {sgn}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(6*sqrt(3*x - sqrt(x))*(6*sqrt(x) - 1) + sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*sqrt(x) - sqrt(3*x - sqrt(x)
)) + 1)))*sgn(x)

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maple [A]  time = 0.02, size = 91, normalized size = 1.36 \[ -\frac {\sqrt {\frac {3 \sqrt {x}-1}{\sqrt {x}}}\, \left (\sqrt {3}\, \ln \left (\sqrt {3}\, \sqrt {x}-\frac {\sqrt {3}}{6}+\sqrt {3 x -\sqrt {x}}\right )-36 \sqrt {3 x -\sqrt {x}}\, \sqrt {x}+6 \sqrt {3 x -\sqrt {x}}\right ) \sqrt {x}}{36 \sqrt {\left (3 \sqrt {x}-1\right ) \sqrt {x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*((3*x^(1/2)-1)/x^(1/2))^(1/2)*x^(1/2)*(ln(-1/6*3^(1/2)+3^(1/2)*x^(1/2)+(3*x-x^(1/2))^(1/2))*3^(1/2)-36*(
3*x-x^(1/2))^(1/2)*x^(1/2)+6*(3*x-x^(1/2))^(1/2))/((3*x^(1/2)-1)*x^(1/2))^(1/2)

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maxima [A]  time = 1.21, size = 78, normalized size = 1.16 \[ \frac {1}{36} \, \sqrt {3} \log \left (-\frac {\sqrt {3} - \sqrt {-\frac {1}{\sqrt {x}} + 3}}{\sqrt {3} + \sqrt {-\frac {1}{\sqrt {x}} + 3}}\right ) + \frac {{\left (-\frac {1}{\sqrt {x}} + 3\right )}^{\frac {3}{2}} + 3 \, \sqrt {-\frac {1}{\sqrt {x}} + 3}}{6 \, {\left ({\left (\frac {1}{\sqrt {x}} - 3\right )}^{2} + \frac {6}{\sqrt {x}} - 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*sqrt(3)*log(-(sqrt(3) - sqrt(-1/sqrt(x) + 3))/(sqrt(3) + sqrt(-1/sqrt(x) + 3))) + 1/6*((-1/sqrt(x) + 3)^(
3/2) + 3*sqrt(-1/sqrt(x) + 3))/((1/sqrt(x) - 3)^2 + 6/sqrt(x) - 9)

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mupad [B]  time = 1.27, size = 31, normalized size = 0.46 \[ \frac {4\,x\,\sqrt {3-\frac {1}{\sqrt {x}}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{2};\ \frac {5}{2};\ 3\,\sqrt {x}\right )}{3\,\sqrt {1-3\,\sqrt {x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3 - 1/x^(1/2))^(1/2),x)

[Out]

(4*x*(3 - 1/x^(1/2))^(1/2)*hypergeom([-1/2, 3/2], 5/2, 3*x^(1/2)))/(3*(1 - 3*x^(1/2))^(1/2))

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sympy [A]  time = 2.96, size = 165, normalized size = 2.46 \[ \begin {cases} \frac {3 x^{\frac {5}{4}}}{\sqrt {3 \sqrt {x} - 1}} - \frac {3 x^{\frac {3}{4}}}{2 \sqrt {3 \sqrt {x} - 1}} + \frac {\sqrt [4]{x}}{6 \sqrt {3 \sqrt {x} - 1}} - \frac {\sqrt {3} \operatorname {acosh}{\left (\sqrt {3} \sqrt [4]{x} \right )}}{18} & \text {for}\: 3 \left |{\sqrt {x}}\right | > 1 \\- \frac {3 i x^{\frac {5}{4}}}{\sqrt {1 - 3 \sqrt {x}}} + \frac {3 i x^{\frac {3}{4}}}{2 \sqrt {1 - 3 \sqrt {x}}} - \frac {i \sqrt [4]{x}}{6 \sqrt {1 - 3 \sqrt {x}}} + \frac {\sqrt {3} i \operatorname {asin}{\left (\sqrt {3} \sqrt [4]{x} \right )}}{18} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x**(5/4)/sqrt(3*sqrt(x) - 1) - 3*x**(3/4)/(2*sqrt(3*sqrt(x) - 1)) + x**(1/4)/(6*sqrt(3*sqrt(x) -
1)) - sqrt(3)*acosh(sqrt(3)*x**(1/4))/18, 3*Abs(sqrt(x)) > 1), (-3*I*x**(5/4)/sqrt(1 - 3*sqrt(x)) + 3*I*x**(3/
4)/(2*sqrt(1 - 3*sqrt(x))) - I*x**(1/4)/(6*sqrt(1 - 3*sqrt(x))) + sqrt(3)*I*asin(sqrt(3)*x**(1/4))/18, True))

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