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

Optimal. Leaf size=89 \[ \tan ^{-1}\left (\frac {3-\sqrt {x}}{2 \sqrt {x-\sqrt {x}-1}}\right )-2 \tanh ^{-1}\left (\frac {1-2 \sqrt {x}}{2 \sqrt {x-\sqrt {x}-1}}\right )-\tanh ^{-1}\left (\frac {3 \sqrt {x}+1}{2 \sqrt {x-\sqrt {x}-1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {990, 621, 206, 1033, 724, 204} \[ \tan ^{-1}\left (\frac {3-\sqrt {x}}{2 \sqrt {x-\sqrt {x}-1}}\right )-2 \tanh ^{-1}\left (\frac {1-2 \sqrt {x}}{2 \sqrt {x-\sqrt {x}-1}}\right )-\tanh ^{-1}\left (\frac {3 \sqrt {x}+1}{2 \sqrt {x-\sqrt {x}-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 - Sqrt[x] + x]/((-1 + x)*Sqrt[x]),x]

[Out]

ArcTan[(3 - Sqrt[x])/(2*Sqrt[-1 - Sqrt[x] + x])] - 2*ArcTanh[(1 - 2*Sqrt[x])/(2*Sqrt[-1 - Sqrt[x] + x])] - Arc
Tanh[(1 + 3*Sqrt[x])/(2*Sqrt[-1 - Sqrt[x] + x])]

Rule 204

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

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 990

Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sqrt[a + b*x +
c*x^2], x], x] - Dist[1/f, Int[(c*d - a*f - b*f*x)/(Sqrt[a + b*x + c*x^2]*(d + f*x^2)), x], x] /; FreeQ[{a, b,
c, d, f}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1033

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[-(a*c), 2]}, Dist[h/2 + (c*g)/(2*q), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - (c*g)
/(2*q), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[-(a*c)]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1-\sqrt {x}+x}}{(-1+x) \sqrt {x}} \, dx &=2 \operatorname {Subst}\left (\int \frac {\sqrt {-1-x+x^2}}{-1+x^2} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1-x+x^2}} \, dx,x,\sqrt {x}\right )-2 \operatorname {Subst}\left (\int \frac {x}{\left (-1+x^2\right ) \sqrt {-1-x+x^2}} \, dx,x,\sqrt {x}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {-1+2 \sqrt {x}}{\sqrt {-1-\sqrt {x}+x}}\right )-\operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {-1-x+x^2}} \, dx,x,\sqrt {x}\right )-\operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {-1-x+x^2}} \, dx,x,\sqrt {x}\right )\\ &=-2 \tanh ^{-1}\left (\frac {1-2 \sqrt {x}}{2 \sqrt {-1-\sqrt {x}+x}}\right )+2 \operatorname {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,\frac {-3+\sqrt {x}}{\sqrt {-1-\sqrt {x}+x}}\right )+2 \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {-1-3 \sqrt {x}}{\sqrt {-1-\sqrt {x}+x}}\right )\\ &=\tan ^{-1}\left (\frac {3-\sqrt {x}}{2 \sqrt {-1-\sqrt {x}+x}}\right )-2 \tanh ^{-1}\left (\frac {1-2 \sqrt {x}}{2 \sqrt {-1-\sqrt {x}+x}}\right )-\tanh ^{-1}\left (\frac {1+3 \sqrt {x}}{2 \sqrt {-1-\sqrt {x}+x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 89, normalized size = 1.00 \[ \tan ^{-1}\left (\frac {3-\sqrt {x}}{2 \sqrt {x-\sqrt {x}-1}}\right )-2 \tanh ^{-1}\left (\frac {1-2 \sqrt {x}}{2 \sqrt {x-\sqrt {x}-1}}\right )-\tanh ^{-1}\left (\frac {3 \sqrt {x}+1}{2 \sqrt {x-\sqrt {x}-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(3 - Sqrt[x])/(2*Sqrt[-1 - Sqrt[x] + x])] - 2*ArcTanh[(1 - 2*Sqrt[x])/(2*Sqrt[-1 - Sqrt[x] + x])] - Arc
Tanh[(1 + 3*Sqrt[x])/(2*Sqrt[-1 - Sqrt[x] + x])]

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fricas [A]  time = 4.14, size = 87, normalized size = 0.98 \[ -\arctan \left (\frac {{\left ({\left (x - 4\right )} \sqrt {x} - 2 \, x + 3\right )} \sqrt {x - \sqrt {x} - 1}}{2 \, {\left (x^{2} - 3 \, x + 1\right )}}\right ) + \log \left (-\frac {8 \, x^{2} + 2 \, {\left ({\left (4 \, x - 5\right )} \sqrt {x} + 2 \, x - 1\right )} \sqrt {x - \sqrt {x} - 1} - 17 \, x - 2 \, \sqrt {x} + 11}{x - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(1/2*((x - 4)*sqrt(x) - 2*x + 3)*sqrt(x - sqrt(x) - 1)/(x^2 - 3*x + 1)) + log(-(8*x^2 + 2*((4*x - 5)*sq
rt(x) + 2*x - 1)*sqrt(x - sqrt(x) - 1) - 17*x - 2*sqrt(x) + 11)/(x - 1))

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giac [A]  time = 1.24, size = 81, normalized size = 0.91 \[ -2 \, \arctan \left (\sqrt {x - \sqrt {x} - 1} - \sqrt {x} + 1\right ) - \log \left (-\sqrt {x - \sqrt {x} - 1} + \sqrt {x} + 2\right ) + \log \left (-\sqrt {x - \sqrt {x} - 1} + \sqrt {x}\right ) - 2 \, \log \left ({\left | 2 \, \sqrt {x - \sqrt {x} - 1} - 2 \, \sqrt {x} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan(sqrt(x - sqrt(x) - 1) - sqrt(x) + 1) - log(-sqrt(x - sqrt(x) - 1) + sqrt(x) + 2) + log(-sqrt(x - sqr
t(x) - 1) + sqrt(x)) - 2*log(abs(2*sqrt(x - sqrt(x) - 1) - 2*sqrt(x) + 1))

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maple [A]  time = 0.02, size = 130, normalized size = 1.46 \[ \arctanh \left (\frac {-3 \sqrt {x}-1}{2 \sqrt {-3 \sqrt {x}+\left (\sqrt {x}+1\right )^{2}-2}}\right )-\arctan \left (\frac {\sqrt {x}-3}{2 \sqrt {\sqrt {x}+\left (\sqrt {x}-1\right )^{2}-2}}\right )+\frac {3 \ln \left (\sqrt {x}-\frac {1}{2}+\sqrt {-3 \sqrt {x}+\left (\sqrt {x}+1\right )^{2}-2}\right )}{2}+\frac {\ln \left (\sqrt {x}-\frac {1}{2}+\sqrt {\sqrt {x}+\left (\sqrt {x}-1\right )^{2}-2}\right )}{2}+\sqrt {\sqrt {x}+\left (\sqrt {x}-1\right )^{2}-2}-\sqrt {-3 \sqrt {x}+\left (\sqrt {x}+1\right )^{2}-2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x - \sqrt {x} - 1}}{{\left (x - 1\right )} \sqrt {x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {x-\sqrt {x}-1}}{\sqrt {x}\,\left (x-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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