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

Optimal. Leaf size=70 \[ -\frac{2}{3} \left (-x-\sqrt{x}+1\right )^{3/2}-\frac{1}{4} \left (2 \sqrt{x}+1\right ) \sqrt{-x-\sqrt{x}+1}-\frac{5}{8} \sin ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{5}}\right ) \]

[Out]

-((1 + 2*Sqrt[x])*Sqrt[1 - Sqrt[x] - x])/4 - (2*(1 - Sqrt[x] - x)^(3/2))/3 - (5*ArcSin[(1 + 2*Sqrt[x])/Sqrt[5]
])/8

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Rubi [A]  time = 0.0347816, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {1341, 640, 612, 619, 216} \[ -\frac{2}{3} \left (-x-\sqrt{x}+1\right )^{3/2}-\frac{1}{4} \left (2 \sqrt{x}+1\right ) \sqrt{-x-\sqrt{x}+1}-\frac{5}{8} \sin ^{-1}\left (\frac{2 \sqrt{x}+1}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - Sqrt[x] - x],x]

[Out]

-((1 + 2*Sqrt[x])*Sqrt[1 - Sqrt[x] - x])/4 - (2*(1 - Sqrt[x] - x)^(3/2))/3 - (5*ArcSin[(1 + 2*Sqrt[x])/Sqrt[5]
])/8

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 640

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

Rule 612

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

Rule 619

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \sqrt{1-\sqrt{x}-x} \, dx &=2 \operatorname{Subst}\left (\int x \sqrt{1-x-x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2}{3} \left (1-\sqrt{x}-x\right )^{3/2}-\operatorname{Subst}\left (\int \sqrt{1-x-x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{4} \left (1+2 \sqrt{x}\right ) \sqrt{1-\sqrt{x}-x}-\frac{2}{3} \left (1-\sqrt{x}-x\right )^{3/2}-\frac{5}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x-x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{4} \left (1+2 \sqrt{x}\right ) \sqrt{1-\sqrt{x}-x}-\frac{2}{3} \left (1-\sqrt{x}-x\right )^{3/2}+\frac{1}{8} \sqrt{5} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{5}}} \, dx,x,-1-2 \sqrt{x}\right )\\ &=-\frac{1}{4} \left (1+2 \sqrt{x}\right ) \sqrt{1-\sqrt{x}-x}-\frac{2}{3} \left (1-\sqrt{x}-x\right )^{3/2}-\frac{5}{8} \sin ^{-1}\left (\frac{1+2 \sqrt{x}}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0281065, size = 53, normalized size = 0.76 \[ \frac{1}{12} \sqrt{-x-\sqrt{x}+1} \left (8 x+2 \sqrt{x}-11\right )+\frac{5}{8} \sin ^{-1}\left (\frac{-2 \sqrt{x}-1}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - Sqrt[x] - x],x]

[Out]

(Sqrt[1 - Sqrt[x] - x]*(-11 + 2*Sqrt[x] + 8*x))/12 + (5*ArcSin[(-1 - 2*Sqrt[x])/Sqrt[5]])/8

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Maple [A]  time = 0.006, size = 50, normalized size = 0.7 \begin{align*} -{\frac{2}{3} \left ( 1-x-\sqrt{x} \right ) ^{{\frac{3}{2}}}}+{\frac{1}{4} \left ( -2\,\sqrt{x}-1 \right ) \sqrt{1-x-\sqrt{x}}}-{\frac{5}{8}\arcsin \left ({\frac{2\,\sqrt{5}}{5} \left ( \sqrt{x}+{\frac{1}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(1-x-x^(1/2))^(3/2)+1/4*(-2*x^(1/2)-1)*(1-x-x^(1/2))^(1/2)-5/8*arcsin(2/5*5^(1/2)*(x^(1/2)+1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-x - \sqrt{x} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-x - sqrt(x) + 1), x)

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Fricas [A]  time = 7.01037, size = 231, normalized size = 3.3 \begin{align*} \frac{1}{12} \,{\left (8 \, x + 2 \, \sqrt{x} - 11\right )} \sqrt{-x - \sqrt{x} + 1} + \frac{5}{16} \, \arctan \left (-\frac{{\left (8 \, x^{2} -{\left (16 \, x^{2} - 38 \, x + 11\right )} \sqrt{x} - 9 \, x + 3\right )} \sqrt{-x - \sqrt{x} + 1}}{4 \,{\left (4 \, x^{3} - 13 \, x^{2} + 7 \, x - 1\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*x + 2*sqrt(x) - 11)*sqrt(-x - sqrt(x) + 1) + 5/16*arctan(-1/4*(8*x^2 - (16*x^2 - 38*x + 11)*sqrt(x) -
9*x + 3)*sqrt(-x - sqrt(x) + 1)/(4*x^3 - 13*x^2 + 7*x - 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- \sqrt{x} - x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-sqrt(x) - x + 1), x)

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Giac [A]  time = 1.23041, size = 59, normalized size = 0.84 \begin{align*} \frac{1}{12} \,{\left (2 \, \sqrt{x}{\left (4 \, \sqrt{x} + 1\right )} - 11\right )} \sqrt{-x - \sqrt{x} + 1} - \frac{5}{8} \, \arcsin \left (\frac{1}{5} \, \sqrt{5}{\left (2 \, \sqrt{x} + 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*sqrt(x)*(4*sqrt(x) + 1) - 11)*sqrt(-x - sqrt(x) + 1) - 5/8*arcsin(1/5*sqrt(5)*(2*sqrt(x) + 1))