3.692 \(\int \sqrt{\sqrt{x}+x} \, dx\)

Optimal. Leaf size=74 \[ \frac{2}{3} \sqrt{x+\sqrt{x}} x+\frac{1}{6} \sqrt{x+\sqrt{x}} \sqrt{x}-\frac{\sqrt{x+\sqrt{x}}}{4}+\frac{1}{4} \tanh ^{-1}\left (\frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}}\right ) \]

[Out]

-Sqrt[Sqrt[x] + x]/4 + (Sqrt[x]*Sqrt[Sqrt[x] + x])/6 + (2*x*Sqrt[Sqrt[x] + x])/3 + ArcTanh[Sqrt[x]/Sqrt[Sqrt[x
] + x]]/4

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Rubi [A]  time = 0.0451278, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {2004, 2018, 670, 640, 620, 206} \[ \frac{2}{3} \sqrt{x+\sqrt{x}} x+\frac{1}{6} \sqrt{x+\sqrt{x}} \sqrt{x}-\frac{\sqrt{x+\sqrt{x}}}{4}+\frac{1}{4} \tanh ^{-1}\left (\frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sqrt[x] + x],x]

[Out]

-Sqrt[Sqrt[x] + x]/4 + (Sqrt[x]*Sqrt[Sqrt[x] + x])/6 + (2*x*Sqrt[Sqrt[x] + x])/3 + ArcTanh[Sqrt[x]/Sqrt[Sqrt[x
] + x]]/4

Rule 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 670

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

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 620

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

Rubi steps

\begin{align*} \int \sqrt{\sqrt{x}+x} \, dx &=\frac{2}{3} x \sqrt{\sqrt{x}+x}+\frac{1}{6} \int \frac{\sqrt{x}}{\sqrt{\sqrt{x}+x}} \, dx\\ &=\frac{2}{3} x \sqrt{\sqrt{x}+x}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{x+x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{6} \sqrt{x} \sqrt{\sqrt{x}+x}+\frac{2}{3} x \sqrt{\sqrt{x}+x}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{x}{\sqrt{x+x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{4} \sqrt{\sqrt{x}+x}+\frac{1}{6} \sqrt{x} \sqrt{\sqrt{x}+x}+\frac{2}{3} x \sqrt{\sqrt{x}+x}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{x+x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{4} \sqrt{\sqrt{x}+x}+\frac{1}{6} \sqrt{x} \sqrt{\sqrt{x}+x}+\frac{2}{3} x \sqrt{\sqrt{x}+x}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{\sqrt{x}+x}}\right )\\ &=-\frac{1}{4} \sqrt{\sqrt{x}+x}+\frac{1}{6} \sqrt{x} \sqrt{\sqrt{x}+x}+\frac{2}{3} x \sqrt{\sqrt{x}+x}+\frac{1}{4} \tanh ^{-1}\left (\frac{\sqrt{x}}{\sqrt{\sqrt{x}+x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0434416, size = 51, normalized size = 0.69 \[ \frac{1}{12} \sqrt{x+\sqrt{x}} \left (8 x+2 \sqrt{x}+\frac{3 \sinh ^{-1}\left (\sqrt [4]{x}\right )}{\sqrt{\sqrt{x}+1} \sqrt [4]{x}}-3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sqrt[x] + x],x]

[Out]

(Sqrt[Sqrt[x] + x]*(-3 + 2*Sqrt[x] + 8*x + (3*ArcSinh[x^(1/4)])/(Sqrt[1 + Sqrt[x]]*x^(1/4))))/12

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Maple [A]  time = 0.003, size = 42, normalized size = 0.6 \begin{align*}{\frac{2}{3} \left ( x+\sqrt{x} \right ) ^{{\frac{3}{2}}}}-{\frac{1}{4} \left ( 1+2\,\sqrt{x} \right ) \sqrt{x+\sqrt{x}}}+{\frac{1}{8}\ln \left ( \sqrt{x}+{\frac{1}{2}}+\sqrt{x+\sqrt{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x+x^(1/2))^(3/2)-1/4*(1+2*x^(1/2))*(x+x^(1/2))^(1/2)+1/8*ln(x^(1/2)+1/2+(x+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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + sqrt(x)), x)

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Fricas [A]  time = 4.05134, size = 157, normalized size = 2.12 \begin{align*} \frac{1}{12} \,{\left (8 \, x + 2 \, \sqrt{x} - 3\right )} \sqrt{x + \sqrt{x}} + \frac{1}{16} \, \log \left (4 \, \sqrt{x + \sqrt{x}}{\left (2 \, \sqrt{x} + 1\right )} + 8 \, x + 8 \, \sqrt{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*x + 2*sqrt(x) - 3)*sqrt(x + sqrt(x)) + 1/16*log(4*sqrt(x + sqrt(x))*(2*sqrt(x) + 1) + 8*x + 8*sqrt(x)
+ 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sqrt(x) + x), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError