3.714 \(\int \sqrt{2+\sqrt{4+\sqrt{x}}} \, dx\)

Optimal. Leaf size=64 \[ \frac{8}{9} \left (\sqrt{\sqrt{x}+4}+2\right )^{9/2}-\frac{48}{7} \left (\sqrt{\sqrt{x}+4}+2\right )^{7/2}+\frac{64}{5} \left (\sqrt{\sqrt{x}+4}+2\right )^{5/2} \]

[Out]

(64*(2 + Sqrt[4 + Sqrt[x]])^(5/2))/5 - (48*(2 + Sqrt[4 + Sqrt[x]])^(7/2))/7 + (8*(2 + Sqrt[4 + Sqrt[x]])^(9/2)
)/9

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Rubi [A]  time = 0.0491043, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {371, 1398, 772} \[ \frac{8}{9} \left (\sqrt{\sqrt{x}+4}+2\right )^{9/2}-\frac{48}{7} \left (\sqrt{\sqrt{x}+4}+2\right )^{7/2}+\frac{64}{5} \left (\sqrt{\sqrt{x}+4}+2\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + Sqrt[4 + Sqrt[x]]],x]

[Out]

(64*(2 + Sqrt[4 + Sqrt[x]])^(5/2))/5 - (48*(2 + Sqrt[4 + Sqrt[x]])^(7/2))/7 + (8*(2 + Sqrt[4 + Sqrt[x]])^(9/2)
)/9

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sqrt{2+\sqrt{4+\sqrt{x}}} \, dx &=2 \operatorname{Subst}\left (\int x \sqrt{2+\sqrt{4+x}} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \sqrt{2+\sqrt{x}} (-4+x) \, dx,x,4+\sqrt{x}\right )\\ &=4 \operatorname{Subst}\left (\int x \sqrt{2+x} \left (-4+x^2\right ) \, dx,x,\sqrt{4+\sqrt{x}}\right )\\ &=4 \operatorname{Subst}\left (\int \left (8 (2+x)^{3/2}-6 (2+x)^{5/2}+(2+x)^{7/2}\right ) \, dx,x,\sqrt{4+\sqrt{x}}\right )\\ &=\frac{64}{5} \left (2+\sqrt{4+\sqrt{x}}\right )^{5/2}-\frac{48}{7} \left (2+\sqrt{4+\sqrt{x}}\right )^{7/2}+\frac{8}{9} \left (2+\sqrt{4+\sqrt{x}}\right )^{9/2}\\ \end{align*}

Mathematica [A]  time = 0.0291675, size = 43, normalized size = 0.67 \[ -\frac{8}{315} \left (\sqrt{\sqrt{x}+4}+2\right )^{5/2} \left (130 \sqrt{\sqrt{x}+4}-35 \sqrt{x}-244\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + Sqrt[4 + Sqrt[x]]],x]

[Out]

(-8*(2 + Sqrt[4 + Sqrt[x]])^(5/2)*(-244 + 130*Sqrt[4 + Sqrt[x]] - 35*Sqrt[x]))/315

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Maple [A]  time = 0.01, size = 41, normalized size = 0.6 \begin{align*}{\frac{64}{5} \left ( 2+\sqrt{4+\sqrt{x}} \right ) ^{{\frac{5}{2}}}}-{\frac{48}{7} \left ( 2+\sqrt{4+\sqrt{x}} \right ) ^{{\frac{7}{2}}}}+{\frac{8}{9} \left ( 2+\sqrt{4+\sqrt{x}} \right ) ^{{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+(4+x^(1/2))^(1/2))^(1/2),x)

[Out]

64/5*(2+(4+x^(1/2))^(1/2))^(5/2)-48/7*(2+(4+x^(1/2))^(1/2))^(7/2)+8/9*(2+(4+x^(1/2))^(1/2))^(9/2)

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Maxima [A]  time = 1.0407, size = 54, normalized size = 0.84 \begin{align*} \frac{8}{9} \,{\left (\sqrt{\sqrt{x} + 4} + 2\right )}^{\frac{9}{2}} - \frac{48}{7} \,{\left (\sqrt{\sqrt{x} + 4} + 2\right )}^{\frac{7}{2}} + \frac{64}{5} \,{\left (\sqrt{\sqrt{x} + 4} + 2\right )}^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+(4+x^(1/2))^(1/2))^(1/2),x, algorithm="maxima")

[Out]

8/9*(sqrt(sqrt(x) + 4) + 2)^(9/2) - 48/7*(sqrt(sqrt(x) + 4) + 2)^(7/2) + 64/5*(sqrt(sqrt(x) + 4) + 2)^(5/2)

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Fricas [A]  time = 1.46157, size = 134, normalized size = 2.09 \begin{align*} \frac{8}{315} \,{\left (2 \,{\left (5 \, \sqrt{x} - 32\right )} \sqrt{\sqrt{x} + 4} + 35 \, x + 4 \, \sqrt{x} - 128\right )} \sqrt{\sqrt{\sqrt{x} + 4} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+(4+x^(1/2))^(1/2))^(1/2),x, algorithm="fricas")

[Out]

8/315*(2*(5*sqrt(x) - 32)*sqrt(sqrt(x) + 4) + 35*x + 4*sqrt(x) - 128)*sqrt(sqrt(sqrt(x) + 4) + 2)

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Sympy [B]  time = 2.26442, size = 216, normalized size = 3.38 \begin{align*} - \frac{2 \sqrt{2} \sqrt{x} \sqrt{\sqrt{x} + 4} \sqrt{\sqrt{\sqrt{x} + 4} + 2} \Gamma \left (- \frac{1}{4}\right ) \Gamma \left (\frac{1}{4}\right )}{63 \pi } - \frac{4 \sqrt{2} \sqrt{x} \sqrt{\sqrt{\sqrt{x} + 4} + 2} \Gamma \left (- \frac{1}{4}\right ) \Gamma \left (\frac{1}{4}\right )}{315 \pi } - \frac{\sqrt{2} x \sqrt{\sqrt{\sqrt{x} + 4} + 2} \Gamma \left (- \frac{1}{4}\right ) \Gamma \left (\frac{1}{4}\right )}{9 \pi } + \frac{64 \sqrt{2} \sqrt{\sqrt{x} + 4} \sqrt{\sqrt{\sqrt{x} + 4} + 2} \Gamma \left (- \frac{1}{4}\right ) \Gamma \left (\frac{1}{4}\right )}{315 \pi } + \frac{128 \sqrt{2} \sqrt{\sqrt{\sqrt{x} + 4} + 2} \Gamma \left (- \frac{1}{4}\right ) \Gamma \left (\frac{1}{4}\right )}{315 \pi } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+(4+x**(1/2))**(1/2))**(1/2),x)

[Out]

-2*sqrt(2)*sqrt(x)*sqrt(sqrt(x) + 4)*sqrt(sqrt(sqrt(x) + 4) + 2)*gamma(-1/4)*gamma(1/4)/(63*pi) - 4*sqrt(2)*sq
rt(x)*sqrt(sqrt(sqrt(x) + 4) + 2)*gamma(-1/4)*gamma(1/4)/(315*pi) - sqrt(2)*x*sqrt(sqrt(sqrt(x) + 4) + 2)*gamm
a(-1/4)*gamma(1/4)/(9*pi) + 64*sqrt(2)*sqrt(sqrt(x) + 4)*sqrt(sqrt(sqrt(x) + 4) + 2)*gamma(-1/4)*gamma(1/4)/(3
15*pi) + 128*sqrt(2)*sqrt(sqrt(sqrt(x) + 4) + 2)*gamma(-1/4)*gamma(1/4)/(315*pi)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+(4+x^(1/2))^(1/2))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError