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

Optimal. Leaf size=118 \[ \frac{2 \sqrt{2} \sqrt{\sqrt{x}+\sqrt{2} \sqrt{x+\sqrt{2} \sqrt{x}+1}+\sqrt{2}} \left (3 \sqrt{2} x^{3/2}+\sqrt{2} \sqrt{x}-\sqrt{2} \left (2 \sqrt{2}-\sqrt{x}\right ) \sqrt{x+\sqrt{2} \sqrt{x}+1}+4\right )}{15 \sqrt{x}} \]

[Out]

(2*Sqrt[2]*Sqrt[Sqrt[2] + Sqrt[x] + Sqrt[2]*Sqrt[1 + Sqrt[2]*Sqrt[x] + x]]*(4 + Sqrt[2]*Sqrt[x] + 3*Sqrt[2]*x^
(3/2) - Sqrt[2]*(2*Sqrt[2] - Sqrt[x])*Sqrt[1 + Sqrt[2]*Sqrt[x] + x]))/(15*Sqrt[x])

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Rubi [A]  time = 0.190937, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {2115, 2114} \[ \frac{2 \sqrt{2} \sqrt{\sqrt{x}+\sqrt{2} \sqrt{x+\sqrt{2} \sqrt{x}+1}+\sqrt{2}} \left (3 \sqrt{2} x^{3/2}+\sqrt{2} \sqrt{x}-\sqrt{2} \left (2 \sqrt{2}-\sqrt{x}\right ) \sqrt{x+\sqrt{2} \sqrt{x}+1}+4\right )}{15 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sqrt[2] + Sqrt[x] + Sqrt[2 + 2*Sqrt[2]*Sqrt[x] + 2*x]],x]

[Out]

(2*Sqrt[2]*Sqrt[Sqrt[2] + Sqrt[x] + Sqrt[2]*Sqrt[1 + Sqrt[2]*Sqrt[x] + x]]*(4 + Sqrt[2]*Sqrt[x] + 3*Sqrt[2]*x^
(3/2) - Sqrt[2]*(2*Sqrt[2] - Sqrt[x])*Sqrt[1 + Sqrt[2]*Sqrt[x] + x]))/(15*Sqrt[x])

Rule 2115

Int[((u_) + (f_.)*((j_.) + (k_.)*Sqrt[v_]))^(n_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Int[(g + h*x)^m*(Ex
pandToSum[u + f*j, x] + f*k*Sqrt[ExpandToSum[v, x]])^n, x] /; FreeQ[{f, g, h, j, k, m, n}, x] && LinearQ[u, x]
&& QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x] && (EqQ[j, 0] || EqQ[f, 1])) && EqQ[(Co
efficient[u, x, 1]*g - h*(Coefficient[u, x, 0] + f*j))^2 - f^2*k^2*(Coefficient[v, x, 2]*g^2 - Coefficient[v,
x, 1]*g*h + Coefficient[v, x, 0]*h^2), 0]

Rule 2114

Int[((g_.) + (h_.)*(x_))*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]], x_Symbol] :
> Simp[(2*(f*(5*b*c*g^2 - 2*b^2*g*h - 3*a*c*g*h + 2*a*b*h^2) + c*f*(10*c*g^2 - b*g*h + a*h^2)*x + 9*c^2*f*g*h*
x^2 + 3*c^2*f*h^2*x^3 - (e*g - d*h)*(5*c*g - 2*b*h + c*h*x)*Sqrt[a + b*x + c*x^2])*Sqrt[d + e*x + f*Sqrt[a + b
*x + c*x^2]])/(15*c^2*f*(g + h*x)), x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[(e*g - d*h)^2 - f^2*(c*g^2
- b*g*h + a*h^2), 0] && EqQ[2*e^2*g - 2*d*e*h - f^2*(2*c*g - b*h), 0]

Rubi steps

\begin{align*} \int \sqrt{\sqrt{2}+\sqrt{x}+\sqrt{2+2 \sqrt{2} \sqrt{x}+2 x}} \, dx &=2 \operatorname{Subst}\left (\int x \sqrt{x+\sqrt{2} \left (1+\sqrt{1+\sqrt{2} x+x^2}\right )} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int x \sqrt{\sqrt{2}+x+\sqrt{2} \sqrt{1+\sqrt{2} x+x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 \sqrt{2} \sqrt{\sqrt{2}+\sqrt{x}+\sqrt{2} \sqrt{1+\sqrt{2} \sqrt{x}+x}} \left (4+\sqrt{2} \sqrt{x}+3 \sqrt{2} x^{3/2}-\sqrt{2} \left (2 \sqrt{2}-\sqrt{x}\right ) \sqrt{1+\sqrt{2} \sqrt{x}+x}\right )}{15 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0787935, size = 112, normalized size = 0.95 \[ \frac{2 \sqrt{2} \left (3 \sqrt{2} x^{3/2}+\sqrt{2} \sqrt{x}+\sqrt{2} \left (\sqrt{x}-2 \sqrt{2}\right ) \sqrt{x+\sqrt{2} \sqrt{x}+1}+4\right ) \sqrt{\sqrt{2} \left (\sqrt{x+\sqrt{2} \sqrt{x}+1}+1\right )+\sqrt{x}}}{15 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*(4 + Sqrt[2]*Sqrt[x] + 3*Sqrt[2]*x^(3/2) + Sqrt[2]*(-2*Sqrt[2] + Sqrt[x])*Sqrt[1 + Sqrt[2]*Sqrt[x]
+ x])*Sqrt[Sqrt[x] + Sqrt[2]*(1 + Sqrt[1 + Sqrt[2]*Sqrt[x] + x])])/(15*Sqrt[x])

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sqrt{2}+\sqrt{x}+\sqrt{2+2\,x+2\,\sqrt{2}\sqrt{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sqrt{2} + \sqrt{2 \, \sqrt{2} \sqrt{x} + 2 \, x + 2} + \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sqrt(2) + sqrt(2*sqrt(2)*sqrt(x) + 2*x + 2) + sqrt(x)), x)

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Fricas [A]  time = 11.0405, size = 219, normalized size = 1.86 \begin{align*} \frac{2 \,{\left (6 \, x^{2} +{\left (\sqrt{2} x - 4 \, \sqrt{x}\right )} \sqrt{2 \, \sqrt{2} \sqrt{x} + 2 \, x + 2} + 4 \, \sqrt{2} \sqrt{x} + 2 \, x\right )} \sqrt{\sqrt{2} + \sqrt{2 \, \sqrt{2} \sqrt{x} + 2 \, x + 2} + \sqrt{x}}}{15 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(6*x^2 + (sqrt(2)*x - 4*sqrt(x))*sqrt(2*sqrt(2)*sqrt(x) + 2*x + 2) + 4*sqrt(2)*sqrt(x) + 2*x)*sqrt(sqrt(2
) + sqrt(2*sqrt(2)*sqrt(x) + 2*x + 2) + sqrt(x))/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sqrt{x} + \sqrt{2 \sqrt{2} \sqrt{x} + 2 x + 2} + \sqrt{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sqrt(x) + sqrt(2*sqrt(2)*sqrt(x) + 2*x + 2) + sqrt(2)), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out