\(\int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx\) [15]

Optimal result

Integrand size = 27, antiderivative size = 77 \[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\frac {2 \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \left (2+\sqrt {x}+6 x^{3/2}-\left (2-\sqrt {x}\right ) \sqrt {1+2 \sqrt {x}+2 x}\right )}{15 \sqrt {x}} \]

[Out]

2/15*(2+6*x^(3/2)+x^(1/2)-(2-x^(1/2))*(1+2*x+2*x^(1/2))^(1/2))*(1+x^(1/2)+(1+2*x+2*x^(1/2))^(1/2))^(1/2)/x^(1/
2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2139} \[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\frac {2 \sqrt {\sqrt {x}+\sqrt {2 x+2 \sqrt {x}+1}+1} \left (6 x^{3/2}+\sqrt {x}-\left (2-\sqrt {x}\right ) \sqrt {2 x+2 \sqrt {x}+1}+2\right )}{15 \sqrt {x}} \]

[In]

Int[Sqrt[1 + Sqrt[x] + Sqrt[1 + 2*Sqrt[x] + 2*x]],x]

[Out]

(2*Sqrt[1 + Sqrt[x] + Sqrt[1 + 2*Sqrt[x] + 2*x]]*(2 + Sqrt[x] + 6*x^(3/2) - (2 - Sqrt[x])*Sqrt[1 + 2*Sqrt[x] +
2*x]))/(15*Sqrt[x])

Rule 2139

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])/(15*c^2*f*(g + h*x)))*Sqrt[
d + e*x + f*Sqrt[a + b*x + c*x^2]], 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*} \text {integral}& = 2 \text {Subst}\left (\int x \sqrt {1+x+\sqrt {1+2 x+2 x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {2 \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \left (2+\sqrt {x}+6 x^{3/2}-\left (2-\sqrt {x}\right ) \sqrt {1+2 \sqrt {x}+2 x}\right )}{15 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\frac {2 \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \left (2+\sqrt {x}+6 x^{3/2}+\left (-2+\sqrt {x}\right ) \sqrt {1+2 \sqrt {x}+2 x}\right )}{15 \sqrt {x}} \]

[In]

Integrate[Sqrt[1 + Sqrt[x] + Sqrt[1 + 2*Sqrt[x] + 2*x]],x]

[Out]

(2*Sqrt[1 + Sqrt[x] + Sqrt[1 + 2*Sqrt[x] + 2*x]]*(2 + Sqrt[x] + 6*x^(3/2) + (-2 + Sqrt[x])*Sqrt[1 + 2*Sqrt[x]
+ 2*x]))/(15*Sqrt[x])

Maple [F]

\[\int \sqrt {1+\sqrt {x}+\sqrt {1+2 x +2 \sqrt {x}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\frac {2 \, {\left (6 \, x^{2} + \sqrt {2 \, x + 2 \, \sqrt {x} + 1} {\left (x - 2 \, \sqrt {x}\right )} + x + 2 \, \sqrt {x}\right )} \sqrt {\sqrt {2 \, x + 2 \, \sqrt {x} + 1} + \sqrt {x} + 1}}{15 \, x} \]

[In]

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

[Out]

2/15*(6*x^2 + sqrt(2*x + 2*sqrt(x) + 1)*(x - 2*sqrt(x)) + x + 2*sqrt(x))*sqrt(sqrt(2*x + 2*sqrt(x) + 1) + sqrt
(x) + 1)/x

Sympy [F]

\[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\int \sqrt {\sqrt {x} + \sqrt {2 \sqrt {x} + 2 x + 1} + 1}\, dx \]

[In]

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

[Out]

Integral(sqrt(sqrt(x) + sqrt(2*sqrt(x) + 2*x + 1) + 1), x)

Maxima [F]

\[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\int { \sqrt {\sqrt {2 \, x + 2 \, \sqrt {x} + 1} + \sqrt {x} + 1} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\int { \sqrt {\sqrt {2 \, x + 2 \, \sqrt {x} + 1} + \sqrt {x} + 1} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {1+\sqrt {x}+\sqrt {1+2 \sqrt {x}+2 x}} \, dx=\int \sqrt {\sqrt {2\,x+2\,\sqrt {x}+1}+\sqrt {x}+1} \,d x \]

[In]

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

[Out]

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