\(\int \sqrt {\sqrt {2}+\sqrt {x}+\sqrt {2+2 \sqrt {2} \sqrt {x}+2 x}} \, dx\) [16]

Optimal result

Integrand size = 36, antiderivative size = 118 \[ \int \sqrt {\sqrt {2}+\sqrt {x}+\sqrt {2+2 \sqrt {2} \sqrt {x}+2 x}} \, dx=\frac {2 \sqrt {2} \sqrt {\sqrt {2}+\sqrt {x}+\sqrt {2} \sqrt {1+\sqrt {2} \sqrt {x}+x}} \left (4+\sqrt {2} \sqrt {x}+3 \sqrt {2} x^{3/2}-\sqrt {2} \left (2 \sqrt {2}-\sqrt {x}\right ) \sqrt {1+\sqrt {2} \sqrt {x}+x}\right )}{15 \sqrt {x}} \] Output:

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

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.95 \[ \int \sqrt {\sqrt {2}+\sqrt {x}+\sqrt {2+2 \sqrt {2} \sqrt {x}+2 x}} \, dx=\frac {2 \sqrt {2} \left (4+\sqrt {2} \sqrt {x}+3 \sqrt {2} x^{3/2}+\sqrt {2} \left (-2 \sqrt {2}+\sqrt {x}\right ) \sqrt {1+\sqrt {2} \sqrt {x}+x}\right ) \sqrt {\sqrt {x}+\sqrt {2} \left (1+\sqrt {1+\sqrt {2} \sqrt {x}+x}\right )}}{15 \sqrt {x}} \] Input:

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

Output:

(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])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {7267, 2540, 2539}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7267

\(\displaystyle 2 \int \sqrt {x} \sqrt {\sqrt {2} \left (\sqrt {x+\sqrt {2} \sqrt {x}+1}+1\right )+\sqrt {x}}d\sqrt {x}\)

\(\Big \downarrow \) 2540

\(\displaystyle 2 \int \sqrt {x} \sqrt {\sqrt {x}+\sqrt {2} \sqrt {x+\sqrt {2} \sqrt {x}+1}+\sqrt {2}}d\sqrt {x}\)

\(\Big \downarrow \) 2539

\(\displaystyle \frac {2 \sqrt {2} \sqrt {\sqrt {x}+\sqrt {2} \sqrt {x+\sqrt {2} \sqrt {x}+1}+\sqrt {2}} \left (3 \sqrt {2} x^{3/2}+\sqrt {2} \sqrt {x}-\sqrt {2} \left (2 \sqrt {2}-\sqrt {x}\right ) \sqrt {x+\sqrt {2} \sqrt {x}+1}+4\right )}{15 \sqrt {x}}\)

Input:

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

Output:

(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])
 

Defintions of rubi rules used

rule 2539

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] /; Fre 
eQ[{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]
 

rule 2540

Int[((u_) + (f_.)*((j_.) + (k_.)*Sqrt[v_]))^(n_.)*((g_.) + (h_.)*(x_))^(m_. 
), x_Symbol] :> Int[(g + h*x)^m*(ExpandToSum[u + f*j, x] + f*k*Sqrt[ExpandT 
oSum[v, x]])^n, x] /; FreeQ[{f, g, h, j, k, m, n}, x] && LinearQ[u, x] && Q 
uadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x] && (EqQ[j 
, 0] || EqQ[f, 1])) && EqQ[(Coefficient[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 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \sqrt {\sqrt {2}+\sqrt {x}+\sqrt {2+2 \sqrt {2} \sqrt {x}+2 x}} \, dx=\frac {2 \, {\left (6 \, x^{2} + {\left (\sqrt {2} x - 4 \, \sqrt {x}\right )} \sqrt {2 \, \sqrt {2} \sqrt {x} + 2 \, x + 2} + 4 \, \sqrt {2} \sqrt {x} + 2 \, x\right )} \sqrt {\sqrt {2} + \sqrt {2 \, \sqrt {2} \sqrt {x} + 2 \, x + 2} + \sqrt {x}}}{15 \, x} \] Input:

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

Output:

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
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \sqrt {\sqrt {2}+\sqrt {x}+\sqrt {2+2 \sqrt {2} \sqrt {x}+2 x}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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