\(\int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx\) [2538]

Optimal result

Integrand size = 31, antiderivative size = 213 \[ \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx=-\frac {64}{315} \sqrt {1-\sqrt {1-\sqrt {\frac {-1+x}{x}}}}+\frac {64}{315} \sqrt {1-\sqrt {1-\sqrt {\frac {-1+x}{x}}}} \sqrt {1-\sqrt {\frac {-1+x}{x}}}+\left (-\frac {8}{315} \sqrt {1-\sqrt {1-\sqrt {\frac {-1+x}{x}}}}+\frac {8}{63} \sqrt {1-\sqrt {1-\sqrt {\frac {-1+x}{x}}}} \sqrt {1-\sqrt {\frac {-1+x}{x}}}\right ) \sqrt {\frac {-1+x}{x}}+\frac {8 \sqrt {1-\sqrt {1-\sqrt {\frac {-1+x}{x}}}} (-1+x)}{9 x} \] Output:

-64/315*(1-(1-((-1+x)/x)^(1/2))^(1/2))^(1/2)+64/315*(1-(1-((-1+x)/x)^(1/2) 
)^(1/2))^(1/2)*(1-((-1+x)/x)^(1/2))^(1/2)+(-8/315*(1-(1-((-1+x)/x)^(1/2))^ 
(1/2))^(1/2)+8/63*(1-(1-((-1+x)/x)^(1/2))^(1/2))^(1/2)*(1-((-1+x)/x)^(1/2) 
)^(1/2))*((-1+x)/x)^(1/2)+8/9*(1-(1-((-1+x)/x)^(1/2))^(1/2))^(1/2)*(-1+x)/ 
x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx=\frac {8 \sqrt {1-\sqrt {1-\sqrt {\frac {-1+x}{x}}}} \left (-35+\left (27+8 \sqrt {1-\sqrt {\frac {-1+x}{x}}}-\sqrt {\frac {-1+x}{x}}+5 \sqrt {1-\sqrt {\frac {-1+x}{x}}} \sqrt {\frac {-1+x}{x}}\right ) x\right )}{315 x} \] Input:

Integrate[Sqrt[1 - Sqrt[1 - Sqrt[1 - x^(-1)]]]/x^2,x]
 

Output:

(8*Sqrt[1 - Sqrt[1 - Sqrt[(-1 + x)/x]]]*(-35 + (27 + 8*Sqrt[1 - Sqrt[(-1 + 
 x)/x]] - Sqrt[(-1 + x)/x] + 5*Sqrt[1 - Sqrt[(-1 + x)/x]]*Sqrt[(-1 + x)/x] 
)*x))/(315*x)
 

Rubi [A] (warning: unable to verify)

Time = 0.54 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {7266, 7267, 896, 25, 1388, 900, 86, 2009}

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.

\(\displaystyle \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx\)

\(\Big \downarrow \) 7266

\(\displaystyle -\int \sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}d\frac {1}{x}\)

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 896

\(\displaystyle 2 \int -\sqrt {1-\sqrt [4]{1-\frac {1}{x}}} \sqrt {1-\frac {1}{x}}d\left (1-\sqrt {1-\frac {1}{x}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int \sqrt {1-\sqrt [4]{1-\frac {1}{x}}} \sqrt {1-\frac {1}{x}}d\left (1-\sqrt {1-\frac {1}{x}}\right )\)

\(\Big \downarrow \) 1388

\(\displaystyle -2 \int \left (1-\sqrt [4]{1-\frac {1}{x}}\right )^{3/2} \left (\sqrt [4]{1-\frac {1}{x}}+1\right )d\left (1-\sqrt {1-\frac {1}{x}}\right )\)

\(\Big \downarrow \) 900

\(\displaystyle -4 \int \left (2-\sqrt {1-\frac {1}{x}}\right ) \left (1-\frac {1}{x}\right )d\sqrt [4]{1-\frac {1}{x}}\)

\(\Big \downarrow \) 86

\(\displaystyle -4 \int \left (\left (1-\frac {1}{x}\right )^{7/4}-3 \left (1-\frac {1}{x}\right )^{5/4}+2 \left (1-\frac {1}{x}\right )^{3/4}\right )d\sqrt [4]{1-\frac {1}{x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \left (-\frac {2}{9} \left (1-\frac {1}{x}\right )^{9/4}+\frac {6}{7} \left (1-\frac {1}{x}\right )^{7/4}-\frac {4}{5} \left (1-\frac {1}{x}\right )^{5/4}\right )\)

Input:

Int[Sqrt[1 - Sqrt[1 - Sqrt[1 - x^(-1)]]]/x^2,x]
 

Output:

-4*((-4*(1 - x^(-1))^(5/4))/5 + (6*(1 - x^(-1))^(7/4))/7 - (2*(1 - x^(-1)) 
^(9/4))/9)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(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 900

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol 
] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(a + b*x^(g*n) 
)^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}, x] 
 && NeQ[b*c - a*d, 0] && FractionQ[n]
 

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7266

Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1)   Subst[Int[SubstFor[x^(m 
+ 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function 
OfQ[x^(m + 1), u, x]
 

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 [A] (verified)

Time = 1.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.33

method	result	size
derivativedivides	\(\frac {8 \left (1-\sqrt {1-\sqrt {1-\frac {1}{x}}}\right )^{\frac {9}{2}}}{9}-\frac {24 \left (1-\sqrt {1-\sqrt {1-\frac {1}{x}}}\right )^{\frac {7}{2}}}{7}+\frac {16 \left (1-\sqrt {1-\sqrt {1-\frac {1}{x}}}\right )^{\frac {5}{2}}}{5}\)	\(71\)
default	\(\frac {8 \left (1-\sqrt {1-\sqrt {1-\frac {1}{x}}}\right )^{\frac {9}{2}}}{9}-\frac {24 \left (1-\sqrt {1-\sqrt {1-\frac {1}{x}}}\right )^{\frac {7}{2}}}{7}+\frac {16 \left (1-\sqrt {1-\sqrt {1-\frac {1}{x}}}\right )^{\frac {5}{2}}}{5}\)	\(71\)

Input:

int((1-(1-(1-1/x)^(1/2))^(1/2))^(1/2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

8/9*(1-(1-(1-1/x)^(1/2))^(1/2))^(9/2)-24/7*(1-(1-(1-1/x)^(1/2))^(1/2))^(7/ 
2)+16/5*(1-(1-(1-1/x)^(1/2))^(1/2))^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.35 \[ \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx=\frac {8 \, {\left ({\left (5 \, x \sqrt {\frac {x - 1}{x}} + 8 \, x\right )} \sqrt {-\sqrt {\frac {x - 1}{x}} + 1} - x \sqrt {\frac {x - 1}{x}} + 27 \, x - 35\right )} \sqrt {-\sqrt {-\sqrt {\frac {x - 1}{x}} + 1} + 1}}{315 \, x} \] Input:

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

Output:

8/315*((5*x*sqrt((x - 1)/x) + 8*x)*sqrt(-sqrt((x - 1)/x) + 1) - x*sqrt((x 
- 1)/x) + 27*x - 35)*sqrt(-sqrt(-sqrt((x - 1)/x) + 1) + 1)/x
 

Sympy [F]

\[ \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx=\int \frac {\sqrt {1 - \sqrt {1 - \sqrt {1 - \frac {1}{x}}}}}{x^{2}}\, dx \] Input:

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

Output:

Integral(sqrt(1 - sqrt(1 - sqrt(1 - 1/x)))/x**2, x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(-sqrt(-sqrt(-1/x + 1) + 1) + 1)/x^2, x)
 

Giac [F]

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

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

Output:

integrate(sqrt(-sqrt(-sqrt(-1/x + 1) + 1) + 1)/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx=\int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {1-\sqrt {1-\sqrt {1-\frac {1}{x}}}}}{x^2} \, dx=\int \frac {\sqrt {-\sqrt {-\sqrt {x -1}+\sqrt {x}}+x^{\frac {1}{4}}}}{x^{\frac {17}{8}}}d x \] Input:

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

Output:

int((x**(7/8)*sqrt( - sqrt( - sqrt(x - 1) + sqrt(x)) + x**(1/4)))/x**3,x)