3.8.13 \(\int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx\) [713]

3.8.13.1 Optimal result
3.8.13.2 Mathematica [A] (verified)
3.8.13.3 Rubi [A] (warning: unable to verify)
3.8.13.4 Maple [A] (verified)
3.8.13.5 Fricas [A] (verification not implemented)
3.8.13.6 Sympy [A] (verification not implemented)
3.8.13.7 Maxima [A] (verification not implemented)
3.8.13.8 Giac [A] (verification not implemented)
3.8.13.9 Mupad [F(-1)]

3.8.13.1 Optimal result

Integrand size = 17, antiderivative size = 83 \[ \int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx=-48 \sqrt {2+\sqrt {1+\sqrt {x}}}+\frac {88}{3} \left (2+\sqrt {1+\sqrt {x}}\right )^{3/2}-\frac {48}{5} \left (2+\sqrt {1+\sqrt {x}}\right )^{5/2}+\frac {8}{7} \left (2+\sqrt {1+\sqrt {x}}\right )^{7/2} \]

output
88/3*(2+(1+x^(1/2))^(1/2))^(3/2)-48/5*(2+(1+x^(1/2))^(1/2))^(5/2)+8/7*(2+( 
1+x^(1/2))^(1/2))^(7/2)-48*(2+(1+x^(1/2))^(1/2))^(1/2)
 
3.8.13.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx=\frac {8}{105} \sqrt {2+\sqrt {1+\sqrt {x}}} \left (-280+76 \sqrt {1+\sqrt {x}}+3 \left (-12+5 \sqrt {1+\sqrt {x}}\right ) \sqrt {x}\right ) \]

input
Integrate[1/Sqrt[2 + Sqrt[1 + Sqrt[x]]],x]
 
output
(8*Sqrt[2 + Sqrt[1 + Sqrt[x]]]*(-280 + 76*Sqrt[1 + Sqrt[x]] + 3*(-12 + 5*S 
qrt[1 + Sqrt[x]])*Sqrt[x]))/105
 
3.8.13.3 Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {7267, 896, 25, 1732, 522, 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.

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

\(\displaystyle -4 \int \frac {(1-x) \sqrt [4]{x}}{\sqrt {\sqrt {x}+3}}d\sqrt [4]{x}\)

\(\Big \downarrow \) 522

\(\displaystyle -4 \int \left (-\left (\sqrt {x}+3\right )^{5/2}+6 \left (\sqrt {x}+3\right )^{3/2}-11 \sqrt {\sqrt {x}+3}+\frac {6}{\sqrt {\sqrt {x}+3}}\right )d\sqrt [4]{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \left (-\frac {2}{7} \left (\sqrt {x}+3\right )^{7/2}+\frac {12}{5} \left (\sqrt {x}+3\right )^{5/2}-\frac {22}{3} \left (\sqrt {x}+3\right )^{3/2}+12 \sqrt {\sqrt {x}+3}\right )\)

input
Int[1/Sqrt[2 + Sqrt[1 + Sqrt[x]]],x]
 
output
-4*(12*Sqrt[3 + Sqrt[x]] - (22*(3 + Sqrt[x])^(3/2))/3 + (12*(3 + Sqrt[x])^ 
(5/2))/5 - (2*(3 + Sqrt[x])^(7/2))/7)
 

3.8.13.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 522
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 1732
Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[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 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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]]
 
3.8.13.4 Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65

method result size
derivativedivides \(\frac {88 \left (2+\sqrt {1+\sqrt {x}}\right )^{\frac {3}{2}}}{3}-\frac {48 \left (2+\sqrt {1+\sqrt {x}}\right )^{\frac {5}{2}}}{5}+\frac {8 \left (2+\sqrt {1+\sqrt {x}}\right )^{\frac {7}{2}}}{7}-48 \sqrt {2+\sqrt {1+\sqrt {x}}}\) \(54\)
default \(\frac {88 \left (2+\sqrt {1+\sqrt {x}}\right )^{\frac {3}{2}}}{3}-\frac {48 \left (2+\sqrt {1+\sqrt {x}}\right )^{\frac {5}{2}}}{5}+\frac {8 \left (2+\sqrt {1+\sqrt {x}}\right )^{\frac {7}{2}}}{7}-48 \sqrt {2+\sqrt {1+\sqrt {x}}}\) \(54\)

input
int(1/(2+(1+x^(1/2))^(1/2))^(1/2),x,method=_RETURNVERBOSE)
 
output
88/3*(2+(1+x^(1/2))^(1/2))^(3/2)-48/5*(2+(1+x^(1/2))^(1/2))^(5/2)+8/7*(2+( 
1+x^(1/2))^(1/2))^(7/2)-48*(2+(1+x^(1/2))^(1/2))^(1/2)
 
3.8.13.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx=\frac {8}{105} \, {\left ({\left (15 \, \sqrt {x} + 76\right )} \sqrt {\sqrt {x} + 1} - 36 \, \sqrt {x} - 280\right )} \sqrt {\sqrt {\sqrt {x} + 1} + 2} \]

input
integrate(1/(2+(1+x^(1/2))^(1/2))^(1/2),x, algorithm="fricas")
 
output
8/105*((15*sqrt(x) + 76)*sqrt(sqrt(x) + 1) - 36*sqrt(x) - 280)*sqrt(sqrt(s 
qrt(x) + 1) + 2)
 
3.8.13.6 Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx=\frac {8 \left (\sqrt {\sqrt {x} + 1} + 2\right )^{\frac {7}{2}}}{7} - \frac {48 \left (\sqrt {\sqrt {x} + 1} + 2\right )^{\frac {5}{2}}}{5} + \frac {88 \left (\sqrt {\sqrt {x} + 1} + 2\right )^{\frac {3}{2}}}{3} - 48 \sqrt {\sqrt {\sqrt {x} + 1} + 2} \]

input
integrate(1/(2+(1+x**(1/2))**(1/2))**(1/2),x)
 
output
8*(sqrt(sqrt(x) + 1) + 2)**(7/2)/7 - 48*(sqrt(sqrt(x) + 1) + 2)**(5/2)/5 + 
 88*(sqrt(sqrt(x) + 1) + 2)**(3/2)/3 - 48*sqrt(sqrt(sqrt(x) + 1) + 2)
 
3.8.13.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx=\frac {8}{7} \, {\left (\sqrt {\sqrt {x} + 1} + 2\right )}^{\frac {7}{2}} - \frac {48}{5} \, {\left (\sqrt {\sqrt {x} + 1} + 2\right )}^{\frac {5}{2}} + \frac {88}{3} \, {\left (\sqrt {\sqrt {x} + 1} + 2\right )}^{\frac {3}{2}} - 48 \, \sqrt {\sqrt {\sqrt {x} + 1} + 2} \]

input
integrate(1/(2+(1+x^(1/2))^(1/2))^(1/2),x, algorithm="maxima")
 
output
8/7*(sqrt(sqrt(x) + 1) + 2)^(7/2) - 48/5*(sqrt(sqrt(x) + 1) + 2)^(5/2) + 8 
8/3*(sqrt(sqrt(x) + 1) + 2)^(3/2) - 48*sqrt(sqrt(sqrt(x) + 1) + 2)
 
3.8.13.8 Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt {2+\sqrt {1+\sqrt {x}}}} \, dx=\frac {8 \, {\left (15 \, {\left (\sqrt {\sqrt {x} + 1} + 2\right )}^{\frac {7}{2}} - 126 \, {\left (\sqrt {\sqrt {x} + 1} + 2\right )}^{\frac {5}{2}} + 385 \, {\left (\sqrt {\sqrt {x} + 1} + 2\right )}^{\frac {3}{2}} - 630 \, \sqrt {\sqrt {\sqrt {x} + 1} + 2}\right )}}{105 \, \mathrm {sgn}\left (4 \, {\left (\sqrt {x} + 1\right )}^{2} - 8 \, \sqrt {x} - 7\right ) \mathrm {sgn}\left (4 \, x - 3\right )} \]

input
integrate(1/(2+(1+x^(1/2))^(1/2))^(1/2),x, algorithm="giac")
 
output
8/105*(15*(sqrt(sqrt(x) + 1) + 2)^(7/2) - 126*(sqrt(sqrt(x) + 1) + 2)^(5/2 
) + 385*(sqrt(sqrt(x) + 1) + 2)^(3/2) - 630*sqrt(sqrt(sqrt(x) + 1) + 2))/( 
sgn(4*(sqrt(x) + 1)^2 - 8*sqrt(x) - 7)*sgn(4*x - 3))
 
3.8.13.9 Mupad [F(-1)]

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

input
int(1/((x^(1/2) + 1)^(1/2) + 2)^(1/2),x)
 
output
int(1/((x^(1/2) + 1)^(1/2) + 2)^(1/2), x)