Integrand size = 19, antiderivative size = 128 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {32}{105} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {32}{105} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\sqrt {1+x} \left (-\frac {48}{35} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{7} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \] Output:
32/105*(1+(1+(1+x)^(1/2))^(1/2))^(1/2)+32/105*(1+(1+x)^(1/2))^(1/2)*(1+(1+ (1+x)^(1/2))^(1/2))^(1/2)+(1+x)^(1/2)*(-48/35*(1+(1+(1+x)^(1/2))^(1/2))^(1 /2)+8/7*(1+(1+x)^(1/2))^(1/2)*(1+(1+(1+x)^(1/2))^(1/2))^(1/2))
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{105} \sqrt {1+\sqrt {1+\sqrt {1+x}}} \left (4-18 \sqrt {1+x}+4 \sqrt {1+\sqrt {1+x}}+15 \sqrt {1+x} \sqrt {1+\sqrt {1+x}}\right ) \] Input:
Integrate[1/Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]],x]
Output:
(8*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]*(4 - 18*Sqrt[1 + x] + 4*Sqrt[1 + Sqrt[1 + x]] + 15*Sqrt[1 + x]*Sqrt[1 + Sqrt[1 + x]]))/105
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.42, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {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.
\(\displaystyle \int \frac {1}{\sqrt {\sqrt {\sqrt {x+1}+1}+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int \frac {\sqrt {x+1}}{\sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 896 |
\(\displaystyle 2 \int \frac {\sqrt {x+1}}{\sqrt {\sqrt [4]{x+1}+1}}d\left (\sqrt {x+1}+1\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int -\frac {\sqrt {x+1}}{\sqrt {\sqrt [4]{x+1}+1}}d\left (\sqrt {x+1}+1\right )\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle -2 \int \left (1-\sqrt [4]{x+1}\right ) \sqrt {\sqrt [4]{x+1}+1}d\left (\sqrt {x+1}+1\right )\) |
\(\Big \downarrow \) 900 |
\(\displaystyle -4 \int -(x+1)^{3/4} \sqrt {\sqrt {x+1}+2}d\sqrt [4]{x+1}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -4 \int \left (-\left (\sqrt {x+1}+2\right )^{5/2}+3 \left (\sqrt {x+1}+2\right )^{3/2}-2 \sqrt {\sqrt {x+1}+2}\right )d\sqrt [4]{x+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (-\frac {2}{7} \left (\sqrt {x+1}+2\right )^{7/2}+\frac {6}{5} \left (\sqrt {x+1}+2\right )^{5/2}-\frac {4}{3} \left (\sqrt {x+1}+2\right )^{3/2}\right )\) |
Input:
Int[1/Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]],x]
Output:
-4*((-4*(2 + Sqrt[1 + x])^(3/2))/3 + (6*(2 + Sqrt[1 + x])^(5/2))/5 - (2*(2 + Sqrt[1 + x])^(7/2))/7)
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]))))
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]
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]
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]))
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]]
Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.37
method | result | size |
derivativedivides | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {7}{2}}}{7}-\frac {24 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}+\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}\) | \(47\) |
default | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {7}{2}}}{7}-\frac {24 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}+\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}\) | \(47\) |
Input:
int(1/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
8/7*(1+(1+(1+x)^(1/2))^(1/2))^(7/2)-24/5*(1+(1+(1+x)^(1/2))^(1/2))^(5/2)+1 6/3*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{105} \, {\left ({\left (15 \, \sqrt {x + 1} + 4\right )} \sqrt {\sqrt {x + 1} + 1} - 18 \, \sqrt {x + 1} + 4\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} \] Input:
integrate(1/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, algorithm="fricas")
Output:
8/105*((15*sqrt(x + 1) + 4)*sqrt(sqrt(x + 1) + 1) - 18*sqrt(x + 1) + 4)*sq rt(sqrt(sqrt(x + 1) + 1) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 445 vs. \(2 (110) = 220\).
Time = 1.07 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.48 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx =\text {Too large to display} \] Input:
integrate(1/(1+(1+(1+x)**(1/2))**(1/2))**(1/2),x)
Output:
-336*(x + 1)**(13/4)*(sqrt(x + 1) + 1)**(1/4)*sin(5*atan((x + 1)**(1/4))/2 )*gamma(1/4)*gamma(3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) - 112* (x + 1)**(11/4)*(sqrt(x + 1) + 1)**(1/4)*sin(5*atan((x + 1)**(1/4))/2)*gam ma(1/4)*gamma(3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) + 224*(x + 1)**(9/4)*(sqrt(x + 1) + 1)**(1/4)*sin(5*atan((x + 1)**(1/4))/2)*gamma(1/4 )*gamma(3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) - 152*(x + 1)**(5 /2)*(sqrt(x + 1) + 1)**(3/4)*cos(7*atan((x + 1)**(1/4))/2)*gamma(1/4)*gamm a(3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) - 64*(x + 1)**(5/2)*gam ma(1/4)*gamma(3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) - 216*(x + 1)**3*(sqrt(x + 1) + 1)**(3/4)*cos(7*atan((x + 1)**(1/4))/2)*gamma(1/4)*ga mma(3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) + 64*(x + 1)**2*(sqrt (x + 1) + 1)**(3/4)*cos(7*atan((x + 1)**(1/4))/2)*gamma(1/4)*gamma(3/4)/(1 05*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2) - 64*(x + 1)**2*gamma(1/4)*gamma (3/4)/(105*pi*(x + 1)**(5/2) + 105*pi*(x + 1)**2)
Result contains higher order function than in optimal. Order 3 vs. order 2.
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{7} \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {7}{2}} - \frac {24}{5} \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {5}{2}} + \frac {16}{3} \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {3}{2}} \] Input:
integrate(1/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, algorithm="maxima")
Output:
8/7*(sqrt(sqrt(x + 1) + 1) + 1)^(7/2) - 24/5*(sqrt(sqrt(x + 1) + 1) + 1)^( 5/2) + 16/3*(sqrt(sqrt(x + 1) + 1) + 1)^(3/2)
Result contains higher order function than in optimal. Order 3 vs. order 2.
Time = 0.42 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8 \, {\left (15 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {7}{2}} - 63 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {5}{2}} + 70 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {3}{2}}\right )}}{105 \, \mathrm {sgn}\left (4 \, {\left (\sqrt {x + 1} + 1\right )}^{2} - 8 \, \sqrt {x + 1} - 7\right ) \mathrm {sgn}\left (4 \, x + 1\right )} \] Input:
integrate(1/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, algorithm="giac")
Output:
8/105*(15*(sqrt(sqrt(x + 1) + 1) + 1)^(7/2) - 63*(sqrt(sqrt(x + 1) + 1) + 1)^(5/2) + 70*(sqrt(sqrt(x + 1) + 1) + 1)^(3/2))/(sgn(4*(sqrt(x + 1) + 1)^ 2 - 8*sqrt(x + 1) - 7)*sgn(4*x + 1))
Timed out. \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {1}{\sqrt {\sqrt {\sqrt {x+1}+1}+1}} \,d x \] Input:
int(1/(((x + 1)^(1/2) + 1)^(1/2) + 1)^(1/2),x)
Output:
int(1/(((x + 1)^(1/2) + 1)^(1/2) + 1)^(1/2), x)
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8 \sqrt {\sqrt {\sqrt {x +1}+1}+1}\, \left (15 \sqrt {x +1}\, \sqrt {\sqrt {x +1}+1}+4 \sqrt {\sqrt {x +1}+1}-18 \sqrt {x +1}+4\right )}{105} \] Input:
int(1/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x)
Output:
(8*sqrt(sqrt(sqrt(x + 1) + 1) + 1)*(15*sqrt(x + 1)*sqrt(sqrt(x + 1) + 1) + 4*sqrt(sqrt(x + 1) + 1) - 18*sqrt(x + 1) + 4))/105