Integrand size = 19, antiderivative size = 128 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {32}{105} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {32}{105} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\sqrt {1+x} \left (-\frac {48}{35} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{7} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \]
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 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{105} \sqrt {1+\sqrt {1+\sqrt {1+x}}} \left (4-18 \sqrt {1+x}+4 \sqrt {1+\sqrt {1+x}}+15 \sqrt {1+x} \sqrt {1+\sqrt {1+x}}\right ) \]
(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 |
\(\displaystyle -4 \left (-\frac {2}{7} \left (\sqrt {x+1}+2\right )^{7/2}+\frac {6}{5} \left (\sqrt {x+1}+2\right )^{5/2}-\frac {4}{3} \left (\sqrt {x+1}+2\right )^{3/2}\right )\) |
-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)
3.19.70.3.1 Defintions of rubi rules used
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 = 0.12 (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 | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {7}{2}}}{7}-\frac {24 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}+\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}\) | \(47\) |
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 = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{105} \, {\left ({\left (15 \, \sqrt {x + 1} + 4\right )} \sqrt {\sqrt {x + 1} + 1} - 18 \, \sqrt {x + 1} + 4\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} \]
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 = 0.98 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.48 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=- \frac {336 \left (x + 1\right )^{\frac {13}{4}} \sqrt [4]{\sqrt {x + 1} + 1} \sin {\left (\frac {5 \operatorname {atan}{\left (\sqrt [4]{x + 1} \right )}}{2} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} - \frac {112 \left (x + 1\right )^{\frac {11}{4}} \sqrt [4]{\sqrt {x + 1} + 1} \sin {\left (\frac {5 \operatorname {atan}{\left (\sqrt [4]{x + 1} \right )}}{2} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} + \frac {224 \left (x + 1\right )^{\frac {9}{4}} \sqrt [4]{\sqrt {x + 1} + 1} \sin {\left (\frac {5 \operatorname {atan}{\left (\sqrt [4]{x + 1} \right )}}{2} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} - \frac {152 \left (x + 1\right )^{\frac {5}{2}} \left (\sqrt {x + 1} + 1\right )^{\frac {3}{4}} \cos {\left (\frac {7 \operatorname {atan}{\left (\sqrt [4]{x + 1} \right )}}{2} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} - \frac {64 \left (x + 1\right )^{\frac {5}{2}} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} - \frac {216 \left (x + 1\right )^{3} \left (\sqrt {x + 1} + 1\right )^{\frac {3}{4}} \cos {\left (\frac {7 \operatorname {atan}{\left (\sqrt [4]{x + 1} \right )}}{2} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} + \frac {64 \left (x + 1\right )^{2} \left (\sqrt {x + 1} + 1\right )^{\frac {3}{4}} \cos {\left (\frac {7 \operatorname {atan}{\left (\sqrt [4]{x + 1} \right )}}{2} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} - \frac {64 \left (x + 1\right )^{2} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right )}{105 \pi \left (x + 1\right )^{\frac {5}{2}} + 105 \pi \left (x + 1\right )^{2}} \]
-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 = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{7} \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {7}{2}} - \frac {24}{5} \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {5}{2}} + \frac {16}{3} \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {3}{2}} \]
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 = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8 \, {\left (15 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {7}{2}} - 63 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {5}{2}} + 70 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {3}{2}}\right )}}{105 \, \mathrm {sgn}\left (4 \, {\left (\sqrt {x + 1} + 1\right )}^{2} - 8 \, \sqrt {x + 1} - 7\right ) \mathrm {sgn}\left (4 \, x + 1\right )} \]
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. \[ \int \frac {1}{\sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {1}{\sqrt {\sqrt {\sqrt {x+1}+1}+1}} \,d x \]