Integrand size = 43, antiderivative size = 184 \[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {88}{15} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{5} \sqrt {1+x} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {32}{15} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-2 \sqrt {2 \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {-1+\sqrt {2}}}\right )-2 \sqrt {2 \left (-1+\sqrt {2}\right )} \text {arctanh}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right ) \]
88/15*(1+(1+(1+x)^(1/2))^(1/2))^(1/2)+8/5*(1+x)^(1/2)*(1+(1+(1+x)^(1/2))^( 1/2))^(1/2)-32/15*(1+(1+x)^(1/2))^(1/2)*(1+(1+(1+x)^(1/2))^(1/2))^(1/2)-2* (2+2*2^(1/2))^(1/2)*arctan((1+(1+(1+x)^(1/2))^(1/2))^(1/2)/(2^(1/2)-1)^(1/ 2))-2*(-2+2*2^(1/2))^(1/2)*arctanh((1+(1+(1+x)^(1/2))^(1/2))^(1/2)/(1+2^(1 /2))^(1/2))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{15} \left (11+3 \sqrt {1+x}-4 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}-2 \sqrt {2 \left (1+\sqrt {2}\right )} \arctan \left (\sqrt {1+\sqrt {2}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-2 \sqrt {2 \left (-1+\sqrt {2}\right )} \text {arctanh}\left (\sqrt {-1+\sqrt {2}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \]
(8*(11 + 3*Sqrt[1 + x] - 4*Sqrt[1 + Sqrt[1 + x]])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/15 - 2*Sqrt[2*(1 + Sqrt[2])]*ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[1 + Sq rt[1 + Sqrt[1 + x]]]] - 2*Sqrt[2*(-1 + Sqrt[2])]*ArcTanh[Sqrt[-1 + Sqrt[2] ]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]]
Time = 1.74 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {7267, 25, 2003, 7267, 25, 2003, 656, 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.
| \(\displaystyle \int \frac {\sqrt {x+1} \sqrt {\sqrt {x+1}+1}}{x \sqrt {\sqrt {\sqrt {x+1}+1}+1}} \, dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 2 \int \frac {(x+1) \sqrt {\sqrt {x+1}+1}}{x \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int -\frac {(x+1) \sqrt {\sqrt {x+1}+1}}{x \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 2003 |
| \(\displaystyle -2 \int \frac {x+1}{\left (1-\sqrt {x+1}\right ) \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 4 \int -\frac {x^2}{(1-x) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {\sqrt {x+1}+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 \int \frac {x^2}{(1-x) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {\sqrt {x+1}+1}\) |
| \(\Big \downarrow \) 2003 |
| \(\displaystyle -4 \int \frac {\left (1-\sqrt {\sqrt {x+1}+1}\right )^2 \left (\sqrt {\sqrt {x+1}+1}+1\right )^{3/2}}{1-x}d\sqrt {\sqrt {x+1}+1}\) |
| \(\Big \downarrow \) 656 |
| \(\displaystyle -8 \int \left (-(x+1)^2+2 (x+1)+\frac {1}{-(x+1)^2+2 (x+1)+1}-1\right )d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -8 \left (\frac {\arctan \left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {2}-1}}\right )}{2 \sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\text {arctanh}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {1+\sqrt {2}}}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {1}{5} (x+1)^{5/2}+\frac {2}{3} (x+1)^{3/2}-\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )\) |
-8*((2*(1 + x)^(3/2))/3 - (1 + x)^(5/2)/5 - Sqrt[1 + Sqrt[1 + Sqrt[1 + x]] ] + ArcTan[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[-1 + Sqrt[2]]]/(2*Sqrt[2*( -1 + Sqrt[2])]) + ArcTanh[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[1 + Sqrt[2] ]]/(2*Sqrt[2*(1 + Sqrt[2])]))
3.24.36.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_ )^2), x_Symbol] :> With[{q = Denominator[m]}, Simp[q/e Subst[Int[ExpandIn tegrand[x^(q*(m + 1) - 1)*(((e*f - d*g)/e + g*(x^q/e))^n/((c*d^2 + a*e^2)/e ^2 - 2*c*d*(x^q/e^2) + c*(x^(2*q)/e^2))), x], x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && IntegerQ[n] && FractionQ[m]
Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : > Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} , x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
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 = 0.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.62
| method | result | size |
| derivativedivides | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}+8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {\sqrt {2}-1}}\right )}{\sqrt {\sqrt {2}-1}}-\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\) | \(115\) |
| default | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}+8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {\sqrt {2}-1}}\right )}{\sqrt {\sqrt {2}-1}}-\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\) | \(115\) |
int((1+x)^(1/2)*(1+(1+x)^(1/2))^(1/2)/x/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, method=_RETURNVERBOSE)
8/5*(1+(1+(1+x)^(1/2))^(1/2))^(5/2)-16/3*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)+8 *(1+(1+(1+x)^(1/2))^(1/2))^(1/2)-2*2^(1/2)/(2^(1/2)-1)^(1/2)*arctan((1+(1+ (1+x)^(1/2))^(1/2))^(1/2)/(2^(1/2)-1)^(1/2))-2*2^(1/2)/(1+2^(1/2))^(1/2)*a rctanh((1+(1+(1+x)^(1/2))^(1/2))^(1/2)/(1+2^(1/2))^(1/2))
Time = 0.26 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=-\sqrt {2} \sqrt {\sqrt {2} - 1} \log \left (2 \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \sqrt {\sqrt {2} - 1} + 4 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) + \sqrt {2} \sqrt {\sqrt {2} - 1} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \sqrt {\sqrt {2} - 1} + 4 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) + \frac {8}{15} \, {\left (3 \, \sqrt {x + 1} - 4 \, \sqrt {\sqrt {x + 1} + 1} + 11\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} + \frac {1}{2} \, \sqrt {-8 \, \sqrt {2} - 8} \log \left ({\left (\sqrt {2} - 2\right )} \sqrt {-8 \, \sqrt {2} - 8} + 4 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) - \frac {1}{2} \, \sqrt {-8 \, \sqrt {2} - 8} \log \left (-{\left (\sqrt {2} - 2\right )} \sqrt {-8 \, \sqrt {2} - 8} + 4 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) \]
integrate((1+x)^(1/2)*(1+(1+x)^(1/2))^(1/2)/x/(1+(1+(1+x)^(1/2))^(1/2))^(1 /2),x, algorithm="fricas")
-sqrt(2)*sqrt(sqrt(2) - 1)*log(2*sqrt(2)*(sqrt(2) + 2)*sqrt(sqrt(2) - 1) + 4*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) + sqrt(2)*sqrt(sqrt(2) - 1)*log(-2*sqr t(2)*(sqrt(2) + 2)*sqrt(sqrt(2) - 1) + 4*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) + 8/15*(3*sqrt(x + 1) - 4*sqrt(sqrt(x + 1) + 1) + 11)*sqrt(sqrt(sqrt(x + 1 ) + 1) + 1) + 1/2*sqrt(-8*sqrt(2) - 8)*log((sqrt(2) - 2)*sqrt(-8*sqrt(2) - 8) + 4*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) - 1/2*sqrt(-8*sqrt(2) - 8)*log(-( sqrt(2) - 2)*sqrt(-8*sqrt(2) - 8) + 4*sqrt(sqrt(sqrt(x + 1) + 1) + 1))
Timed out. \[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {\sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}}{x \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
integrate((1+x)^(1/2)*(1+(1+x)^(1/2))^(1/2)/x/(1+(1+(1+x)^(1/2))^(1/2))^(1 /2),x, algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (126) = 252\).
Time = 8.39 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=-\frac {\frac {30 \, \sqrt {2 \, \sqrt {2} + 2} \arctan \left (\frac {\sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}}{\sqrt {\sqrt {2} - 1}}\right )}{\mathrm {sgn}\left (4 \, {\left (\sqrt {x + 1} + 1\right )}^{2} - 8 \, \sqrt {x + 1} - 7\right )} + \frac {15 \, \sqrt {2 \, \sqrt {2} - 2} \log \left (\sqrt {\sqrt {2} + 1} + \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right )}{\mathrm {sgn}\left (4 \, {\left (\sqrt {x + 1} + 1\right )}^{2} - 8 \, \sqrt {x + 1} - 7\right )} - \frac {15 \, \sqrt {2 \, \sqrt {2} - 2} \log \left ({\left | -\sqrt {\sqrt {2} + 1} + \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} \right |}\right )}{\mathrm {sgn}\left (4 \, {\left (\sqrt {x + 1} + 1\right )}^{2} - 8 \, \sqrt {x + 1} - 7\right )} - \frac {8 \, {\left (3 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {5}{2}} - 10 \, {\left (\sqrt {\sqrt {x + 1} + 1} + 1\right )}^{\frac {3}{2}} + 15 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right )}}{\mathrm {sgn}\left (4 \, {\left (\sqrt {x + 1} + 1\right )}^{2} - 8 \, \sqrt {x + 1} - 7\right )}}{15 \, \mathrm {sgn}\left (4 \, x + 1\right )} \]
integrate((1+x)^(1/2)*(1+(1+x)^(1/2))^(1/2)/x/(1+(1+(1+x)^(1/2))^(1/2))^(1 /2),x, algorithm="giac")
-1/15*(30*sqrt(2*sqrt(2) + 2)*arctan(sqrt(sqrt(sqrt(x + 1) + 1) + 1)/sqrt( sqrt(2) - 1))/sgn(4*(sqrt(x + 1) + 1)^2 - 8*sqrt(x + 1) - 7) + 15*sqrt(2*s qrt(2) - 2)*log(sqrt(sqrt(2) + 1) + sqrt(sqrt(sqrt(x + 1) + 1) + 1))/sgn(4 *(sqrt(x + 1) + 1)^2 - 8*sqrt(x + 1) - 7) - 15*sqrt(2*sqrt(2) - 2)*log(abs (-sqrt(sqrt(2) + 1) + sqrt(sqrt(sqrt(x + 1) + 1) + 1)))/sgn(4*(sqrt(x + 1) + 1)^2 - 8*sqrt(x + 1) - 7) - 8*(3*(sqrt(sqrt(x + 1) + 1) + 1)^(5/2) - 10 *(sqrt(sqrt(x + 1) + 1) + 1)^(3/2) + 15*sqrt(sqrt(sqrt(x + 1) + 1) + 1))/s gn(4*(sqrt(x + 1) + 1)^2 - 8*sqrt(x + 1) - 7))/sgn(4*x + 1)
Timed out. \[ \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}}{x\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}} \,d x \]