Integrand size = 19, antiderivative size = 73 \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (1-\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )+\frac {2}{5} \left (5-\sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right ) \]
2/5*ln(1+5^(1/2)-2*(1+(1+x)^(1/2))^(1/2))*(5-5^(1/2))+2/5*ln(1-5^(1/2)-2*( 1+(1+x)^(1/2))^(1/2))*(5+5^(1/2))
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=-\frac {2}{5} \left (-5+\sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )+\frac {2}{5} \left (5+\sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right ) \]
(-2*(-5 + Sqrt[5])*Log[1 + Sqrt[5] - 2*Sqrt[1 + Sqrt[1 + x]]])/5 + (2*(5 + Sqrt[5])*Log[-1 + Sqrt[5] + 2*Sqrt[1 + Sqrt[1 + x]]])/5
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {7267, 25, 7267, 25, 1141, 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.
\(\displaystyle \int \frac {1}{x-\sqrt {\sqrt {x+1}+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {\sqrt {x+1}}{\sqrt {\sqrt {x+1}+1}-x}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\sqrt {x+1}}{\sqrt {\sqrt {x+1}+1}-x}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle -4 \int -\frac {1-\sqrt {\sqrt {x+1}+1}}{\sqrt {\sqrt {x+1}+1}-x}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 4 \int \frac {1-\sqrt {\sqrt {x+1}+1}}{\sqrt {\sqrt {x+1}+1}-x}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle -4 \int \left (\frac {5+\sqrt {5}}{5 \left (-2 \sqrt {\sqrt {x+1}+1}-\sqrt {5}+1\right )}-\frac {1-\sqrt {5}}{-2 \sqrt {5} \sqrt {\sqrt {x+1}+1}+\sqrt {5}+5}\right )d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (-\frac {1}{10} \left (5+\sqrt {5}\right ) \log \left (-2 \sqrt {\sqrt {x+1}+1}-\sqrt {5}+1\right )-\frac {1}{10} \left (5-\sqrt {5}\right ) \log \left (-2 \sqrt {\sqrt {x+1}+1}+\sqrt {5}+1\right )\right )\) |
-4*(-1/10*((5 + Sqrt[5])*Log[1 - Sqrt[5] - 2*Sqrt[1 + Sqrt[1 + x]]]) - ((5 - Sqrt[5])*Log[1 + Sqrt[5] - 2*Sqrt[1 + Sqrt[1 + x]]])/10)
3.1.11.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
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 = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(2 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )+\frac {4 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{5}\) | \(46\) |
default | \(\frac {\ln \left (x^{2}-x -1\right )}{2}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x -1\right ) \sqrt {5}}{5}\right )}{5}-\ln \left (\sqrt {1+x}+\sqrt {1+\sqrt {1+x}}\right )+\frac {2 \,\operatorname {arctanh}\left (\frac {\left (1+2 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{5}+\ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )+\frac {2 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{5}+\frac {2 \left (7+3 \sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \sqrt {1+\sqrt {1+x}}}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}-\frac {2 \sqrt {5}\, \left (-7+3 \sqrt {5}\right ) \arctan \left (\frac {2 \sqrt {1+\sqrt {1+x}}}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}-\frac {\ln \left (x +\sqrt {1+x}\right )}{2}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {5}}{5}\right )}{5}+\frac {\ln \left (x -\sqrt {1+x}\right )}{2}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+x}-1\right ) \sqrt {5}}{5}\right )}{5}-\frac {2 \left (3+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \sqrt {1+\sqrt {1+x}}}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}+\frac {2 \left (\sqrt {5}-3\right ) \sqrt {5}\, \arctan \left (\frac {2 \sqrt {1+\sqrt {1+x}}}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}-\frac {4 \left (2+\sqrt {5}\right ) \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \sqrt {1+\sqrt {1+x}}}{\sqrt {2+2 \sqrt {5}}}\right )}{5 \sqrt {2+2 \sqrt {5}}}+\frac {4 \left (-2+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {2 \sqrt {1+\sqrt {1+x}}}{\sqrt {-2+2 \sqrt {5}}}\right )}{5 \sqrt {-2+2 \sqrt {5}}}\) | \(419\) |
2*ln((1+x)^(1/2)-(1+(1+x)^(1/2))^(1/2))+4/5*5^(1/2)*arctanh(1/5*(2*(1+(1+x )^(1/2))^(1/2)-1)*5^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (51) = 102\).
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.53 \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=\frac {2}{5} \, \sqrt {5} \log \left (\frac {2 \, x^{2} + \sqrt {5} {\left (3 \, x + 1\right )} + {\left (\sqrt {5} {\left (x + 2\right )} + 5 \, x\right )} \sqrt {x + 1} + {\left (\sqrt {5} {\left (x + 2\right )} + {\left (\sqrt {5} {\left (2 \, x - 1\right )} + 5\right )} \sqrt {x + 1} + 5 \, x\right )} \sqrt {\sqrt {x + 1} + 1} + 3 \, x + 3}{x^{2} - x - 1}\right ) + 2 \, \log \left (\sqrt {x + 1} - \sqrt {\sqrt {x + 1} + 1}\right ) \]
2/5*sqrt(5)*log((2*x^2 + sqrt(5)*(3*x + 1) + (sqrt(5)*(x + 2) + 5*x)*sqrt( x + 1) + (sqrt(5)*(x + 2) + (sqrt(5)*(2*x - 1) + 5)*sqrt(x + 1) + 5*x)*sqr t(sqrt(x + 1) + 1) + 3*x + 3)/(x^2 - x - 1)) + 2*log(sqrt(x + 1) - sqrt(sq rt(x + 1) + 1))
Time = 2.49 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=- \frac {2 \sqrt {5} \left (- \log {\left (\sqrt {\sqrt {x + 1} + 1} - \frac {1}{2} + \frac {\sqrt {5}}{2} \right )} + \log {\left (\sqrt {\sqrt {x + 1} + 1} - \frac {\sqrt {5}}{2} - \frac {1}{2} \right )}\right )}{5} + 2 \log {\left (\sqrt {x + 1} - \sqrt {\sqrt {x + 1} + 1} \right )} \]
-2*sqrt(5)*(-log(sqrt(sqrt(x + 1) + 1) - 1/2 + sqrt(5)/2) + log(sqrt(sqrt( x + 1) + 1) - sqrt(5)/2 - 1/2))/5 + 2*log(sqrt(x + 1) - sqrt(sqrt(x + 1) + 1))
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=-\frac {2}{5} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - 2 \, \sqrt {\sqrt {x + 1} + 1} + 1}{\sqrt {5} + 2 \, \sqrt {\sqrt {x + 1} + 1} - 1}\right ) + 2 \, \log \left (\sqrt {x + 1} - \sqrt {\sqrt {x + 1} + 1}\right ) \]
-2/5*sqrt(5)*log(-(sqrt(5) - 2*sqrt(sqrt(x + 1) + 1) + 1)/(sqrt(5) + 2*sqr t(sqrt(x + 1) + 1) - 1)) + 2*log(sqrt(x + 1) - sqrt(sqrt(x + 1) + 1))
Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=-\frac {2}{5} \, \sqrt {5} \log \left (\frac {{\left | -\sqrt {5} + 2 \, \sqrt {\sqrt {x + 1} + 1} - 1 \right |}}{{\left | \sqrt {5} + 2 \, \sqrt {\sqrt {x + 1} + 1} - 1 \right |}}\right ) + 2 \, \log \left ({\left | \sqrt {x + 1} - \sqrt {\sqrt {x + 1} + 1} \right |}\right ) \]
-2/5*sqrt(5)*log(abs(-sqrt(5) + 2*sqrt(sqrt(x + 1) + 1) - 1)/abs(sqrt(5) + 2*sqrt(sqrt(x + 1) + 1) - 1)) + 2*log(abs(sqrt(x + 1) - sqrt(sqrt(x + 1) + 1)))
Timed out. \[ \int \frac {1}{x-\sqrt {1+\sqrt {1+x}}} \, dx=\int \frac {1}{x-\sqrt {\sqrt {x+1}+1}} \,d x \]