Integrand size = 30, antiderivative size = 245 \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\frac {2}{55} \left (15+\sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )-\frac {2}{55} \left (-15+\sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right )+\frac {4}{11} \text {RootSum}\left [-1+\text {$\#$1}-\text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4+\text {$\#$1}^5\&,\frac {\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )-5 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-12 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2+6 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3+8 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4}{1-2 \text {$\#$1}-6 \text {$\#$1}^2+4 \text {$\#$1}^3+5 \text {$\#$1}^4}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.99 \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\frac {2}{55} \left (\left (15+\sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )-\left (-15+\sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right )+10 \text {RootSum}\left [-1+\text {$\#$1}-\text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4+\text {$\#$1}^5\&,\frac {\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )-5 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-12 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2+6 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3+8 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4}{1-2 \text {$\#$1}-6 \text {$\#$1}^2+4 \text {$\#$1}^3+5 \text {$\#$1}^4}\&\right ]\right ) \]
(2*((15 + Sqrt[5])*Log[1 + Sqrt[5] - 2*Sqrt[1 + Sqrt[1 + x]]] - (-15 + Sqr t[5])*Log[-1 + Sqrt[5] + 2*Sqrt[1 + Sqrt[1 + x]]] + 10*RootSum[-1 + #1 - # 1^2 - 2*#1^3 + #1^4 + #1^5 & , (Log[Sqrt[1 + Sqrt[1 + x]] - #1] - 5*Log[Sq rt[1 + Sqrt[1 + x]] - #1]*#1 - 12*Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^2 + 6 *Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^3 + 8*Log[Sqrt[1 + Sqrt[1 + x]] - #1]* #1^4)/(1 - 2*#1 - 6*#1^2 + 4*#1^3 + 5*#1^4) & ]))/55
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.
\(\displaystyle \int \frac {x}{x^2-\sqrt {x+1} \sqrt {\sqrt {x+1}+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {x \sqrt {x+1}}{\sqrt {x+1} \sqrt {\sqrt {x+1}+1}-x^2}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 4 \int \frac {(x+1) \left ((x+1)^2-3 (x+1)+2\right )}{(x+1)^{7/2}-4 (x+1)^{5/2}+4 (x+1)^{3/2}-x}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle 4 \int \left (\frac {8 (x+1)^2+6 (x+1)^{3/2}-12 (x+1)-5 \sqrt {\sqrt {x+1}+1}+1}{11 \left ((x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2\right )}+\frac {3 \sqrt {\sqrt {x+1}+1}-1}{11 \left (x-\sqrt {\sqrt {x+1}+1}\right )}\right )d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (-\frac {3}{55} \int \frac {1}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}-\frac {12}{55} \int \frac {x+1}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}-\frac {2}{55} \int \frac {(x+1)^{3/2}}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}-\frac {9}{55} \int \frac {\sqrt {\sqrt {x+1}+1}}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}+\frac {1}{110} \left (15-\sqrt {5}\right ) \log \left (-2 \sqrt {\sqrt {x+1}+1}-\sqrt {5}+1\right )+\frac {8}{55} \log \left ((x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2\right )+\frac {1}{110} \left (15+\sqrt {5}\right ) \log \left (-2 \sqrt {5} \sqrt {\sqrt {x+1}+1}+\sqrt {5}+5\right )\right )\) |
3.28.3.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
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 = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {6 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )}{11}-\frac {4 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{55}-\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\frac {\left (-8 \textit {\_R}^{4}-6 \textit {\_R}^{3}+12 \textit {\_R}^{2}+5 \textit {\_R} -1\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{5 \textit {\_R}^{4}+4 \textit {\_R}^{3}-6 \textit {\_R}^{2}-2 \textit {\_R} +1}\right )}{11}\) | \(128\) |
default | \(\frac {6 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )}{11}-\frac {4 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{55}-\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\frac {\left (-8 \textit {\_R}^{4}-6 \textit {\_R}^{3}+12 \textit {\_R}^{2}+5 \textit {\_R} -1\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{5 \textit {\_R}^{4}+4 \textit {\_R}^{3}-6 \textit {\_R}^{2}-2 \textit {\_R} +1}\right )}{11}\) | \(128\) |
6/11*ln((1+x)^(1/2)-(1+(1+x)^(1/2))^(1/2))-4/55*5^(1/2)*arctanh(1/5*(2*(1+ (1+x)^(1/2))^(1/2)-1)*5^(1/2))-4/11*sum((-8*_R^4-6*_R^3+12*_R^2+5*_R-1)/(5 *_R^4+4*_R^3-6*_R^2-2*_R+1)*ln((1+(1+x)^(1/2))^(1/2)-_R),_R=RootOf(_Z^5+_Z ^4-2*_Z^3-_Z^2+_Z-1))
Timed out. \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\text {Timed out} \]
Not integrable
Time = 147.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.10 \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int \frac {x}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}}\, dx \]
Not integrable
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.11 \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {x}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}} \,d x } \]
Not integrable
Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.11 \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {x}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}} \,d x } \]
Not integrable
Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.12 \[ \int \frac {x}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=-\int \frac {x}{\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}-x^2} \,d x \]