Integrand size = 32, antiderivative size = 252 \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=x+\frac {4}{55} \left (5+4 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )-\frac {4}{55} \left (-5+4 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right )-\frac {4}{11} \text {RootSum}\left [-1+\text {$\#$1}-\text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4+\text {$\#$1}^5\&,\frac {3 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )+7 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-3 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2-4 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3+2 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4}{1-2 \text {$\#$1}-6 \text {$\#$1}^2+4 \text {$\#$1}^3+5 \text {$\#$1}^4}\&\right ] \]
Time = 0.15 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=x+\frac {4}{55} \left (5+4 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )-\frac {4}{55} \left (-5+4 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right )-\frac {4}{11} \text {RootSum}\left [-1+\text {$\#$1}-\text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4+\text {$\#$1}^5\&,\frac {3 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )+7 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-3 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2-4 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3+2 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4}{1-2 \text {$\#$1}-6 \text {$\#$1}^2+4 \text {$\#$1}^3+5 \text {$\#$1}^4}\&\right ] \]
x + (4*(5 + 4*Sqrt[5])*Log[1 + Sqrt[5] - 2*Sqrt[1 + Sqrt[1 + x]]])/55 - (4 *(-5 + 4*Sqrt[5])*Log[-1 + Sqrt[5] + 2*Sqrt[1 + Sqrt[1 + x]]])/55 - (4*Roo tSum[-1 + #1 - #1^2 - 2*#1^3 + #1^4 + #1^5 & , (3*Log[Sqrt[1 + Sqrt[1 + x] ] - #1] + 7*Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1 - 3*Log[Sqrt[1 + Sqrt[1 + x ]] - #1]*#1^2 - 4*Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^3 + 2*Log[Sqrt[1 + Sq rt[1 + x]] - #1]*#1^4)/(1 - 2*#1 - 6*#1^2 + 4*#1^3 + 5*#1^4) & ])/11
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.
\(\displaystyle \int \frac {x^2}{x^2-\sqrt {x+1} \sqrt {\sqrt {x+1}+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {x^2 \sqrt {x+1}}{\sqrt {x+1} \sqrt {\sqrt {x+1}+1}-x^2}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {x^2 \sqrt {x+1}}{\sqrt {x+1} \sqrt {\sqrt {x+1}+1}-x^2}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 4 \int \frac {(1-x)^2 x (x+1)^2}{(x+1)^{7/2}-4 (x+1)^{5/2}+4 (x+1)^{3/2}-x}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int -\frac {(1-x)^2 x (x+1)^2}{(x+1)^{7/2}-4 (x+1)^{5/2}+4 (x+1)^{3/2}-x}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle -4 \int \left (-(x+1)^{3/2}+\frac {2 (x+1)^2-4 (x+1)^{3/2}-3 (x+1)+7 \sqrt {\sqrt {x+1}+1}+3}{11 \left ((x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2\right )}+\sqrt {\sqrt {x+1}+1}+\frac {-2 \sqrt {\sqrt {x+1}+1}-3}{11 \left (x-\sqrt {\sqrt {x+1}+1}\right )}\right )d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (-\frac {13}{55} \int \frac {1}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}+\frac {3}{55} \int \frac {x+1}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}+\frac {28}{55} \int \frac {(x+1)^{3/2}}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}-\frac {39}{55} \int \frac {\sqrt {\sqrt {x+1}+1}}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}+\frac {1}{4} (x+1)^2+\frac {1}{2} (-x-1)+\frac {1}{55} \left (5-4 \sqrt {5}\right ) \log \left (-2 \sqrt {\sqrt {x+1}+1}-\sqrt {5}+1\right )-\frac {2}{55} \log \left ((x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2\right )+\frac {1}{55} \left (5+4 \sqrt {5}\right ) \log \left (-2 \sqrt {5} \sqrt {\sqrt {x+1}+1}+\sqrt {5}+5\right )\right )\) |
3.28.34.3.1 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 = 0.05 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.58
method | result | size |
derivativedivides | \(\left (1+\sqrt {1+x}\right )^{2}-2-2 \sqrt {1+x}+\frac {4 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )}{11}-\frac {32 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{55}-\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\frac {\left (2 \textit {\_R}^{4}-4 \textit {\_R}^{3}-3 \textit {\_R}^{2}+7 \textit {\_R} +3\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{5 \textit {\_R}^{4}+4 \textit {\_R}^{3}-6 \textit {\_R}^{2}-2 \textit {\_R} +1}\right )}{11}\) | \(145\) |
default | \(\left (1+\sqrt {1+x}\right )^{2}-2-2 \sqrt {1+x}+\frac {4 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )}{11}-\frac {32 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{55}-\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\frac {\left (2 \textit {\_R}^{4}-4 \textit {\_R}^{3}-3 \textit {\_R}^{2}+7 \textit {\_R} +3\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{5 \textit {\_R}^{4}+4 \textit {\_R}^{3}-6 \textit {\_R}^{2}-2 \textit {\_R} +1}\right )}{11}\) | \(145\) |
(1+(1+x)^(1/2))^2-2-2*(1+x)^(1/2)+4/11*ln((1+x)^(1/2)-(1+(1+x)^(1/2))^(1/2 ))-32/55*5^(1/2)*arctanh(1/5*(2*(1+(1+x)^(1/2))^(1/2)-1)*5^(1/2))-4/11*sum ((2*_R^4-4*_R^3-3*_R^2+7*_R+3)/(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. \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\text {Timed out} \]
Not integrable
Time = 162.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.10 \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int \frac {x^{2}}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}}\, dx \]
Not integrable
Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.11 \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {x^{2}}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}} \,d x } \]
Not integrable
Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.11 \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {x^{2}}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}} \,d x } \]
Not integrable
Time = 7.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.12 \[ \int \frac {x^2}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=-\int \frac {x^2}{\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}-x^2} \,d x \]