Integrand size = 39, antiderivative size = 275 \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\frac {2}{3} (1+x)^{3/2}+4 \sqrt {1+\sqrt {1+x}}+\frac {2}{55} \left (25+9 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )-\frac {2}{55} \left (-25+9 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right )-\frac {4}{11} \text {RootSum}\left [-1+\text {$\#$1}-\text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4+\text {$\#$1}^5\&,\frac {-9 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )+\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-2 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2+\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3+5 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4}{1-2 \text {$\#$1}-6 \text {$\#$1}^2+4 \text {$\#$1}^3+5 \text {$\#$1}^4}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\frac {2}{165} \left (330 \sqrt {1+\sqrt {1+x}}+55 \left (-2+(1+x)^{3/2}\right )+3 \left (25+9 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 \sqrt {1+\sqrt {1+x}}\right )-3 \left (-25+9 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 \sqrt {1+\sqrt {1+x}}\right )-30 \text {RootSum}\left [-1+\text {$\#$1}-\text {$\#$1}^2-2 \text {$\#$1}^3+\text {$\#$1}^4+\text {$\#$1}^5\&,\frac {-9 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )+\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-2 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2+\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3+5 \log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4}{1-2 \text {$\#$1}-6 \text {$\#$1}^2+4 \text {$\#$1}^3+5 \text {$\#$1}^4}\&\right ]\right ) \]
(2*(330*Sqrt[1 + Sqrt[1 + x]] + 55*(-2 + (1 + x)^(3/2)) + 3*(25 + 9*Sqrt[5 ])*Log[1 + Sqrt[5] - 2*Sqrt[1 + Sqrt[1 + x]]] - 3*(-25 + 9*Sqrt[5])*Log[-1 + Sqrt[5] + 2*Sqrt[1 + Sqrt[1 + x]]] - 30*RootSum[-1 + #1 - #1^2 - 2*#1^3 + #1^4 + #1^5 & , (-9*Log[Sqrt[1 + Sqrt[1 + x]] - #1] + Log[Sqrt[1 + Sqrt [1 + x]] - #1]*#1 - 2*Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^2 + Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^3 + 5*Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^4)/(1 - 2*# 1 - 6*#1^2 + 4*#1^3 + 5*#1^4) & ]))/165
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.
| \(\displaystyle \int \frac {x^2 \sqrt {x+1}}{x^2-\sqrt {x+1} \sqrt {\sqrt {x+1}+1}} \, dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 2 \int -\frac {x^2 (x+1)}{\sqrt {x+1} \sqrt {\sqrt {x+1}+1}-x^2}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {x^2 (x+1)}{\sqrt {x+1} \sqrt {\sqrt {x+1}+1}-x^2}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 4 \int \frac {(x+1)^2 \left ((x+1)^2-3 (x+1)+2\right )^2}{(x+1)^{7/2}-4 (x+1)^{5/2}+4 (x+1)^{3/2}-x}d\sqrt {\sqrt {x+1}+1}\) |
| \(\Big \downarrow \) 2462 |
| \(\displaystyle 4 \int \left ((x+1)^{5/2}-2 (x+1)^{3/2}+\frac {5 \sqrt {\sqrt {x+1}+1}+2}{11 \left (x-\sqrt {\sqrt {x+1}+1}\right )}+\sqrt {\sqrt {x+1}+1}+\frac {-5 (x+1)^2-(x+1)^{3/2}+2 (x+1)-\sqrt {\sqrt {x+1}+1}+9}{11 \left ((x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2\right )}+1\right )d\sqrt {\sqrt {x+1}+1}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \left (\frac {10}{11} \int \frac {1}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}-\frac {4}{11} \int \frac {x+1}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}+\frac {3}{11} \int \frac {(x+1)^{3/2}}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}-\frac {3}{11} \int \frac {\sqrt {\sqrt {x+1}+1}}{(x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2}d\sqrt {\sqrt {x+1}+1}+\frac {1}{6} (x+1)^3-\frac {1}{2} (x+1)^2+\frac {x+1}{2}+\sqrt {\sqrt {x+1}+1}+\frac {1}{110} \left (25-9 \sqrt {5}\right ) \log \left (-2 \sqrt {\sqrt {x+1}+1}-\sqrt {5}+1\right )-\frac {1}{11} \log \left ((x+1)^{5/2}+(x+1)^2-2 (x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-2\right )+\frac {1}{110} \left (25+9 \sqrt {5}\right ) \log \left (-2 \sqrt {5} \sqrt {\sqrt {x+1}+1}+\sqrt {5}+5\right )\right )\) |
3.29.10.3.1 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 = 0.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.60
| method | result | size |
| derivativedivides | \(\frac {2 \left (1+\sqrt {1+x}\right )^{3}}{3}-2 \left (1+\sqrt {1+x}\right )^{2}+2 \sqrt {1+x}+2+4 \sqrt {1+\sqrt {1+x}}-\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\frac {\left (5 \textit {\_R}^{4}+\textit {\_R}^{3}-2 \textit {\_R}^{2}+\textit {\_R} -9\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{5 \textit {\_R}^{4}+4 \textit {\_R}^{3}-6 \textit {\_R}^{2}-2 \textit {\_R} +1}\right )}{11}+\frac {10 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )}{11}-\frac {36 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{55}\) | \(165\) |
| default | \(\frac {2 \left (1+\sqrt {1+x}\right )^{3}}{3}-2 \left (1+\sqrt {1+x}\right )^{2}+2 \sqrt {1+x}+2+4 \sqrt {1+\sqrt {1+x}}-\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{4}-2 \textit {\_Z}^{3}-\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\frac {\left (5 \textit {\_R}^{4}+\textit {\_R}^{3}-2 \textit {\_R}^{2}+\textit {\_R} -9\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{5 \textit {\_R}^{4}+4 \textit {\_R}^{3}-6 \textit {\_R}^{2}-2 \textit {\_R} +1}\right )}{11}+\frac {10 \ln \left (\sqrt {1+x}-\sqrt {1+\sqrt {1+x}}\right )}{11}-\frac {36 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 \sqrt {1+\sqrt {1+x}}-1\right ) \sqrt {5}}{5}\right )}{55}\) | \(165\) |
2/3*(1+(1+x)^(1/2))^3-2*(1+(1+x)^(1/2))^2+2*(1+x)^(1/2)+2+4*(1+(1+x)^(1/2) )^(1/2)-4/11*sum((5*_R^4+_R^3-2*_R^2+_R-9)/(5*_R^4+4*_R^3-6*_R^2-2*_R+1)*l n((1+(1+x)^(1/2))^(1/2)-_R),_R=RootOf(_Z^5+_Z^4-2*_Z^3-_Z^2+_Z-1))+10/11*l n((1+x)^(1/2)-(1+(1+x)^(1/2))^(1/2))-36/55*5^(1/2)*arctanh(1/5*(2*(1+(1+x) ^(1/2))^(1/2)-1)*5^(1/2))
Timed out. \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\text {Timed out} \]
Not integrable
Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.12 \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {\sqrt {x + 1} x^{2}}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}} \,d x } \]
Not integrable
Time = 0.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.12 \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=\int { \frac {\sqrt {x + 1} x^{2}}{x^{2} - \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}} \,d x } \]
Not integrable
Time = 0.00 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.13 \[ \int \frac {x^2 \sqrt {1+x}}{x^2-\sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx=-\int \frac {x^2\,\sqrt {x+1}}{\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}-x^2} \,d x \]