Integrand size = 19, antiderivative size = 155 \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=x-2 \sqrt {x+\sqrt {1+x}}-\log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right )+4 \text {RootSum}\left [1+3 \text {$\#$1}-5 \text {$\#$1}^2+2 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {\log \left (\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}+\log \left (\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^3}{3-10 \text {$\#$1}+6 \text {$\#$1}^2+4 \text {$\#$1}^3}\&\right ] \]
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\[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x \left (-1+x^2\right )}{-1+x^2+\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = 2 \text {Subst}\left (\int \left (-\frac {x}{-1+x^2+\sqrt {-1+x+x^2}}+\frac {x^3}{-1+x^2+\sqrt {-1+x+x^2}}\right ) \, dx,x,\sqrt {1+x}\right ) \\ & = -\left (2 \text {Subst}\left (\int \frac {x}{-1+x^2+\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\right )+2 \text {Subst}\left (\int \frac {x^3}{-1+x^2+\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = -\left (2 \text {Subst}\left (\int \left (\frac {x \left (-1+x^2\right )}{2-x-3 x^2+x^4}-\frac {x \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4}\right ) \, dx,x,\sqrt {1+x}\right )\right )+2 \text {Subst}\left (\int \left (x-\frac {x^3 \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4}+\frac {x \left (-2+x+2 x^2\right )}{2-x-3 x^2+x^4}\right ) \, dx,x,\sqrt {1+x}\right ) \\ & = x-2 \text {Subst}\left (\int \frac {x \left (-1+x^2\right )}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )-2 \text {Subst}\left (\int \frac {x^3 \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x \left (-2+x+2 x^2\right )}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right ) \\ & = x+\frac {1}{2} \log \left (2-\sqrt {1+x}-3 (1+x)+(1+x)^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1+2 x}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {2+4 x+4 x^2}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )-2 \text {Subst}\left (\int \frac {x^3 \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right ) \\ & = x+\frac {1}{2} \log \left (2-\sqrt {1+x}-3 (1+x)+(1+x)^2\right )-\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{2-x-3 x^2+x^4}+\frac {2 x}{2-x-3 x^2+x^4}\right ) \, dx,x,\sqrt {1+x}\right )+\frac {1}{2} \text {Subst}\left (\int \left (\frac {2}{2-x-3 x^2+x^4}+\frac {4 x}{2-x-3 x^2+x^4}+\frac {4 x^2}{2-x-3 x^2+x^4}\right ) \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )-2 \text {Subst}\left (\int \frac {x^3 \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right ) \\ & = x+\frac {1}{2} \log \left (2-\sqrt {1+x}-3 (1+x)+(1+x)^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x^2}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {x \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )-2 \text {Subst}\left (\int \frac {x^3 \sqrt {-1+x+x^2}}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )+\text {Subst}\left (\int \frac {1}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right )-\text {Subst}\left (\int \frac {x}{2-x-3 x^2+x^4} \, dx,x,\sqrt {1+x}\right ) \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03 \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=x-2 \sqrt {x+\sqrt {1+x}}-\log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right )+4 \text {RootSum}\left [1+3 \text {$\#$1}-5 \text {$\#$1}^2+2 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}+\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3}{3-10 \text {$\#$1}+6 \text {$\#$1}^2+4 \text {$\#$1}^3}\&\right ] \]
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Time = 0.18 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.24
method | result | size |
derivativedivides | \(-\sqrt {x +\sqrt {1+x}}+\sqrt {1+x}+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3}-5 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{\sum }\frac {\left (2 \textit {\_R}^{3}+\textit {\_R}^{2}-\textit {\_R} \right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}+6 \textit {\_R}^{2}-10 \textit {\_R} +3}\right )-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+\textit {\_Z}^{2}+5 \textit {\_Z} -1\right )}{\sum }\frac {\left (2 \textit {\_R}^{3}-3 \textit {\_R}^{2}-\textit {\_R} +6\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R}^{2}+2 \textit {\_R} +5}\right )+\frac {5}{2 \left (-1-2 \sqrt {1+x}+2 \sqrt {x +\sqrt {1+x}}\right )}+\ln \left (-1-2 \sqrt {1+x}+2 \sqrt {x +\sqrt {1+x}}\right )+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-3 \textit {\_Z}^{2}-\textit {\_Z} +2\right )}{\sum }\frac {\textit {\_R} \ln \left (\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R} -1}\right )-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-3 \textit {\_Z}^{2}-\textit {\_Z} +2\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R} -1}\right )+1+x +2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-3 \textit {\_Z}^{2}-\textit {\_Z} +2\right )}{\sum }\frac {\left (3 \textit {\_R}^{3}+\textit {\_R}^{2}-2 \textit {\_R} \right ) \ln \left (\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R} -1}\right )-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3}-5 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R}^{2}-2 \textit {\_R} \right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}+6 \textit {\_R}^{2}-10 \textit {\_R} +3}\right )+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+\textit {\_Z}^{2}+5 \textit {\_Z} -1\right )}{\sum }\frac {\left (\textit {\_R}^{3}-\textit {\_R}^{2}+2\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R}^{2}+2 \textit {\_R} +5}\right )\) | \(502\) |
default | \(-\sqrt {x +\sqrt {1+x}}+\sqrt {1+x}+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3}-5 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{\sum }\frac {\left (2 \textit {\_R}^{3}+\textit {\_R}^{2}-\textit {\_R} \right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}+6 \textit {\_R}^{2}-10 \textit {\_R} +3}\right )-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+\textit {\_Z}^{2}+5 \textit {\_Z} -1\right )}{\sum }\frac {\left (2 \textit {\_R}^{3}-3 \textit {\_R}^{2}-\textit {\_R} +6\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R}^{2}+2 \textit {\_R} +5}\right )+\frac {5}{2 \left (-1-2 \sqrt {1+x}+2 \sqrt {x +\sqrt {1+x}}\right )}+\ln \left (-1-2 \sqrt {1+x}+2 \sqrt {x +\sqrt {1+x}}\right )+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-3 \textit {\_Z}^{2}-\textit {\_Z} +2\right )}{\sum }\frac {\textit {\_R} \ln \left (\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R} -1}\right )-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-3 \textit {\_Z}^{2}-\textit {\_Z} +2\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R} -1}\right )+1+x +2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-3 \textit {\_Z}^{2}-\textit {\_Z} +2\right )}{\sum }\frac {\left (3 \textit {\_R}^{3}+\textit {\_R}^{2}-2 \textit {\_R} \right ) \ln \left (\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R} -1}\right )-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+2 \textit {\_Z}^{3}-5 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R}^{2}-2 \textit {\_R} \right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}+6 \textit {\_R}^{2}-10 \textit {\_R} +3}\right )+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+\textit {\_Z}^{2}+5 \textit {\_Z} -1\right )}{\sum }\frac {\left (\textit {\_R}^{3}-\textit {\_R}^{2}+2\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{3}-6 \textit {\_R}^{2}+2 \textit {\_R} +5}\right )\) | \(502\) |
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Timed out. \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=\text {Timed out} \]
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Not integrable
Time = 2.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.10 \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {x}{x + \sqrt {x + \sqrt {x + 1}}}\, dx \]
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Not integrable
Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11 \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {x}{x + \sqrt {x + \sqrt {x + 1}}} \,d x } \]
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Not integrable
Time = 1.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11 \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {x}{x + \sqrt {x + \sqrt {x + 1}}} \,d x } \]
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Not integrable
Time = 5.70 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11 \[ \int \frac {x}{x+\sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {x}{x+\sqrt {x+\sqrt {x+1}}} \,d x \]
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