Integrand size = 45, antiderivative size = 196 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=-\frac {32}{15} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{5} \sqrt {1+x} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {32}{15} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{\text {$\#$1}-\text {$\#$1}^3-4 \text {$\#$1}^5+8 \text {$\#$1}^7-5 \text {$\#$1}^9+\text {$\#$1}^{11}}\&\right ] \]
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\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^3 \left (-2+x^2\right )}{\sqrt {1+x} \left (2-2 x^2+x^4\right ) \sqrt {1+\sqrt {1+x}}} \, dx,x,\sqrt {1+x}\right ) \\ & = 4 \text {Subst}\left (\int \frac {\left (-1+x^2\right )^3 \left (-1-2 x^2+x^4\right )}{\sqrt {1+x} \left (1+4 x^4-4 x^6+x^8\right )} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = 4 \text {Subst}\left (\int \frac {(-1+x)^3 (1+x)^{5/2} \left (-1-2 x^2+x^4\right )}{1+4 x^4-4 x^6+x^8} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = 8 \text {Subst}\left (\int \frac {x^6 \left (-2+x^2\right )^3 \left (-2+4 x^4-4 x^6+x^8\right )}{1+4 \left (-1+x^2\right )^4-4 \left (-1+x^2\right )^6+\left (-1+x^2\right )^8} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = 8 \text {Subst}\left (\int \left (-2 x^2+x^4-\frac {2 x^2 \left (-2+x^2\right )}{1+4 \left (-1+x^2\right )^4-4 \left (-1+x^2\right )^6+\left (-1+x^2\right )^8}\right ) \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = -\frac {16}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+\frac {8}{5} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{5/2}-16 \text {Subst}\left (\int \frac {x^2 \left (-2+x^2\right )}{1+4 \left (-1+x^2\right )^4-4 \left (-1+x^2\right )^6+\left (-1+x^2\right )^8} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = -\frac {16}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+\frac {8}{5} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{5/2}-16 \text {Subst}\left (\int \left (-\frac {2 x^2}{2-8 x^4+8 x^6+14 x^8-32 x^{10}+24 x^{12}-8 x^{14}+x^{16}}+\frac {x^4}{2-8 x^4+8 x^6+14 x^8-32 x^{10}+24 x^{12}-8 x^{14}+x^{16}}\right ) \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = -\frac {16}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+\frac {8}{5} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{5/2}-16 \text {Subst}\left (\int \frac {x^4}{2-8 x^4+8 x^6+14 x^8-32 x^{10}+24 x^{12}-8 x^{14}+x^{16}} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )+32 \text {Subst}\left (\int \frac {x^2}{2-8 x^4+8 x^6+14 x^8-32 x^{10}+24 x^{12}-8 x^{14}+x^{16}} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.80 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{15} \left (-4+3 \sqrt {1+x}-4 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{\text {$\#$1}-\text {$\#$1}^3-4 \text {$\#$1}^5+8 \text {$\#$1}^7-5 \text {$\#$1}^9+\text {$\#$1}^{11}}\&\right ] \]
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Time = 0.17 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(137\) |
default | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(137\) |
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Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
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Not integrable
Time = 0.61 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {\sqrt {x + 1} + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
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Not integrable
Time = 1.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {\sqrt {x + 1} + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
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Not integrable
Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}\,\sqrt {\sqrt {x+1}+1}} \,d x \]
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