Integrand size = 39, antiderivative size = 170 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=-\frac {16}{3} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{3} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{2 \text {$\#$1}^3-\text {$\#$1}^5-8 \text {$\#$1}^7+12 \text {$\#$1}^9-6 \text {$\#$1}^{11}+\text {$\#$1}^{13}}\&\right ] \]
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\[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {(-1+x) \sqrt {1+x}}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx \\ & = 2 \text {Subst}\left (\int \frac {x^2 \left (-2+x^2\right )}{\sqrt {1+\sqrt {1+x}} \left (1+\left (-1+x^2\right )^2\right )} \, dx,x,\sqrt {1+x}\right ) \\ & = 4 \text {Subst}\left (\int \frac {x \left (-1+x^2\right )^2 \left (-2+\left (-1+x^2\right )^2\right )}{\sqrt {1+x} \left (1+x^4 \left (-2+x^2\right )^2\right )} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = 4 \text {Subst}\left (\int \frac {(-1+x)^2 x (1+x)^{3/2} \left (-2+\left (-1+x^2\right )^2\right )}{1+x^4 \left (-2+x^2\right )^2} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = 8 \text {Subst}\left (\int \frac {x^4 \left (-2+x^2\right )^2 \left (-1+x^2\right ) \left (-2+x^4 \left (-2+x^2\right )^2\right )}{1+\left (-1+x^2\right )^4 \left (-2+\left (-1+x^2\right )^2\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = 8 \text {Subst}\left (\int \left (-1+x^2+\frac {2 \left (1-x^2\right )}{1+\left (-1+x^2\right )^4 \left (-2+\left (-1+x^2\right )^2\right )^2}\right ) \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = -8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+16 \text {Subst}\left (\int \frac {1-x^2}{1+\left (-1+x^2\right )^4 \left (-2+\left (-1+x^2\right )^2\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \\ & = -8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+16 \text {Subst}\left (\int \left (\frac {1}{2-8 x^4+8 x^6+14 x^8-32 x^{10}+24 x^{12}-8 x^{14}+x^{16}}-\frac {x^2}{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 ) \\ & = -8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+16 \text {Subst}\left (\int \frac {1}{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 )-16 \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 = 0.00 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{3} \left (-2+\sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{2 \text {$\#$1}^3-\text {$\#$1}^5-8 \text {$\#$1}^7+12 \text {$\#$1}^9-6 \text {$\#$1}^{11}+\text {$\#$1}^{13}}\&\right ] \]
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Time = 0.15 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(133\) |
default | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(133\) |
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Timed out. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
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Not integrable
Time = 0.60 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {x + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
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Not integrable
Time = 1.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {x + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
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Not integrable
Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}\,\sqrt {x+1}} \,d x \]
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