Integrand size = 23, antiderivative size = 20 \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=2 \arctan \left (\frac {x}{\sqrt {1+\sqrt {1+x^2}}}\right ) \]
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\[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=\int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {i \sqrt {1+\sqrt {1+x^2}}}{2 (i-x)}+\frac {i \sqrt {1+\sqrt {1+x^2}}}{2 (i+x)}\right ) \, dx \\ & = \frac {1}{2} i \int \frac {\sqrt {1+\sqrt {1+x^2}}}{i-x} \, dx+\frac {1}{2} i \int \frac {\sqrt {1+\sqrt {1+x^2}}}{i+x} \, dx \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=2 \arctan \left (\frac {x}{\sqrt {1+\sqrt {1+x^2}}}\right ) \]
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\[\int \frac {\sqrt {1+\sqrt {x^{2}+1}}}{x^{2}+1}d x\]
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Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (16) = 32\).
Time = 0.96 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=-\frac {1}{2} \, \arctan \left (\frac {4 \, {\left (x^{4} - 12 \, x^{2} + {\left (5 \, x^{2} - 3\right )} \sqrt {x^{2} + 1} + 3\right )} \sqrt {\sqrt {x^{2} + 1} + 1}}{x^{5} - 46 \, x^{3} + 17 \, x}\right ) \]
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\[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=\int \frac {\sqrt {\sqrt {x^{2} + 1} + 1}}{x^{2} + 1}\, dx \]
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\[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=\int { \frac {\sqrt {\sqrt {x^{2} + 1} + 1}}{x^{2} + 1} \,d x } \]
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\[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=\int { \frac {\sqrt {\sqrt {x^{2} + 1} + 1}}{x^{2} + 1} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{1+x^2} \, dx=\int \frac {\sqrt {\sqrt {x^2+1}+1}}{x^2+1} \,d x \]
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