Integrand size = 19, antiderivative size = 25 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \]
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Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {385, 209} \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \]
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Rule 209
Rule 385
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1-x^2}}\right ) \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \]
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Time = 2.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}}{2 x}\right )}{2}\) | \(24\) |
default | \(-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}\, x}{x^{2}-1}\right )}{2}\) | \(28\) |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+4 x \sqrt {-x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{x^{2}+1}\right )}{4}\) | \(50\) |
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none
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-x^{2} + 1}}{2 \, x}\right ) \]
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\[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\int \frac {1}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 1\right )}\, dx \]
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\[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )} \sqrt {-x^{2} + 1}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {1}{4} \, \sqrt {2} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (-\frac {\sqrt {2} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{4 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} \]
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Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.16 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (-1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\mathrm {i}}\right )\,1{}\mathrm {i}}{4}-\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+1{}\mathrm {i}}\right )\,1{}\mathrm {i}}{4} \]
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