Integrand size = 21, antiderivative size = 8 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\frac {\arctan (x)}{\sqrt {2}} \]
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Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {22, 209} \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\frac {\arctan (x)}{\sqrt {2}} \]
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Rule 22
Rule 209
Rubi steps \begin{align*} \text {integral}& = \sqrt {2} \int \frac {1}{2+2 x^2} \, dx \\ & = \frac {\tan ^{-1}(x)}{\sqrt {2}} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\frac {\arctan (x)}{\sqrt {2}} \]
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Time = 2.48 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (7) = 14\).
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 4.25 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {2 \, x^{2} + 2} \sqrt {x^{2} + 1} x}{x^{4} - 1}\right ) \]
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Time = 0.97 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\frac {\sqrt {2} \operatorname {atan}{\left (x \right )}}{2} \]
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\[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{2} + 2} \sqrt {x^{2} + 1}} \,d x } \]
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\[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{2} + 2} \sqrt {x^{2} + 1}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+2 x^2}} \, dx=\int \frac {1}{\sqrt {x^2+1}\,\sqrt {2\,x^2+2}} \,d x \]
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