Integrand size = 17, antiderivative size = 37 \[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+a x^n}}\right )}{\sqrt {b} (2-n)} \]
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Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2004, 2033, 212} \[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x^n+b x^2}}\right )}{\sqrt {b} (2-n)} \]
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Rule 212
Rule 2004
Rule 2033
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {b x^2+a x^n}} \, dx \\ & = \frac {2 \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+a x^n}}\right )}{2-n} \\ & = \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+a x^n}}\right )}{\sqrt {b} (2-n)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(78\) vs. \(2(37)=74\).
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.11 \[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=-\frac {2 \sqrt {a} x^{n/2} \sqrt {1+\frac {b x^{2-n}}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^{1-\frac {n}{2}}}{\sqrt {a}}\right )}{\sqrt {b} (-2+n) \sqrt {b x^2+a x^n}} \]
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\[\int \frac {1}{\sqrt {x^{2} \left (b +a \,x^{-2+n}\right )}}d x\]
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none
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.95 \[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\left [\frac {\sqrt {b} \log \left (\frac {a x x^{n - 2} + 2 \, b x - 2 \, \sqrt {a x^{2} x^{n - 2} + b x^{2}} \sqrt {b}}{x x^{n - 2}}\right )}{b n - 2 \, b}, \frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {a x^{2} x^{n - 2} + b x^{2}} \sqrt {-b}}{b x}\right )}{b n - 2 \, b}\right ] \]
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\[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\int \frac {1}{\sqrt {x^{2} \left (a x^{n - 2} + b\right )}}\, dx \]
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\[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\int { \frac {1}{\sqrt {{\left (a x^{n - 2} + b\right )} x^{2}}} \,d x } \]
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\[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\int { \frac {1}{\sqrt {{\left (a x^{n - 2} + b\right )} x^{2}}} \,d x } \]
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Time = 9.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.81 \[ \int \frac {1}{\sqrt {x^2 \left (b+a x^{-2+n}\right )}} \, dx=\frac {\sqrt {a}\,x^{n/2}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x^{1-\frac {n}{2}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\sqrt {\frac {b\,x^{2-n}}{a}+1}\,1{}\mathrm {i}}{\sqrt {b}\,\left (\frac {n}{2}-1\right )\,\sqrt {a\,x^n+b\,x^2}} \]
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