Integrand size = 15, antiderivative size = 144 \[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=-\frac {2 \sqrt {1-x^2} \coth ^{-1}(x) \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right )}{\sqrt {a-a x^2}}-\frac {i \sqrt {1-x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-x}}{\sqrt {1+x}}\right )}{\sqrt {a-a x^2}}+\frac {i \sqrt {1-x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-x}}{\sqrt {1+x}}\right )}{\sqrt {a-a x^2}} \]
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Time = 0.04 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6102, 6098} \[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=-\frac {2 \sqrt {1-x^2} \arctan \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \coth ^{-1}(x)}{\sqrt {a-a x^2}}-\frac {i \sqrt {1-x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}}+\frac {i \sqrt {1-x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-x}}{\sqrt {x+1}}\right )}{\sqrt {a-a x^2}} \]
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Rule 6098
Rule 6102
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-x^2} \int \frac {\coth ^{-1}(x)}{\sqrt {1-x^2}} \, dx}{\sqrt {a-a x^2}} \\ & = -\frac {2 \sqrt {1-x^2} \coth ^{-1}(x) \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right )}{\sqrt {a-a x^2}}-\frac {i \sqrt {1-x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-x}}{\sqrt {1+x}}\right )}{\sqrt {a-a x^2}}+\frac {i \sqrt {1-x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-x}}{\sqrt {1+x}}\right )}{\sqrt {a-a x^2}} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.53 \[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=\frac {\sqrt {a-a x^2} \left (\coth ^{-1}(x) \left (\log \left (1-e^{-\coth ^{-1}(x)}\right )-\log \left (1+e^{-\coth ^{-1}(x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-\coth ^{-1}(x)}\right )-\operatorname {PolyLog}\left (2,e^{-\coth ^{-1}(x)}\right )\right )}{a \sqrt {1-\frac {1}{x^2}} x} \]
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Time = 0.42 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.32
method | result | size |
default | \(\frac {\ln \left (1-\frac {1}{\sqrt {\frac {x -1}{1+x}}}\right ) \operatorname {arccoth}\left (x \right ) \sqrt {\frac {x -1}{1+x}}\, \sqrt {-\left (1+x \right ) \left (x -1\right ) a}}{\left (x -1\right ) a}+\frac {\operatorname {polylog}\left (2, \frac {1}{\sqrt {\frac {x -1}{1+x}}}\right ) \sqrt {\frac {x -1}{1+x}}\, \sqrt {-\left (1+x \right ) \left (x -1\right ) a}}{\left (x -1\right ) a}-\frac {\ln \left (\frac {1}{\sqrt {\frac {x -1}{1+x}}}+1\right ) \operatorname {arccoth}\left (x \right ) \sqrt {\frac {x -1}{1+x}}\, \sqrt {-\left (1+x \right ) \left (x -1\right ) a}}{\left (x -1\right ) a}-\frac {\operatorname {polylog}\left (2, -\frac {1}{\sqrt {\frac {x -1}{1+x}}}\right ) \sqrt {\frac {x -1}{1+x}}\, \sqrt {-\left (1+x \right ) \left (x -1\right ) a}}{\left (x -1\right ) a}\) | \(190\) |
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\[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=\int { \frac {\operatorname {arcoth}\left (x\right )}{\sqrt {-a x^{2} + a}} \,d x } \]
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\[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=\int \frac {\operatorname {acoth}{\left (x \right )}}{\sqrt {- a \left (x - 1\right ) \left (x + 1\right )}}\, dx \]
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\[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=\int { \frac {\operatorname {arcoth}\left (x\right )}{\sqrt {-a x^{2} + a}} \,d x } \]
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\[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=\int { \frac {\operatorname {arcoth}\left (x\right )}{\sqrt {-a x^{2} + a}} \,d x } \]
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Timed out. \[ \int \frac {\coth ^{-1}(x)}{\sqrt {a-a x^2}} \, dx=\int \frac {\mathrm {acoth}\left (x\right )}{\sqrt {a-a\,x^2}} \,d x \]
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