Integrand size = 14, antiderivative size = 41 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\frac {x^{1+m} \operatorname {AppellF1}\left (-1-m,-\frac {5}{4},\frac {5}{4},-m,\frac {1}{a x},-\frac {1}{a x}\right )}{1+m} \]
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Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6308, 138} \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\frac {x^{m+1} \operatorname {AppellF1}\left (-m-1,-\frac {5}{4},\frac {5}{4},-m,\frac {1}{a x},-\frac {1}{a x}\right )}{m+1} \]
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Rule 138
Rule 6308
Rubi steps \begin{align*} \text {integral}& = -\left (\left (\left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-2-m} \left (1-\frac {x}{a}\right )^{5/4}}{\left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )\right ) \\ & = \frac {x^{1+m} \operatorname {AppellF1}\left (-1-m,-\frac {5}{4},\frac {5}{4},-m,\frac {1}{a x},-\frac {1}{a x}\right )}{1+m} \\ \end{align*}
\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx \]
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\[\int x^{m} \left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}d x\]
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\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\int { x^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} \,d x } \]
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Timed out. \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\text {Timed out} \]
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\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\int { x^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} \,d x } \]
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\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\int { x^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} \,d x } \]
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Timed out. \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx=\int x^m\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4} \,d x \]
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