Integrand size = 12, antiderivative size = 151 \[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=-\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}-\frac {x^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m}+\frac {4 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}+\frac {4 x^m \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m} \]
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Time = 0.91 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6307, 6874, 371, 864, 822} \[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}+\frac {4 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}-\frac {x^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m}+\frac {4 x^m \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m} \]
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Rule 371
Rule 822
Rule 864
Rule 6307
Rule 6874
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 )^2}{\left (1-\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\right ) \\ & = -\left (\left (\left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \left (-\frac {3 x^{-2-m}}{\sqrt {1-\frac {x^2}{a^2}}}-\frac {x^{-1-m}}{a \sqrt {1-\frac {x^2}{a^2}}}+\frac {4 x^{-2-m}}{\left (1-\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}}\right ) \, dx,x,\frac {1}{x}\right )\right ) \\ & = \left (3 \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )-\left (4 \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{\left (1-\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )+\frac {\left (\left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-1-m}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a} \\ & = -\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}-\frac {x^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m}-\left (4 \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-2-m} \left (1+\frac {x}{a}\right )}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}-\frac {x^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m}-\left (4 \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )-\frac {\left (4 \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-1-m}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{a} \\ & = -\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}-\frac {x^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m}+\frac {4 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}+\frac {4 x^m \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.35 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.51 \[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=\frac {x^{1+m} \left (3 (1+m) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1-a x} \sqrt {\frac {1+a x}{a^2}} \operatorname {AppellF1}\left (m,-\frac {1}{2},\frac {1}{2},1+m,-a x,a x\right )-2 (1+m) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1-a x} \sqrt {\frac {1+a x}{a^2}} \operatorname {AppellF1}\left (m,-\frac {1}{2},\frac {3}{2},1+m,-a x,a x\right )+m \sqrt {-1+a x} \sqrt {1+a x} \sqrt {-\frac {1}{a^2}+x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-\frac {m}{2},\frac {1}{2}-\frac {m}{2},\frac {1}{a^2 x^2}\right )\right )}{m (1+m) \sqrt {-1+a x} \sqrt {1+a x} \sqrt {-\frac {1}{a^2}+x^2}} \]
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\[\int \frac {x^{m}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}d x\]
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\[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=\int { \frac {x^{m}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=\int \frac {x^{m}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=\int { \frac {x^{m}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
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Exception generated. \[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int e^{3 \coth ^{-1}(a x)} x^m \, dx=\int \frac {x^m}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]
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