Integrand size = 22, antiderivative size = 118 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} \left (1-\frac {1}{a x}\right )^{-\frac {3}{2}+p} \left (1+\frac {1}{a x}\right )^{\frac {5}{2}+p} x \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,\frac {3}{2}-p,-2 p,\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{1+2 p} \]
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Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6327, 6331, 134} \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\frac {x \left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} \left (1-\frac {1}{a x}\right )^{p-\frac {3}{2}} \left (\frac {1}{a x}+1\right )^{p+\frac {5}{2}} \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-2 p-1,\frac {3}{2}-p,-2 p,\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{2 p+1} \]
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Rule 134
Rule 6327
Rule 6331
Rubi steps \begin{align*} \text {integral}& = \left (\left (1-\frac {1}{a^2 x^2}\right )^{-p} x^{-2 p} \left (c-a^2 c x^2\right )^p\right ) \int e^{3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^p x^{2 p} \, dx \\ & = -\left (\left (\left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {1}{x}\right )^{2 p} \left (c-a^2 c x^2\right )^p\right ) \text {Subst}\left (\int x^{-2-2 p} \left (1-\frac {x}{a}\right )^{-\frac {3}{2}+p} \left (1+\frac {x}{a}\right )^{\frac {3}{2}+p} \, dx,x,\frac {1}{x}\right )\right ) \\ & = \frac {\left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} \left (1-\frac {1}{a x}\right )^{-\frac {3}{2}+p} \left (1+\frac {1}{a x}\right )^{\frac {5}{2}+p} x \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,\frac {3}{2}-p,-2 p,\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{1+2 p} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=-\frac {4^{1+p} e^{5 \coth ^{-1}(a x)} \left (1-e^{2 \coth ^{-1}(a x)}\right )^{2 p} \left (\frac {e^{\coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}\right )^{2 p} \left (a \sqrt {1-\frac {1}{a^2 x^2}} x\right )^{-2 p} \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2}+p,2+2 p,\frac {7}{2}+p,e^{2 \coth ^{-1}(a x)}\right )}{5 a+2 a p} \]
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\[\int \frac {\left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}d x\]
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\[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{p}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{p}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{p}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^p}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]
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