Integrand size = 22, antiderivative size = 114 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\frac {\left (c-\frac {c}{a x}\right )^{2+p} x}{c^2}+\frac {\left (c-\frac {c}{a x}\right )^{2+p} \operatorname {Hypergeometric2F1}\left (1,2+p,3+p,\frac {a-\frac {1}{x}}{2 a}\right )}{2 a c^2 (2+p)}-\frac {\left (c-\frac {c}{a x}\right )^{2+p} \operatorname {Hypergeometric2F1}\left (1,2+p,3+p,1-\frac {1}{a x}\right )}{a c^2} \]
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Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {6302, 6268, 25, 528, 382, 105, 162, 67, 70} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\frac {\left (c-\frac {c}{a x}\right )^{p+2} \operatorname {Hypergeometric2F1}\left (1,p+2,p+3,\frac {a-\frac {1}{x}}{2 a}\right )}{2 a c^2 (p+2)}-\frac {\left (c-\frac {c}{a x}\right )^{p+2} \operatorname {Hypergeometric2F1}\left (1,p+2,p+3,1-\frac {1}{a x}\right )}{a c^2}+\frac {x \left (c-\frac {c}{a x}\right )^{p+2}}{c^2} \]
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Rule 25
Rule 67
Rule 70
Rule 105
Rule 162
Rule 382
Rule 528
Rule 6268
Rule 6302
Rubi steps \begin{align*} \text {integral}& = -\int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx \\ & = -\int \frac {\left (c-\frac {c}{a x}\right )^p (1-a x)}{1+a x} \, dx \\ & = \frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{1+p} x}{1+a x} \, dx}{c} \\ & = \frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{1+p}}{a+\frac {1}{x}} \, dx}{c} \\ & = -\frac {a \text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{1+p}}{x^2 (a+x)} \, dx,x,\frac {1}{x}\right )}{c} \\ & = \frac {\left (c-\frac {c}{a x}\right )^{2+p} x}{c^2}+\frac {\text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{1+p} \left (c (2+p)+\frac {c (1+p) x}{a}\right )}{x (a+x)} \, dx,x,\frac {1}{x}\right )}{c^2} \\ & = \frac {\left (c-\frac {c}{a x}\right )^{2+p} x}{c^2}-\frac {\text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{1+p}}{a+x} \, dx,x,\frac {1}{x}\right )}{a c}+\frac {(2+p) \text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{1+p}}{x} \, dx,x,\frac {1}{x}\right )}{a c} \\ & = \frac {\left (c-\frac {c}{a x}\right )^{2+p} x}{c^2}+\frac {\left (c-\frac {c}{a x}\right )^{2+p} \operatorname {Hypergeometric2F1}\left (1,2+p,3+p,\frac {a-\frac {1}{x}}{2 a}\right )}{2 a c^2 (2+p)}-\frac {\left (c-\frac {c}{a x}\right )^{2+p} \operatorname {Hypergeometric2F1}\left (1,2+p,3+p,1-\frac {1}{a x}\right )}{a c^2} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\frac {\left (c-\frac {c}{a x}\right )^p (-1+a x)^2 \left (\operatorname {Hypergeometric2F1}\left (1,2+p,3+p,\frac {a-\frac {1}{x}}{2 a}\right )+2 (2+p) \left (a x-\operatorname {Hypergeometric2F1}\left (1,2+p,3+p,1-\frac {1}{a x}\right )\right )\right )}{2 a^3 (2+p) x^2} \]
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\[\int \frac {\left (c -\frac {c}{a x}\right )^{p} \left (a x -1\right )}{a x +1}d x\]
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\[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\int { \frac {{\left (a x - 1\right )} {\left (c - \frac {c}{a x}\right )}^{p}}{a x + 1} \,d x } \]
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\[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\int \frac {\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{p} \left (a x - 1\right )}{a x + 1}\, dx \]
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\[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\int { \frac {{\left (a x - 1\right )} {\left (c - \frac {c}{a x}\right )}^{p}}{a x + 1} \,d x } \]
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\[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\int { \frac {{\left (a x - 1\right )} {\left (c - \frac {c}{a x}\right )}^{p}}{a x + 1} \,d x } \]
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Timed out. \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx=\int \frac {{\left (c-\frac {c}{a\,x}\right )}^p\,\left (a\,x-1\right )}{a\,x+1} \,d x \]
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