Integrand size = 23, antiderivative size = 62 \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=-\frac {4 b^2 (a-b x)^{1-n} (a+b x)^{-1+n} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{a (1-n)} \]
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Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {133} \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=-\frac {4 b^2 (a-b x)^{1-n} (a+b x)^{n-1} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{a (1-n)} \]
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Rule 133
Rubi steps \begin{align*} \text {integral}& = -\frac {4 b^2 (a-b x)^{1-n} (a+b x)^{-1+n} \, _2F_1\left (3,1-n;2-n;\frac {a-b x}{a+b x}\right )}{a (1-n)} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=-\frac {4 b^2 (a-b x)^{1-n} (a+b x)^{-1+n} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{a (1-n)} \]
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\[\int \frac {\left (b x +a \right )^{1+n} \left (-b x +a \right )^{-n}}{x^{3}}d x\]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{3}} \,d x } \]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=\int \frac {\left (a - b x\right )^{- n} \left (a + b x\right )^{n + 1}}{x^{3}}\, dx \]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{3}} \,d x } \]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{3}} \,d x } \]
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Timed out. \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{n+1}}{x^3\,{\left (a-b\,x\right )}^n} \,d x \]
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