Integrand size = 18, antiderivative size = 44 \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=-\frac {2 (1-x)^{1+n} (1+x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {1-x}{1+x}\right )}{1+n} \]
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Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {133} \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=-\frac {2 (1-x)^{n+1} (x+1)^{-n-1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {1-x}{x+1}\right )}{n+1} \]
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Rule 133
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (1-x)^{1+n} (1+x)^{-1-n} \, _2F_1\left (2,1+n;2+n;\frac {1-x}{1+x}\right )}{1+n} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=-\frac {2 (1-x)^{1+n} (1+x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {1-x}{1+x}\right )}{1+n} \]
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\[\int \frac {\left (1-x \right )^{n} \left (1+x \right )^{-n}}{x^{2}}d x\]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=\int { \frac {{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{2}} \,d x } \]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=\int \frac {\left (1 - x\right )^{n} \left (x + 1\right )^{- n}}{x^{2}}\, dx \]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=\int { \frac {{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{2}} \,d x } \]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=\int { \frac {{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx=\int \frac {{\left (1-x\right )}^n}{x^2\,{\left (x+1\right )}^n} \,d x \]
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