Integrand size = 18, antiderivative size = 71 \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=-\frac {(1-x)^{1+n} (1+x)^{1-n}}{2 x^2}+\frac {2 n (1-x)^{1+n} (1+x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {1-x}{1+x}\right )}{1+n} \]
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Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {98, 133} \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\frac {2 n (1-x)^{n+1} (x+1)^{-n-1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {1-x}{x+1}\right )}{n+1}-\frac {(1-x)^{n+1} (x+1)^{1-n}}{2 x^2} \]
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Rule 98
Rule 133
Rubi steps \begin{align*} \text {integral}& = -\frac {(1-x)^{1+n} (1+x)^{1-n}}{2 x^2}-n \int \frac {(1-x)^n (1+x)^{-n}}{x^2} \, dx \\ & = -\frac {(1-x)^{1+n} (1+x)^{1-n}}{2 x^2}+\frac {2 n (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 = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\frac {(1-x)^{1+n} (1+x)^{-1-n} \left (-\left ((1+n) (1+x)^2\right )+4 n x^2 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {1-x}{1+x}\right )\right )}{2 (1+n) x^2} \]
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\[\int \frac {\left (1-x \right )^{n} \left (1+x \right )^{-n}}{x^{3}}d x\]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\int { \frac {{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{3}} \,d x } \]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\int \frac {\left (1 - x\right )^{n} \left (x + 1\right )^{- n}}{x^{3}}\, dx \]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\int { \frac {{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{3}} \,d x } \]
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\[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\int { \frac {{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{3}} \,d x } \]
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Timed out. \[ \int \frac {(1-x)^n (1+x)^{-n}}{x^3} \, dx=\int \frac {{\left (1-x\right )}^n}{x^3\,{\left (x+1\right )}^n} \,d x \]
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