Integrand size = 20, antiderivative size = 62 \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\frac {(b c-a d) (a+b x)^{1+n} (c+d x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c (a+b x)}{a (c+d x)}\right )}{a^2 (1+n)} \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {133} \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\frac {(b c-a d) (a+b x)^{n+1} (c+d x)^{-n-1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {c (a+b x)}{a (c+d x)}\right )}{a^2 (n+1)} \]
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Rule 133
Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d) (a+b x)^{1+n} (c+d x)^{-1-n} \, _2F_1\left (2,1+n;2+n;\frac {c (a+b x)}{a (c+d x)}\right )}{a^2 (1+n)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\frac {(b c-a d) (a+b x)^{1+n} (c+d x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c (a+b x)}{a (c+d x)}\right )}{a^2 (1+n)} \]
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\[\int \frac {\left (b x +a \right )^{n} \left (d x +c \right )^{-n}}{x^{2}}d x\]
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\[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx=\int \frac {{\left (a+b\,x\right )}^n}{x^2\,{\left (c+d\,x\right )}^n} \,d x \]
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