Integrand size = 23, antiderivative size = 139 \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=-\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{4 a^2 x^4}-\frac {b (1+2 n) (a-b x)^{1-n} (a+b x)^{2+n}}{12 a^3 x^3}-\frac {4 b^4 \left (1+n+n^2\right ) (a-b x)^{1-n} (a+b x)^{-1+n} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{3 a^3 (1-n)} \]
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Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {105, 156, 12, 133} \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=-\frac {4 b^4 \left (n^2+n+1\right ) (a+b x)^{n-1} (a-b x)^{1-n} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{3 a^3 (1-n)}-\frac {b (2 n+1) (a+b x)^{n+2} (a-b x)^{1-n}}{12 a^3 x^3}-\frac {(a+b x)^{n+2} (a-b x)^{1-n}}{4 a^2 x^4} \]
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Rule 12
Rule 105
Rule 133
Rule 156
Rubi steps \begin{align*} \text {integral}& = -\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{4 a^2 x^4}-\frac {\int \frac {(a-b x)^{-n} (a+b x)^{1+n} \left (-a b (1+2 n)-b^2 x\right )}{x^4} \, dx}{4 a^2} \\ & = -\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{4 a^2 x^4}-\frac {b (1+2 n) (a-b x)^{1-n} (a+b x)^{2+n}}{12 a^3 x^3}+\frac {\int \frac {4 a^2 b^2 \left (1+n+n^2\right ) (a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx}{12 a^4} \\ & = -\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{4 a^2 x^4}-\frac {b (1+2 n) (a-b x)^{1-n} (a+b x)^{2+n}}{12 a^3 x^3}+\frac {\left (b^2 \left (1+n+n^2\right )\right ) \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx}{3 a^2} \\ & = -\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{4 a^2 x^4}-\frac {b (1+2 n) (a-b x)^{1-n} (a+b x)^{2+n}}{12 a^3 x^3}-\frac {4 b^4 \left (1+n+n^2\right ) (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 )}{3 a^3 (1-n)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.73 \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=\frac {(a-b x)^{1-n} (a+b x)^{-1+n} \left (-\left ((-1+n) (a+b x)^3 (3 a+b (1+2 n) x)\right )+16 b^4 \left (1+n+n^2\right ) x^4 \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )\right )}{12 a^3 (-1+n) x^4} \]
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\[\int \frac {\left (b x +a \right )^{1+n} \left (-b x +a \right )^{-n}}{x^{5}}d x\]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{5}} \,d x } \]
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Timed out. \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=\text {Timed out} \]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{5}} \,d x } \]
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\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{5}} \,d x } \]
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Timed out. \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^5} \, dx=\int \frac {{\left (a+b\,x\right )}^{n+1}}{x^5\,{\left (a-b\,x\right )}^n} \,d x \]
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