Integrand size = 20, antiderivative size = 194 \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=-\frac {(a+b x)^{1+n} (c+d x)^{1-n}}{3 a c x^3}+\frac {(b c (2-n)+a d (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 a^2 c^2 x^2}+\frac {(b c-a d) \left (2 a b c d \left (1-n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) (a+b x)^{1+n} (c+d x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c (a+b x)}{a (c+d x)}\right )}{6 a^4 c^2 (1+n)} \]
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Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {105, 156, 12, 133} \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\frac {(a+b x)^{n+1} (c+d x)^{1-n} (a d (n+2)+b c (2-n))}{6 a^2 c^2 x^2}+\frac {(b c-a d) (a+b x)^{n+1} (c+d x)^{-n-1} \left (a^2 d^2 \left (n^2+3 n+2\right )+2 a b c d \left (1-n^2\right )+b^2 c^2 \left (n^2-3 n+2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {c (a+b x)}{a (c+d x)}\right )}{6 a^4 c^2 (n+1)}-\frac {(a+b x)^{n+1} (c+d x)^{1-n}}{3 a c x^3} \]
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Rule 12
Rule 105
Rule 133
Rule 156
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{1+n} (c+d x)^{1-n}}{3 a c x^3}-\frac {\int \frac {(a+b x)^n (c+d x)^{-n} (b c (2-n)+a d (2+n)+b d x)}{x^3} \, dx}{3 a c} \\ & = -\frac {(a+b x)^{1+n} (c+d x)^{1-n}}{3 a c x^3}+\frac {(b c (2-n)+a d (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 a^2 c^2 x^2}+\frac {\int \frac {\left (2 a b c d \left (1-n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) (a+b x)^n (c+d x)^{-n}}{x^2} \, dx}{6 a^2 c^2} \\ & = -\frac {(a+b x)^{1+n} (c+d x)^{1-n}}{3 a c x^3}+\frac {(b c (2-n)+a d (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 a^2 c^2 x^2}+\frac {\left (2 a b c d \left (1-n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) \int \frac {(a+b x)^n (c+d x)^{-n}}{x^2} \, dx}{6 a^2 c^2} \\ & = -\frac {(a+b x)^{1+n} (c+d x)^{1-n}}{3 a c x^3}+\frac {(b c (2-n)+a d (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 a^2 c^2 x^2}+\frac {(b c-a d) \left (2 a b c d \left (1-n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) (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 )}{6 a^4 c^2 (1+n)} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\frac {(a+b x)^{1+n} (c+d x)^{-1-n} \left (-\frac {2 a^3 c (c+d x)^2}{x^3}+\frac {a^2 (-b c (-2+n)+a d (2+n)) (c+d x)^2}{x^2}+\frac {(b c-a d) \left (-2 a b c d \left (-1+n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c (a+b x)}{a (c+d x)}\right )}{1+n}\right )}{6 a^4 c^2} \]
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\[\int \frac {\left (b x +a \right )^{n} \left (d x +c \right )^{-n}}{x^{4}}d x\]
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\[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{4}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{4}} \,d x } \]
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\[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{4}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\int \frac {{\left (a+b\,x\right )}^n}{x^4\,{\left (c+d\,x\right )}^n} \,d x \]
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