Integrand size = 18, antiderivative size = 190 \[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=-\frac {d (b c-2 a d) (a+b x)^{1+n}}{a c^2 (b c-a d) (c+d x)}-\frac {(a+b x)^{1+n}}{a c x (c+d x)}-\frac {d^2 (2 a d-b c (2-n)) (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {d (a+b x)}{b c-a d}\right )}{c^3 (b c-a d)^2 (1+n)}+\frac {(2 a d-b c n) (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )}{a^2 c^3 (1+n)} \]
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Time = 0.14 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {105, 156, 162, 67, 70} \[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=\frac {(a+b x)^{n+1} (2 a d-b c n) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b x}{a}+1\right )}{a^2 c^3 (n+1)}-\frac {d^2 (a+b x)^{n+1} (2 a d-b c (2-n)) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {d (a+b x)}{b c-a d}\right )}{c^3 (n+1) (b c-a d)^2}-\frac {d (b c-2 a d) (a+b x)^{n+1}}{a c^2 (c+d x) (b c-a d)}-\frac {(a+b x)^{n+1}}{a c x (c+d x)} \]
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Rule 67
Rule 70
Rule 105
Rule 156
Rule 162
Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{1+n}}{a c x (c+d x)}-\frac {\int \frac {(a+b x)^n (2 a d-b c n+b d (1-n) x)}{x (c+d x)^2} \, dx}{a c} \\ & = -\frac {d (b c-2 a d) (a+b x)^{1+n}}{a c^2 (b c-a d) (c+d x)}-\frac {(a+b x)^{1+n}}{a c x (c+d x)}+\frac {\int \frac {(a+b x)^n (-((b c-a d) (2 a d-b c n))+b d (b c-2 a d) n x)}{x (c+d x)} \, dx}{a c^2 (b c-a d)} \\ & = -\frac {d (b c-2 a d) (a+b x)^{1+n}}{a c^2 (b c-a d) (c+d x)}-\frac {(a+b x)^{1+n}}{a c x (c+d x)}-\frac {\left (d^2 (2 a d-b c (2-n))\right ) \int \frac {(a+b x)^n}{c+d x} \, dx}{c^3 (b c-a d)}-\frac {(2 a d-b c n) \int \frac {(a+b x)^n}{x} \, dx}{a c^3} \\ & = -\frac {d (b c-2 a d) (a+b x)^{1+n}}{a c^2 (b c-a d) (c+d x)}-\frac {(a+b x)^{1+n}}{a c x (c+d x)}-\frac {d^2 (2 a d-b c (2-n)) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;-\frac {d (a+b x)}{b c-a d}\right )}{c^3 (b c-a d)^2 (1+n)}+\frac {(2 a d-b c n) (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {b x}{a}\right )}{a^2 c^3 (1+n)} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=-\frac {(a+b x)^{1+n} \left (a c^2 (b c-a d)^2 (1+n)+a c d (-b c+a d) (-b c+2 a d) (1+n) x-x (c+d x) \left (-a^2 d^2 (2 a d+b c (-2+n)) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )+(b c-a d)^2 (2 a d-b c n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b x}{a}\right )\right )\right )}{a^2 c^3 (b c-a d)^2 (1+n) x (c+d x)} \]
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\[\int \frac {\left (b x +a \right )^{n}}{x^{2} \left (d x +c \right )^{2}}d x\]
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\[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=\int \frac {\left (a + b x\right )^{n}}{x^{2} \left (c + d x\right )^{2}}\, dx \]
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\[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{2} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^n}{x^2 (c+d x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^n}{x^2\,{\left (c+d\,x\right )}^2} \,d x \]
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