Integrand size = 31, antiderivative size = 268 \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=-\frac {d^2 (A b (1+m+n)-a B (1+m+2 n)) x^{1+n} (e x)^m}{a b^2 n (1+m+n)}-\frac {d (A b (2 b c (1+m)-a d (1+m+n))-a B (2 b c (1+m+n)-a d (1+m+2 n))) (e x)^{1+m}}{a b^3 e (1+m) n}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )}-\frac {(b c-a d) (A b (b c (1+m-n)-a d (1+m+n))-a B (b c (1+m)-a d (1+m+2 n))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2 b^3 e (1+m) n} \]
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Time = 0.41 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {608, 584, 20, 30, 371} \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=-\frac {(e x)^{m+1} (b c-a d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right ) (A b (b c (m-n+1)-a d (m+n+1))-a B (b c (m+1)-a d (m+2 n+1)))}{a^2 b^3 e (m+1) n}-\frac {d (e x)^{m+1} (A b (2 b c (m+1)-a d (m+n+1))-a B (2 b c (m+n+1)-a d (m+2 n+1)))}{a b^3 e (m+1) n}-\frac {d^2 x^{n+1} (e x)^m (A b (m+n+1)-a B (m+2 n+1))}{a b^2 n (m+n+1)}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )} \]
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Rule 20
Rule 30
Rule 371
Rule 584
Rule 608
Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )}-\frac {\int \frac {(e x)^m \left (c+d x^n\right ) \left (-c (a B (1+m)-A b (1+m-n))+d (A b (1+m+n)-a B (1+m+2 n)) x^n\right )}{a+b x^n} \, dx}{a b n} \\ & = \frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )}-\frac {\int \left (\frac {d (A b (2 b c (1+m)-a d (1+m+n))-a B (2 b c (1+m+n)-a d (1+m+2 n))) (e x)^m}{b^2}+\frac {d^2 (A b (1+m+n)-a B (1+m+2 n)) x^n (e x)^m}{b}+\frac {(b c-a d) (A b (b c (1+m-n)-a d (1+m+n))-a B (b c (1+m)-a d (1+m+2 n))) (e x)^m}{b^2 \left (a+b x^n\right )}\right ) \, dx}{a b n} \\ & = -\frac {d (A b (2 b c (1+m)-a d (1+m+n))-a B (2 b c (1+m+n)-a d (1+m+2 n))) (e x)^{1+m}}{a b^3 e (1+m) n}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )}-\frac {\left (d^2 (A b (1+m+n)-a B (1+m+2 n))\right ) \int x^n (e x)^m \, dx}{a b^2 n}-\frac {((b c-a d) (A b (b c (1+m-n)-a d (1+m+n))-a B (b c (1+m)-a d (1+m+2 n)))) \int \frac {(e x)^m}{a+b x^n} \, dx}{a b^3 n} \\ & = -\frac {d (A b (2 b c (1+m)-a d (1+m+n))-a B (2 b c (1+m+n)-a d (1+m+2 n))) (e x)^{1+m}}{a b^3 e (1+m) n}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )}-\frac {(b c-a d) (A b (b c (1+m-n)-a d (1+m+n))-a B (b c (1+m)-a d (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a^2 b^3 e (1+m) n}-\frac {\left (d^2 (A b (1+m+n)-a B (1+m+2 n)) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{a b^2 n} \\ & = -\frac {d^2 (A b (1+m+n)-a B (1+m+2 n)) x^{1+n} (e x)^m}{a b^2 n (1+m+n)}-\frac {d (A b (2 b c (1+m)-a d (1+m+n))-a B (2 b c (1+m+n)-a d (1+m+2 n))) (e x)^{1+m}}{a b^3 e (1+m) n}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )^2}{a b e n \left (a+b x^n\right )}-\frac {(b c-a d) (A b (b c (1+m-n)-a d (1+m+n))-a B (b c (1+m)-a d (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a^2 b^3 e (1+m) n} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.59 \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=\frac {x (e x)^m \left (\frac {d (2 b B c+A b d-2 a B d)}{1+m}+\frac {b B d^2 x^n}{1+m+n}+\frac {(b c-a d) (b B c+2 A b d-3 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a (1+m)}+\frac {(A b-a B) (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2 (1+m)}\right )}{b^3} \]
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\[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right ) \left (c +d \,x^{n}\right )^{2}}{\left (a +b \,x^{n}\right )^{2}}d x\]
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\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (d x^{n} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=\int \frac {\left (e x\right )^{m} \left (A + B x^{n}\right ) \left (c + d x^{n}\right )^{2}}{\left (a + b x^{n}\right )^{2}}\, dx \]
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\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (d x^{n} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (d x^{n} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2}{\left (a+b x^n\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (c+d\,x^n\right )}^2}{{\left (a+b\,x^n\right )}^2} \,d x \]
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