Integrand size = 31, antiderivative size = 322 \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\frac {b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) (e x)^{1+m}}{2 c^2 d^3 e (1+m) n^2}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac {(a d (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))-b c (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{2 c^3 d^3 e (1+m) n^2} \]
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Time = 0.35 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {608, 470, 371} \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right ) (a d (b c (m+1)-a d (m-n+1)) (B c (m+1)-A d (m-2 n+1))-b c (a d (m+1)-b c (m+n+1)) (A d (m+1)-B c (m+2 n+1)))}{2 c^3 d^3 e (m+1) n^2}+\frac {b (e x)^{m+1} (a d (m+1)-b c (m+n+1)) (A d (m+1)-B c (m+2 n+1))}{2 c^2 d^3 e (m+1) n^2}-\frac {(e x)^{m+1} (b c-a d) \left (a (B c (m+1)-A d (m-2 n+1))-b x^n (A d (m+1)-B c (m+2 n+1))\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}-\frac {(e x)^{m+1} \left (a+b x^n\right )^2 (B c-A d)}{2 c d e n \left (c+d x^n\right )^2} \]
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Rule 371
Rule 470
Rule 608
Rubi steps \begin{align*} \text {integral}& = -\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac {\int \frac {(e x)^m \left (a+b x^n\right ) \left (-a (B c (1+m)-A d (1+m-2 n))+b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{\left (c+d x^n\right )^2} \, dx}{2 c d n} \\ & = -\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac {\int \frac {(e x)^m \left (a (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))+b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) x^n\right )}{c+d x^n} \, dx}{2 c^2 d^2 n^2} \\ & = \frac {b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) (e x)^{1+m}}{2 c^2 d^3 e (1+m) n^2}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac {\left (a (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))-\frac {b c (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n))}{d}\right ) \int \frac {(e x)^m}{c+d x^n} \, dx}{2 c^2 d^2 n^2} \\ & = \frac {b (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n)) (e x)^{1+m}}{2 c^2 d^3 e (1+m) n^2}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{2 c d e n \left (c+d x^n\right )^2}-\frac {(b c-a d) (e x)^{1+m} \left (a (B c (1+m)-A d (1+m-2 n))-b (A d (1+m)-B c (1+m+2 n)) x^n\right )}{2 c^2 d^2 e n^2 \left (c+d x^n\right )}+\frac {\left (a (B c (1+m)-A d (1+m-2 n)) (b c (1+m)-a d (1+m-n))-\frac {b c (a d (1+m)-b c (1+m+n)) (A d (1+m)-B c (1+m+2 n))}{d}\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{2 c^3 d^2 e (1+m) n^2} \\ \end{align*}
Time = 0.70 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.53 \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\frac {x (e x)^m \left (b^2 B-\frac {b (3 b B c-A b d-2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c}+\frac {(b c-a d) (3 b B c-2 A b d-a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2}-\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^3}\right )}{d^3 (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{\left (c +d \,x^{n}\right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^2}{{\left (c+d\,x^n\right )}^3} \,d x \]
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