Integrand size = 31, antiderivative size = 366 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a (b c-a d)^3 e (1+m)}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{2 c^3 (b c-a d)^3 e (1+m) n^2} \]
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Time = 0.77 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {609, 611, 371} \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=-\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )\right )+b^2 c^2 (m-2 n+1) (A d (m-3 n+1)-B c (m-n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^3}+\frac {b^2 (e x)^{m+1} (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{2 c^2 e n^2 (b c-a d)^2 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2} \]
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Rule 371
Rule 609
Rule 611
Rubi steps \begin{align*} \text {integral}& = \frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {\int \frac {(e x)^m \left (-a B c (1+m)+a A d (1+m-2 n)+2 A b c n-b (B c-A d) (1+m-2 n) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^2} \, dx}{2 c (b c-a d) n} \\ & = \frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {\int \frac {(e x)^m \left (-a (1+m) (b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n)))-(b c-a d) n (a B c (1+m)-a A d (1+m-2 n)-2 A b c n)-b (b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (1+m-n) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{2 c^2 (b c-a d)^2 n^2} \\ & = \frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {\int \left (\frac {2 b^2 (A b-a B) c^2 n^2 (e x)^m}{(b c-a d) \left (a+b x^n\right )}+\frac {\left (-b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)+a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)-2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^n\right )}\right ) \, dx}{2 c^2 (b c-a d)^2 n^2} \\ & = \frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {\left (b^2 (A b-a B)\right ) \int \frac {(e x)^m}{a+b x^n} \, dx}{(b c-a d)^3}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) \int \frac {(e x)^m}{c+d x^n} \, dx}{2 c^2 (b c-a d)^3 n^2} \\ & = \frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a (b c-a d)^3 e (1+m)}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\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 (b c-a d)^3 e (1+m) n^2} \\ \end{align*}
Time = 0.50 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.55 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a}-\frac {b (A b-a B) d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c}-\frac {(A b-a B) d (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2}+\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^3}\right )}{(b c-a d)^3 (1+m)} \]
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\[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right )}{\left (a +b \,x^{n}\right ) \left (c +d \,x^{n}\right )^{3}}d x\]
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\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}} \,d x } \]
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Exception generated. \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{\left (a+b\,x^n\right )\,{\left (c+d\,x^n\right )}^3} \,d x \]
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