Integrand size = 33, antiderivative size = 279 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=-\frac {n (A-2 A n+2 B (1+n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(1+n) (B+2 A (1-n)+2 B n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{3 a^2 d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2} \]
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Time = 0.36 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3057, 2827, 2722} \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\frac {(n+1) (2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{3 a^2 d^2 f (n+2) \sqrt {\cos ^2(e+f x)}}-\frac {n (-2 A n+A+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{3 a^2 d f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 a^2 d f (\sin (e+f x)+1)}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 d f (a \sin (e+f x)+a)^2} \]
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Rule 2722
Rule 2827
Rule 3057
Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac {\int \frac {(d \sin (e+f x))^n (a d (2 A+B-A n+B n)+a (A-B) d n \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{3 a^2 d} \\ & = \frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac {\int (d \sin (e+f x))^n \left (-a^2 d^2 n (A-2 A n+2 B (1+n))+a^2 d^2 (1+n) (2 A (1-n)+B (1+2 n)) \sin (e+f x)\right ) \, dx}{3 a^4 d^2} \\ & = \frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac {((1+n) (B+2 A (1-n)+2 B n)) \int (d \sin (e+f x))^{1+n} \, dx}{3 a^2 d}-\frac {(n (A-2 A n+2 B (1+n))) \int (d \sin (e+f x))^n \, dx}{3 a^2} \\ & = -\frac {n (A-2 A n+2 B (1+n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(1+n) (B+2 A (1-n)+2 B n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{3 a^2 d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2} \\ \end{align*}
Time = 2.58 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.79 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\frac {(d \sin (e+f x))^n \left ((A-B) \sin (2 (e+f x))-\frac {2 (1+\sin (e+f x)) \left ((1+n) (2+n) (2 A (-1+n)-B (1+2 n)) \cos ^2(e+f x)+\sqrt {\cos ^2(e+f x)} (1+\sin (e+f x)) \left (n (2+n) (A-2 A n+2 B (1+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right )+(1+n)^2 (2 A (-1+n)-B (1+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right ) \tan (e+f x)}{(1+n) (2+n)}\right )}{6 a^2 f (1+\sin (e+f x))^2} \]
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\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )\right )}{\left (a +a \sin \left (f x +e \right )\right )^{2}}d x\]
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\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]
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