Integrand size = 33, antiderivative size = 373 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=-\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}+\frac {a^3 \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (2+n) (4+n) \sqrt {\cos ^2(e+f x)}}+\frac {a^3 (B (9+4 n)+A (11+4 n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{d^2 f (2+n) (3+n) \sqrt {\cos ^2(e+f x)}}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)} \]
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Time = 0.59 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3055, 3047, 3102, 2827, 2722} \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {a^3 (A (4 n+11)+B (4 n+9)) \cos (e+f x) (d \sin (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{d^2 f (n+2) (n+3) \sqrt {\cos ^2(e+f x)}}+\frac {a^3 \left (A \left (4 n^2+21 n+20\right )+B \left (4 n^2+19 n+15\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{d f (n+1) (n+2) (n+4) \sqrt {\cos ^2(e+f x)}}-\frac {a^3 \left (A \left (2 n^2+15 n+28\right )+B \left (2 n^2+14 n+27\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2) (n+3) (n+4)}-\frac {(A (n+4)+B (n+6)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right ) (d \sin (e+f x))^{n+1}}{d f (n+3) (n+4)}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2 (d \sin (e+f x))^{n+1}}{d f (n+4)} \]
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Rule 2722
Rule 2827
Rule 3047
Rule 3055
Rule 3102
Rubi steps \begin{align*} \text {integral}& = -\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}+\frac {\int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (a d (B (1+n)+A (4+n))+a d (A (4+n)+B (6+n)) \sin (e+f x)) \, dx}{d (4+n)} \\ & = -\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\int (d \sin (e+f x))^n (a+a \sin (e+f x)) \left (a^2 d^2 \left (2 A \left (8+6 n+n^2\right )+B \left (9+11 n+2 n^2\right )\right )+a^2 d^2 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \sin (e+f x)\right ) \, dx}{d^2 (3+n) (4+n)} \\ & = -\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\int (d \sin (e+f x))^n \left (a^3 d^2 \left (2 A \left (8+6 n+n^2\right )+B \left (9+11 n+2 n^2\right )\right )+\left (a^3 d^2 \left (2 A \left (8+6 n+n^2\right )+B \left (9+11 n+2 n^2\right )\right )+a^3 d^2 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right )\right ) \sin (e+f x)+a^3 d^2 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \sin ^2(e+f x)\right ) \, dx}{d^2 (3+n) (4+n)} \\ & = -\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\int (d \sin (e+f x))^n \left (a^3 d^3 (3+n) \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right )+a^3 d^3 (2+n) (4+n) (B (9+4 n)+A (11+4 n)) \sin (e+f x)\right ) \, dx}{d^3 (2+n) (3+n) (4+n)} \\ & = -\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\left (a^3 (B (9+4 n)+A (11+4 n))\right ) \int (d \sin (e+f x))^{1+n} \, dx}{d (3+n)}+\frac {\left (a^3 \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right )\right ) \int (d \sin (e+f x))^n \, dx}{(2+n) (4+n)} \\ & = -\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}+\frac {a^3 \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right ) \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}}{d f (1+n) (2+n) (4+n) \sqrt {\cos ^2(e+f x)}}+\frac {a^3 (B (9+4 n)+A (11+4 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}}{d^2 f (2+n) (3+n) \sqrt {\cos ^2(e+f x)}}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)} \\ \end{align*}
Time = 1.52 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.66 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {a^3 \cos (e+f x) \sin (e+f x) (d \sin (e+f x))^n \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right )}{1+n}+\sin (e+f x) \left (\frac {(3 A+B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right )}{2+n}+\sin (e+f x) \left (\frac {3 (A+B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(e+f x)\right )}{3+n}+\sin (e+f x) \left (\frac {(A+3 B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(e+f x)\right )}{4+n}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+n}{2},\frac {7+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)}{5+n}\right )\right )\right )\right )}{f \sqrt {\cos ^2(e+f x)}} \]
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\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{3} \left (A +B \sin \left (f x +e \right )\right )d x\]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3 \,d x \]
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