Integrand size = 35, antiderivative size = 351 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {2^{\frac {1}{2}+m} \left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)} \]
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Time = 0.67 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3062, 3047, 3102, 2830, 2731, 2730} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=-\frac {2^{m+\frac {1}{2}} \cos (e+f x) \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac {\cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1) (m+2) (m+3)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{a f (m+2) (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)} \]
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Rule 2730
Rule 2731
Rule 2830
Rule 3047
Rule 3062
Rule 3102
Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\int (a+a \sin (e+f x))^m (c+d \sin (e+f x)) (a (A c (3+m)+B (2 d+c m))+a (A d (3+m)+B (2 c+d m)) \sin (e+f x)) \, dx}{a (3+m)} \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\int (a+a \sin (e+f x))^m \left (a c (A c (3+m)+B (2 d+c m))+(a d (A c (3+m)+B (2 d+c m))+a c (A d (3+m)+B (2 c+d m))) \sin (e+f x)+a d (A d (3+m)+B (2 c+d m)) \sin ^2(e+f x)\right ) \, dx}{a (3+m)} \\ & = -\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\int (a+a \sin (e+f x))^m \left (a^2 (c (2+m) (A c (3+m)+B (2 d+c m))+d (1+m) (A d (3+m)+B (2 c+d m)))-a^2 \left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \sin (e+f x)\right ) \, dx}{a^2 (2+m) (3+m)} \\ & = \frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)} \\ & = \frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\left (\left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)} \\ & = \frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {2^{\frac {1}{2}+m} \left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)} \\ \end{align*}
Result contains complex when optimal does not.
Time = 24.35 (sec) , antiderivative size = 854, normalized size of antiderivative = 2.43 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {i (a (1+\sin (e+f x)))^m (1-i \cos (e+f x)+\sin (e+f x))^{-2 m} \left (\frac {8 A c^2 \operatorname {Hypergeometric2F1}(-2 m,-m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}+\frac {8 B c d \operatorname {Hypergeometric2F1}(-2 m,-m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}+\frac {4 A d^2 \operatorname {Hypergeometric2F1}(-2 m,-m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}+\frac {4 B c^2 \operatorname {Hypergeometric2F1}(1-m,-2 m,2-m,i \cos (e+f x)-\sin (e+f x)) (-i \cos (e+f x)+\sin (e+f x))}{-1+m}+\frac {8 A c d \operatorname {Hypergeometric2F1}(1-m,-2 m,2-m,i \cos (e+f x)-\sin (e+f x)) (-i \cos (e+f x)+\sin (e+f x))}{-1+m}+\frac {3 B d^2 \operatorname {Hypergeometric2F1}(1-m,-2 m,2-m,i \cos (e+f x)-\sin (e+f x)) (-i \cos (e+f x)+\sin (e+f x))}{-1+m}+\frac {4 B c^2 \operatorname {Hypergeometric2F1}(-1-m,-2 m,-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (e+f x)+\sin (e+f x))}{1+m}+\frac {8 A c d \operatorname {Hypergeometric2F1}(-1-m,-2 m,-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (e+f x)+\sin (e+f x))}{1+m}+\frac {3 B d^2 \operatorname {Hypergeometric2F1}(-1-m,-2 m,-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (e+f x)+\sin (e+f x))}{1+m}-\frac {2 d (2 B c+A d) \operatorname {Hypergeometric2F1}(-2-m,-2 m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}-\frac {4 B c d \operatorname {Hypergeometric2F1}(2-m,-2 m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}-\frac {2 A d^2 \operatorname {Hypergeometric2F1}(2-m,-2 m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}-\frac {i B d^2 \operatorname {Hypergeometric2F1}(-3-m,-2 m,-2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))-i \sin (3 (e+f x)))}{3+m}+\frac {i B d^2 \operatorname {Hypergeometric2F1}(3-m,-2 m,4-m,i \cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))}{-3+m}\right )}{8 f} \]
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\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{2}d x\]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]
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