Integrand size = 27, antiderivative size = 598 \[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {2 b (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 (1+m) (3+m)}+\frac {a (e \cos (c+d x))^{-3-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (3+m)}+\frac {a (3 b+a (2+m)) (e \cos (c+d x))^{-3-m} (1-\sin (c+d x)) (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e (1+m) (3+m)}-\frac {2^{\frac {3}{2}-\frac {m}{2}} a b (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 (a+b) d e^3 (1+m) (3+m)}-\frac {2^{-\frac {1}{2}-\frac {m}{2}} a \left (2 a b-b^2+a^2 (2+m)\right ) (e \cos (c+d x))^{-3-m} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {3+m}{2},\frac {3-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x))^2 \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {3+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^3 d e (1-m) (3+m)} \]
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Time = 0.67 (sec) , antiderivative size = 598, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2779, 2778, 2999, 134, 136, 160, 12} \[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=-\frac {a 2^{-\frac {m}{2}-\frac {1}{2}} \left (a^2 (m+2)+2 a b-b^2\right ) (1-\sin (c+d x))^2 (e \cos (c+d x))^{-m-3} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+3}{2}} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+3}{2},\frac {3-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e (1-m) (m+3) (a-b) (a+b)^3}+\frac {a (\sin (c+d x)+1) (e \cos (c+d x))^{-m-3} (a+b \sin (c+d x))^{m+1}}{d e (m+3) \left (a^2-b^2\right )}-\frac {a b 2^{\frac {3}{2}-\frac {m}{2}} (e \cos (c+d x))^{-m-1} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+1}{2}} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-m-1),\frac {m+1}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e^3 (m+1) (m+3) (a-b)^2 (a+b)}+\frac {2 b (e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1}}{d e^3 (m+1) (m+3) (a-b)^2}+\frac {a (a (m+2)+3 b) (1-\sin (c+d x)) (\sin (c+d x)+1) (e \cos (c+d x))^{-m-3} (a+b \sin (c+d x))^{m+1}}{d e (m+1) (m+3) (a-b) (a+b)^2}-\frac {(e \cos (c+d x))^{-m-3} (a+b \sin (c+d x))^{m+1}}{d e (m+3) (a-b)} \]
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Rule 12
Rule 134
Rule 136
Rule 160
Rule 2778
Rule 2779
Rule 2999
Rubi steps \begin{align*} \text {integral}& = -\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {a \int \frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2}-\frac {(2 b) \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx}{(a-b) e^2 (3+m)} \\ & = -\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {2 b (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 (1+m) (3+m)}-\frac {(2 a b) \int \frac {(e \cos (c+d x))^{-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b)^2 e^4 (3+m)}+\frac {\left (a (e \cos (c+d x))^{-3-m} (1-\sin (c+d x))^{\frac {3+m}{2}} (1+\sin (c+d x))^{\frac {3+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{-1+\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e} \\ & = -\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {2 b (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 (1+m) (3+m)}+\frac {a (e \cos (c+d x))^{-3-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (3+m)}-\frac {\left (2 a b (e \cos (c+d x))^{-1-m} (1-\sin (c+d x))^{\frac {1+m}{2}} (1+\sin (c+d x))^{\frac {1+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{-1+\frac {1}{2} (-1-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b)^2 d e^3 (3+m)}-\frac {\left (a (e \cos (c+d x))^{-3-m} (1-\sin (c+d x))^{\frac {3+m}{2}} (1+\sin (c+d x))^{\frac {3+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-3-m)} (-2 b-a (2+m)-b x) (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) (a+b) d e (3+m)} \\ & = -\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {2 b (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 (1+m) (3+m)}+\frac {a (e \cos (c+d x))^{-3-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (3+m)}+\frac {a (3 b+a (2+m)) (e \cos (c+d x))^{-3-m} (1-\sin (c+d x)) (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e (1+m) (3+m)}-\frac {2^{\frac {3}{2}-\frac {m}{2}} a b (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 (a+b) d e^3 (1+m) (3+m)}+\frac {\left (a (e \cos (c+d x))^{-3-m} (1-\sin (c+d x))^{\frac {3+m}{2}} (1+\sin (c+d x))^{\frac {3+m}{2}}\right ) \text {Subst}\left (\int (1+m) \left (2 a b-b^2+a^2 (2+m)\right ) (1-x)^{1+\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{2 (-a-b) (a-b) (a+b) d e \left (1+\frac {1}{2} (-3-m)\right ) (3+m)} \\ & = -\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {2 b (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 (1+m) (3+m)}+\frac {a (e \cos (c+d x))^{-3-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (3+m)}+\frac {a (3 b+a (2+m)) (e \cos (c+d x))^{-3-m} (1-\sin (c+d x)) (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e (1+m) (3+m)}-\frac {2^{\frac {3}{2}-\frac {m}{2}} a b (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 (a+b) d e^3 (1+m) (3+m)}+\frac {\left (a (1+m) \left (2 a b-b^2+a^2 (2+m)\right ) (e \cos (c+d x))^{-3-m} (1-\sin (c+d x))^{\frac {3+m}{2}} (1+\sin (c+d x))^{\frac {3+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{1+\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{2 (-a-b) (a-b) (a+b) d e \left (1+\frac {1}{2} (-3-m)\right ) (3+m)} \\ & = -\frac {(e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (3+m)}+\frac {2 b (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 (1+m) (3+m)}+\frac {a (e \cos (c+d x))^{-3-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (3+m)}+\frac {a (3 b+a (2+m)) (e \cos (c+d x))^{-3-m} (1-\sin (c+d x)) (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e (1+m) (3+m)}-\frac {2^{\frac {3}{2}-\frac {m}{2}} a b (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b)^2 (a+b) d e^3 (1+m) (3+m)}-\frac {2^{-\frac {1}{2}-\frac {m}{2}} a \left (2 a b-b^2+a^2 (2+m)\right ) (e \cos (c+d x))^{-3-m} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {3+m}{2},\frac {3-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x))^2 \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {3+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^3 d e (1-m) (3+m)} \\ \end{align*}
Time = 6.08 (sec) , antiderivative size = 826, normalized size of antiderivative = 1.38 \[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=\frac {\cos (c+d x) (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d (-3-m)}+\frac {2 b \cos ^{4+m}(c+d x) (e \cos (c+d x))^{-4-m} \left (\frac {\cos ^{-1-m}(c+d x) (a+b \sin (c+d x))^{1+m}}{(a-b) d (-1-m)}+\frac {2^{1+\frac {1}{2} (-1-m)} a \cos ^{-1-m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},1+\frac {1}{2} (-1-m),-\frac {(-a+b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x))^{\frac {1}{2} (-1-m)+\frac {1+m}{2}} (1+\sin (c+d x))^{\frac {1}{2} (-1-m)+\frac {1+m}{2}} \left (-\frac {(-a-b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{(-a-b) (a-b) d (-1-m)}\right )}{(a-b) (-3-m)}+\frac {a \cos (c+d x) (e \cos (c+d x))^{-4-m} (1-\sin (c+d x))^{\frac {3+m}{2}} (1+\sin (c+d x))^{\frac {3+m}{2}} \left (\frac {(1-\sin (c+d x))^{\frac {1}{2} (-3-m)} (1+\sin (c+d x))^{1+\frac {1}{2} (-3-m)} (a+b \sin (c+d x))^{1+m}}{(-a-b) (-3-m)}-\frac {-\frac {(3 b+a (2+m)) (1-\sin (c+d x))^{1+\frac {1}{2} (-3-m)} (1+\sin (c+d x))^{1+\frac {1}{2} (-3-m)} (a+b \sin (c+d x))^{1+m}}{2 (-a-b) \left (1+\frac {1}{2} (-3-m)\right )}-\frac {2^{-1+\frac {1}{2} (-3-m)} (1+m) \left (2 a b-b^2+a^2 (2+m)\right ) \operatorname {Hypergeometric2F1}\left (2+\frac {1}{2} (-3-m),\frac {3+m}{2},3+\frac {1}{2} (-3-m),-\frac {(-a+b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x))^{2+\frac {1}{2} (-3-m)} (1+\sin (c+d x))^{\frac {1}{2} (-3-m)} \left (-\frac {(-a-b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {3+m}{2}} (a+b \sin (c+d x))^{1+m}}{(-a-b)^2 \left (1+\frac {1}{2} (-3-m)\right ) \left (2+\frac {1}{2} (-3-m)\right )}}{(-a-b) (-3-m)}\right )}{(a-b) d} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{-4-m} \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{- m - 4} \left (a + b \sin {\left (c + d x \right )}\right )^{m}\, dx \]
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\[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^{-4-m} (a+b \sin (c+d x))^m \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+4}} \,d x \]
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