Integrand size = 23, antiderivative size = 249 \[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {b^3 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)} \]
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Time = 0.45 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3957, 2991, 2722, 2644, 371, 2657} \[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\frac {a^3 \cos (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {b^3 (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)} \]
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Rule 371
Rule 2644
Rule 2657
Rule 2722
Rule 2991
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int (-b-a \cos (c+d x))^3 \sec ^3(c+d x) (e \sin (c+d x))^m \, dx \\ & = -\int \left (-a^3 (e \sin (c+d x))^m-3 a^2 b \sec (c+d x) (e \sin (c+d x))^m-3 a b^2 \sec ^2(c+d x) (e \sin (c+d x))^m-b^3 \sec ^3(c+d x) (e \sin (c+d x))^m\right ) \, dx \\ & = a^3 \int (e \sin (c+d x))^m \, dx+\left (3 a^2 b\right ) \int \sec (c+d x) (e \sin (c+d x))^m \, dx+\left (3 a b^2\right ) \int \sec ^2(c+d x) (e \sin (c+d x))^m \, dx+b^3 \int \sec ^3(c+d x) (e \sin (c+d x))^m \, dx \\ & = \frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {\left (3 a^2 b\right ) \text {Subst}\left (\int \frac {x^m}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac {b^3 \text {Subst}\left (\int \frac {x^m}{\left (1-\frac {x^2}{e^2}\right )^2} \, dx,x,e \sin (c+d x)\right )}{d e} \\ & = \frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^2 b \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {b^3 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {3 a b^2 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.73 \[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\frac {\left (a^3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )+b \left (3 a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )+b \left (3 a \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )+b \cos (c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )\right )\right )\right ) (e \sin (c+d x))^m \tan (c+d x)}{d (1+m)} \]
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\[\int \left (a +b \sec \left (d x +c \right )\right )^{3} \left (e \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
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\[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\int \left (e \sin {\left (c + d x \right )}\right )^{m} \left (a + b \sec {\left (c + d x \right )}\right )^{3}\, dx \]
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\[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
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\[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^m\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3 \,d x \]
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