Integrand size = 29, antiderivative size = 360 \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (3 a^2-9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^3 d (1+n)}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^3 d (1+n)}-\frac {b^6 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a \left (a^2-b^2\right )^3 d (1+n)}+\frac {(3 a-5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^2 d (1+n)}+\frac {(3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^2 d (1+n)}+\frac {\operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a-b) d (1+n)}+\frac {\operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a+b) d (1+n)} \]
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Time = 0.39 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 975, 66} \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (3 a^2-9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,-\sin (c+d x))}{16 d (n+1) (a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,\sin (c+d x))}{16 d (n+1) (a+b)^3}-\frac {b^6 \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {b \sin (c+d x)}{a}\right )}{a d (n+1) \left (a^2-b^2\right )^3}+\frac {(3 a-5 b) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(2,n+1,n+2,-\sin (c+d x))}{16 d (n+1) (a-b)^2}+\frac {(3 a+5 b) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(2,n+1,n+2,\sin (c+d x))}{16 d (n+1) (a+b)^2}+\frac {\sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(3,n+1,n+2,-\sin (c+d x))}{8 d (n+1) (a-b)}+\frac {\sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(3,n+1,n+2,\sin (c+d x))}{8 d (n+1) (a+b)} \]
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Rule 66
Rule 975
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^5 \text {Subst}\left (\int \left (\frac {\left (\frac {x}{b}\right )^n}{8 b^3 (a+b) (b-x)^3}+\frac {(3 a+5 b) \left (\frac {x}{b}\right )^n}{16 b^4 (a+b)^2 (b-x)^2}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \left (\frac {x}{b}\right )^n}{16 b^5 (a+b)^3 (b-x)}-\frac {\left (\frac {x}{b}\right )^n}{(a-b)^3 (a+b)^3 (a+x)}-\frac {\left (\frac {x}{b}\right )^n}{8 b^3 (-a+b) (b+x)^3}+\frac {(3 a-5 b) \left (\frac {x}{b}\right )^n}{16 (a-b)^2 b^4 (b+x)^2}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \left (\frac {x}{b}\right )^n}{16 (a-b)^3 b^5 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {((3 a-5 b) b) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^2 d}+\frac {b^2 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 (a-b) d}+\frac {b^2 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 (a+b) d}+\frac {(b (3 a+5 b)) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^2 d}-\frac {b^5 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{a+x} \, dx,x,b \sin (c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^3 d}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^3 d} \\ & = \frac {\left (3 a^2-9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^3 d (1+n)}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^3 d (1+n)}-\frac {b^6 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a \left (a^2-b^2\right )^3 d (1+n)}+\frac {(3 a-5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^2 d (1+n)}+\frac {(3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^2 d (1+n)}+\frac {\operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a-b) d (1+n)}+\frac {\operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a+b) d (1+n)} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (\frac {\left (3 a^2-9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x))}{(a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,\sin (c+d x))}{(a+b)^3}-\frac {16 b^6 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{a (a-b)^3 (a+b)^3}+\frac {(3 a-5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x))}{(a-b)^2}+\frac {(3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x))}{(a+b)^2}+\frac {2 \operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x))}{a-b}+\frac {2 \operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x))}{a+b}\right ) \sin ^{1+n}(c+d x)}{16 d (1+n)} \]
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\[\int \frac {\left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
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\[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n}{{\cos \left (c+d\,x\right )}^5\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \,d x \]
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