Integrand size = 21, antiderivative size = 349 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\frac {\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \left (1-4 n^2\right ) (1-\cos (c+d x))^2}-\frac {a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac {a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d \left (3-8 n+4 n^2\right ) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}+\frac {n \left (7-3 n-n^2\right ) \cos (c+d x) \left (\frac {1+\cos (c+d x)}{1-\cos (c+d x)}\right )^{-\frac {1}{2}-n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-n,1-n,2-n,-\frac {2 \cos (c+d x)}{1-\cos (c+d x)}\right ) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1-n) (1+2 n) (1-\cos (c+d x))^2} \]
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Time = 0.69 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3961, 2962, 136, 160, 12, 134} \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a^4 \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d (3-2 n) (a-a \cos (c+d x))^2 (a \cos (c+d x)+a)^2}-\frac {a^3 (4-n) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d \left (4 n^2-8 n+3\right ) (a-a \cos (c+d x))^2 (a \cos (c+d x)+a)}+\frac {n \left (-n^2-3 n+7\right ) \sin (c+d x) \cos (c+d x) \left (\frac {\cos (c+d x)+1}{1-\cos (c+d x)}\right )^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (-n-\frac {1}{2},1-n,2-n,-\frac {2 \cos (c+d x)}{1-\cos (c+d x)}\right )}{d (1-2 n) (3-2 n) (1-n) (2 n+1) (1-\cos (c+d x))^2}+\frac {\left (n^2-n+2\right ) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d (3-2 n) \left (1-4 n^2\right ) (1-\cos (c+d x))^2} \]
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Rule 12
Rule 134
Rule 136
Rule 160
Rule 2962
Rule 3961
Rubi steps \begin{align*} \text {integral}& = \left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n \csc ^4(c+d x) \, dx \\ & = -\frac {\left (a^6 (-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (-a-a x)^{-\frac {5}{2}+n}}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {-a+a \cos (c+d x)}} \\ & = -\frac {a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac {\left (a^3 (-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (-a-a x)^{-\frac {3}{2}+n} \left (-a^2 (2-n)+2 a^2 x\right )}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d (3-2 n) \sqrt {-a+a \cos (c+d x)}} \\ & = -\frac {a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac {a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}-\frac {\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (-a-a x)^{-\frac {1}{2}+n} \left (-a^4 \left (2-n^2\right )-a^4 (4-n) x\right )}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d (1-2 n) (3-2 n) \sqrt {-a+a \cos (c+d x)}} \\ & = \frac {\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1+2 n) (1-\cos (c+d x))^2}-\frac {a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac {a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}-\frac {\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \text {Subst}\left (\int \frac {a^6 n \left (7-3 n-n^2\right ) (-x)^{-n} (-a-a x)^{\frac {1}{2}+n}}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{a^3 d (1-2 n) (3-2 n) (1+2 n) \sqrt {-a+a \cos (c+d x)}} \\ & = \frac {\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1+2 n) (1-\cos (c+d x))^2}-\frac {a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac {a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}-\frac {\left (a^3 n \left (7-3 n-n^2\right ) (-\cos (c+d x))^n (-a-a \cos (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (-a-a x)^{\frac {1}{2}+n}}{(-a+a x)^{5/2}} \, dx,x,\cos (c+d x)\right )}{d (1-2 n) (3-2 n) (1+2 n) \sqrt {-a+a \cos (c+d x)}} \\ & = \frac {\left (2-n+n^2\right ) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1+2 n) (1-\cos (c+d x))^2}-\frac {a^4 \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))^2}-\frac {a^3 (4-n) \cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (a-a \cos (c+d x))^2 (a+a \cos (c+d x))}+\frac {n \left (7-3 n-n^2\right ) \cos (c+d x) \left (\frac {1+\cos (c+d x)}{1-\cos (c+d x)}\right )^{-\frac {1}{2}-n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-n,1-n,2-n,-\frac {2 \cos (c+d x)}{1-\cos (c+d x)}\right ) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) (3-2 n) (1-n) (1+2 n) (1-\cos (c+d x))^2} \\ \end{align*}
Time = 4.57 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\frac {(a (1+\sec (c+d x)))^n \left (-2 \cot ^2\left (\frac {1}{2} (c+d x)\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},n,\frac {1}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^n (1+\sec (c+d x))^{-n} \left (3\ 2^n \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n+2 (1+\sec (c+d x))^n+n (1+\sec (c+d x))^n\right )+\frac {-\cos (c+d x) (4 n \cos (c+d x)+(-3+n) (3+\cos (2 (c+d x)))) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+24 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^n (1+\sec (c+d x))^{-n} \left (-3 2^n \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n-2 (1+\sec (c+d x))^n+n \left (2^{1+n} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n+(1+\sec (c+d x))^n\right )\right )}{4 (-3+2 n)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 d} \]
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\[\int \csc \left (d x +c \right )^{4} \left (a +a \sec \left (d x +c \right )\right )^{n}d x\]
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\[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
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Timed out. \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\text {Timed out} \]
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\[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
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\[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
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Timed out. \[ \int \csc ^4(c+d x) (a+a \sec (c+d x))^n \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^4} \,d x \]
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