Integrand size = 21, antiderivative size = 102 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {7 a^2 x}{16}-\frac {7 a^2 \cos ^5(c+d x)}{30 d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d} \]
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Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 2748, 2715, 8} \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {7 a^2 \cos ^5(c+d x)}{30 d}-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}+\frac {7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7 a^2 x}{16} \]
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Rule 8
Rule 2715
Rule 2748
Rule 2757
Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac {1}{6} (7 a) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx \\ & = -\frac {7 a^2 \cos ^5(c+d x)}{30 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac {1}{6} \left (7 a^2\right ) \int \cos ^4(c+d x) \, dx \\ & = -\frac {7 a^2 \cos ^5(c+d x)}{30 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac {1}{8} \left (7 a^2\right ) \int \cos ^2(c+d x) \, dx \\ & = -\frac {7 a^2 \cos ^5(c+d x)}{30 d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac {1}{16} \left (7 a^2\right ) \int 1 \, dx \\ & = \frac {7 a^2 x}{16}-\frac {7 a^2 \cos ^5(c+d x)}{30 d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.48 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \cos ^5(c+d x) \left (-210 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (-96+231 \sin (c+d x)+57 \sin ^2(c+d x)-182 \sin ^3(c+d x)-106 \sin ^4(c+d x)+56 \sin ^5(c+d x)+40 \sin ^6(c+d x)\right )\right )}{240 d (-1+\sin (c+d x))^3 (1+\sin (c+d x))^{5/2}} \]
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Time = 0.66 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(-\frac {\left (-84 d x +\sin \left (6 d x +6 c \right )+48 \cos \left (d x +c \right )+24 \cos \left (3 d x +3 c \right )+\frac {24 \cos \left (5 d x +5 c \right )}{5}-51 \sin \left (2 d x +2 c \right )-3 \sin \left (4 d x +4 c \right )+\frac {384}{5}\right ) a^{2}}{192 d}\) | \(76\) |
risch | \(\frac {7 a^{2} x}{16}-\frac {a^{2} \cos \left (d x +c \right )}{4 d}-\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{64 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{8 d}+\frac {17 a^{2} \sin \left (2 d x +2 c \right )}{64 d}\) | \(107\) |
derivativedivides | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(109\) |
default | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(109\) |
norman | \(\frac {\frac {7 a^{2} x}{16}-\frac {4 a^{2}}{5 d}+\frac {9 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {89 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {11 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {11 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {89 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {9 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {21 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {105 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {35 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {105 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {21 a^{2} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {7 a^{2} x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(341\) |
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Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {96 \, a^{2} \cos \left (d x + c\right )^{5} - 105 \, a^{2} d x + 5 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 14 \, a^{2} \cos \left (d x + c\right )^{3} - 21 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (95) = 190\).
Time = 0.39 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.81 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {384 \, a^{2} \cos \left (d x + c\right )^{5} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{960 \, d} \]
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Time = 0.36 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {7}{16} \, a^{2} x - \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{4 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {17 \, a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]
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Time = 8.61 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.42 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {7\,a^2\,x}{16}-\frac {\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {a^2\,\left (105\,c+105\,d\,x\right )}{240}-\frac {a^2\,\left (105\,c+105\,d\,x-192\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^2\,\left (630\,c+630\,d\,x-192\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^2\,\left (630\,c+630\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^2\,\left (1575\,c+1575\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^2\,\left (1575\,c+1575\,d\,x-1920\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{12}-\frac {a^2\,\left (2100\,c+2100\,d\,x-1920\right )}{240}\right )-\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]
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