Integrand size = 19, antiderivative size = 65 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a x}{8}-\frac {b \cos ^5(c+d x)}{5 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \]
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Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2748, 2715, 8} \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8}-\frac {b \cos ^5(c+d x)}{5 d} \]
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Rule 8
Rule 2715
Rule 2748
Rubi steps \begin{align*} \text {integral}& = -\frac {b \cos ^5(c+d x)}{5 d}+a \int \cos ^4(c+d x) \, dx \\ & = -\frac {b \cos ^5(c+d x)}{5 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} (3 a) \int \cos ^2(c+d x) \, dx \\ & = -\frac {b \cos ^5(c+d x)}{5 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} (3 a) \int 1 \, dx \\ & = \frac {3 a x}{8}-\frac {b \cos ^5(c+d x)}{5 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a (c+d x)}{8 d}-\frac {b \cos ^5(c+d x)}{5 d}+\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]
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Time = 1.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {-\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(52\) |
default | \(\frac {-\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(52\) |
parallelrisch | \(\frac {60 a x d -10 b \cos \left (3 d x +3 c \right )-20 \cos \left (d x +c \right ) b -2 b \cos \left (5 d x +5 c \right )+5 a \sin \left (4 d x +4 c \right )+40 a \sin \left (2 d x +2 c \right )-32 b}{160 d}\) | \(72\) |
risch | \(\frac {3 a x}{8}-\frac {b \cos \left (d x +c \right )}{8 d}-\frac {b \cos \left (5 d x +5 c \right )}{80 d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}-\frac {b \cos \left (3 d x +3 c \right )}{16 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(78\) |
norman | \(\frac {\frac {3 a x}{8}-\frac {2 b}{5 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {15 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {4 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(203\) |
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Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {8 \, b \cos \left (d x + c\right )^{5} - 15 \, a d x - 5 \, {\left (2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (60) = 120\).
Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.91 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {b \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {32 \, b \cos \left (d x + c\right )^{5} - 5 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{160 \, d} \]
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Time = 0.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3}{8} \, a x - \frac {b \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {b \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac {b \cos \left (d x + c\right )}{8 \, d} + \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]
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Time = 8.38 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.71 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3\,a\,x}{8}-\frac {\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {2\,b}{5}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]
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