Integrand size = 26, antiderivative size = 165 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {4 a b^3 \cos ^3(c+d x)}{3 d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}+\frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {b^4 \sin ^5(c+d x)}{5 d} \]
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Time = 0.21 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3169, 2713, 2645, 30, 2644, 14} \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}-\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}-\frac {4 a b^3 \cos ^3(c+d x)}{3 d}+\frac {b^4 \sin ^5(c+d x)}{5 d} \]
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Rule 14
Rule 30
Rule 2644
Rule 2645
Rule 2713
Rule 3169
Rubi steps \begin{align*} \text {integral}& = \int \left (a^4 \cos ^5(c+d x)+4 a^3 b \cos ^4(c+d x) \sin (c+d x)+6 a^2 b^2 \cos ^3(c+d x) \sin ^2(c+d x)+4 a b^3 \cos ^2(c+d x) \sin ^3(c+d x)+b^4 \cos (c+d x) \sin ^4(c+d x)\right ) \, dx \\ & = a^4 \int \cos ^5(c+d x) \, dx+\left (4 a^3 b\right ) \int \cos ^4(c+d x) \sin (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \cos ^3(c+d x) \sin ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+b^4 \int \cos (c+d x) \sin ^4(c+d x) \, dx \\ & = -\frac {a^4 \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {\left (4 a^3 b\right ) \text {Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (6 a^2 b^2\right ) \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (4 a b^3\right ) \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {b^4 \text {Subst}\left (\int x^4 \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {4 a^3 b \cos ^5(c+d x)}{5 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^5(c+d x)}{5 d}+\frac {b^4 \sin ^5(c+d x)}{5 d}+\frac {\left (6 a^2 b^2\right ) \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (4 a b^3\right ) \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {4 a b^3 \cos ^3(c+d x)}{3 d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}+\frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {b^4 \sin ^5(c+d x)}{5 d} \\ \end{align*}
Time = 1.46 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {-12 a^3 b \cos ^5(c+d x)+15 a^4 \sin (c+d x)-10 a^2 \left (a^2-3 b^2\right ) \sin ^3(c+d x)+3 \left (a^4-6 a^2 b^2+b^4\right ) \sin ^5(c+d x)+4 a b^3 \cos (c+d x) \left (-2+\frac {2}{\sqrt {\cos ^2(c+d x)}}-\sin ^2(c+d x)+3 \sin ^4(c+d x)\right )}{15 d} \]
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Time = 1.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.79
method | result | size |
parts | \(\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {b^{4} \sin \left (d x +c \right )^{5}}{5 d}+\frac {4 a \,b^{3} \left (\frac {\cos \left (d x +c \right )^{5}}{5}-\frac {\cos \left (d x +c \right )^{3}}{3}\right )}{d}-\frac {4 a^{3} b \cos \left (d x +c \right )^{5}}{5 d}+\frac {6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}\) | \(131\) |
derivativedivides | \(\frac {\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {4 a^{3} b \cos \left (d x +c \right )^{5}}{5}+6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+4 a \,b^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+\frac {b^{4} \sin \left (d x +c \right )^{5}}{5}}{d}\) | \(142\) |
default | \(\frac {\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {4 a^{3} b \cos \left (d x +c \right )^{5}}{5}+6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+4 a \,b^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+\frac {b^{4} \sin \left (d x +c \right )^{5}}{5}}{d}\) | \(142\) |
parallelrisch | \(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a^{4}-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{3} b +\frac {8 \left (a^{4}+6 a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3}-16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a \,b^{3}+\frac {4 \left (29 a^{4}-24 a^{2} b^{2}+24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15}+\frac {16 \left (-3 a^{3} b +a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3}+\frac {8 \left (a^{4}+6 a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{3}}{3}+2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 a^{3} b}{5}-\frac {16 a \,b^{3}}{15}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(222\) |
norman | \(\frac {-\frac {24 a^{3} b +16 a \,b^{3}}{15 d}+\frac {2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {16 a \,b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {16 a \,b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {8 a^{3} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {4 \left (29 a^{4}-24 a^{2} b^{2}+24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {2 \left (24 a^{3} b -8 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {8 a^{2} \left (a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 a^{2} \left (a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(252\) |
risch | \(-\frac {a^{3} b \cos \left (d x +c \right )}{2 d}-\frac {a \,b^{3} \cos \left (d x +c \right )}{2 d}+\frac {5 a^{4} \sin \left (d x +c \right )}{8 d}+\frac {3 a^{2} b^{2} \sin \left (d x +c \right )}{4 d}+\frac {b^{4} \sin \left (d x +c \right )}{8 d}-\frac {a^{3} b \cos \left (5 d x +5 c \right )}{20 d}+\frac {a \,b^{3} \cos \left (5 d x +5 c \right )}{20 d}+\frac {\sin \left (5 d x +5 c \right ) a^{4}}{80 d}-\frac {3 \sin \left (5 d x +5 c \right ) a^{2} b^{2}}{40 d}+\frac {\sin \left (5 d x +5 c \right ) b^{4}}{80 d}-\frac {a^{3} b \cos \left (3 d x +3 c \right )}{4 d}-\frac {a \,b^{3} \cos \left (3 d x +3 c \right )}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) a^{4}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) a^{2} b^{2}}{8 d}-\frac {\sin \left (3 d x +3 c \right ) b^{4}}{16 d}\) | \(257\) |
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Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {20 \, a b^{3} \cos \left (d x + c\right )^{3} + 12 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 12 \, a^{2} b^{2} + 3 \, b^{4} + 2 \, {\left (2 \, a^{4} + 3 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]
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Time = 0.26 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.25 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\begin {cases} \frac {8 a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {4 a^{3} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {4 a^{2} b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {4 a b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {8 a b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{4} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {12 \, a^{3} b \cos \left (d x + c\right )^{5} - 3 \, b^{4} \sin \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} + 6 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} b^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b^{3}}{15 \, d} \]
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Time = 0.40 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {{\left (a^{3} b - a b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{20 \, d} - \frac {{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )}{2 \, d} + \frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (5 \, a^{4} - 6 \, a^{2} b^{2} - 3 \, b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \]
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Time = 22.84 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.24 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^4\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^4\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^4-6\,a^3\,b\,{\cos \left (c+d\,x\right )}^5-9\,\sin \left (c+d\,x\right )\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^4+3\,\sin \left (c+d\,x\right )\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^2+6\,\sin \left (c+d\,x\right )\,a^2\,b^2+6\,a\,b^3\,{\cos \left (c+d\,x\right )}^5-10\,a\,b^3\,{\cos \left (c+d\,x\right )}^3+\frac {3\,\sin \left (c+d\,x\right )\,b^4\,{\cos \left (c+d\,x\right )}^4}{2}-3\,\sin \left (c+d\,x\right )\,b^4\,{\cos \left (c+d\,x\right )}^2+\frac {3\,\sin \left (c+d\,x\right )\,b^4}{2}\right )}{15\,d} \]
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