Integrand size = 19, antiderivative size = 127 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5 a x}{16}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d} \]
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Time = 0.18 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3957, 2917, 2672, 308, 212, 2715, 8} \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {5 a \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16} \]
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Rule 8
Rule 212
Rule 308
Rule 2672
Rule 2715
Rule 2917
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int (-a-a \cos (c+d x)) \sin ^5(c+d x) \tan (c+d x) \, dx \\ & = a \int \sin ^6(c+d x) \, dx+a \int \sin ^5(c+d x) \tan (c+d x) \, dx \\ & = -\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{6} (5 a) \int \sin ^4(c+d x) \, dx+\frac {a \text {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{8} (5 a) \int \sin ^2(c+d x) \, dx+\frac {a \text {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {a \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{16} (5 a) \int 1 \, dx+\frac {a \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {5 a x}{16}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d} \\ \end{align*}
Time = 0.67 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.68 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {a \left (960 \text {arctanh}(\sin (c+d x))-960 \sin (c+d x)-320 \sin ^3(c+d x)-192 \sin ^5(c+d x)+5 (60 c+60 d x-45 \sin (2 (c+d x))+9 \sin (4 (c+d x))-\sin (6 (c+d x)))\right )}{960 d} \]
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Time = 1.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(96\) |
default | \(\frac {a \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(96\) |
parts | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {a \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(98\) |
parallelrisch | \(-\frac {a \left (-60 d x +\sin \left (6 d x +6 c \right )+192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+264 \sin \left (d x +c \right )+45 \sin \left (2 d x +2 c \right )-28 \sin \left (3 d x +3 c \right )-9 \sin \left (4 d x +4 c \right )+\frac {12 \sin \left (5 d x +5 c \right )}{5}\right )}{192 d}\) | \(101\) |
risch | \(\frac {5 a x}{16}+\frac {11 i a \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {11 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \sin \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 a \sin \left (3 d x +3 c \right )}{48 d}-\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\) | \(150\) |
norman | \(\frac {\frac {5 a x}{16}-\frac {21 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {389 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {853 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}-\frac {523 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}-\frac {73 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8 d}-\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {75 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {25 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {75 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {5 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(248\) |
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Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {75 \, a d x + 120 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 120 \, a \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a \cos \left (d x + c\right )^{5} + 48 \, a \cos \left (d x + c\right )^{4} - 130 \, a \cos \left (d x + c\right )^{3} - 176 \, a \cos \left (d x + c\right )^{2} + 165 \, a \cos \left (d x + c\right ) + 368 \, a\right )} \sin \left (d x + c\right )}{240 \, d} \]
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\[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=a \left (\int \sin ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{6}{\left (c + d x \right )}\, dx\right ) \]
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Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=-\frac {32 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{960 \, d} \]
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Time = 0.31 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.15 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {75 \, {\left (d x + c\right )} a + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1095 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 3138 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5118 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1945 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
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Time = 13.66 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5\,a\,x}{16}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {15\,a\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {7\,a\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {3\,a\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}-\frac {a\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}-\frac {a\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}-\frac {11\,a\,\sin \left (c+d\,x\right )}{8\,d} \]
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