Integrand size = 21, antiderivative size = 178 \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\frac {3}{8} \left (a^2-4 b^2\right ) x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d} \]
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Time = 0.66 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2973, 3128, 3112, 3102, 2814, 3855} \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (12 a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{12 a b d}-\frac {\left (39 a^2+2 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {3}{8} x \left (a^2-4 b^2\right )+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}+\frac {\sin (c+d x) (a \cos (c+d x)+b)^3}{4 a d}+\frac {\tan (c+d x) (a \cos (c+d x)+b)^3}{b d} \]
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Rule 2814
Rule 2973
Rule 3102
Rule 3112
Rule 3128
Rule 3855
Rule 3957
Rubi steps \begin{align*} \text {integral}& = \int (-b-a \cos (c+d x))^2 \sin ^2(c+d x) \tan ^2(c+d x) \, dx \\ & = \frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int (-b-a \cos (c+d x))^2 \left (-8 a^2+5 a b \cos (c+d x)+\left (12 a^2+b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{4 a b} \\ & = -\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int (-b-a \cos (c+d x)) \left (24 a^2 b-17 a b^2 \cos (c+d x)-b \left (39 a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a b} \\ & = -\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int \left (-48 a^2 b^2-9 a b \left (a^2-4 b^2\right ) \cos (c+d x)+4 b^2 \left (28 a^2+b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a b} \\ & = -\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int \left (-48 a^2 b^2-9 a b \left (a^2-4 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a b} \\ & = \frac {3}{8} \left (a^2-4 b^2\right ) x-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}+(2 a b) \int \sec (c+d x) \, dx \\ & = \frac {3}{8} \left (a^2-4 b^2\right ) x+\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d} \\ \end{align*}
Time = 1.46 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.88 \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\frac {12 \left (3 \left (a^2-4 b^2\right ) (c+d x)-16 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+16 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-208 a b \sin (c+d x)+\left (-6 \left (3 a^2-4 b^2\right ) \cos (2 (c+d x))+16 a b \cos (3 (c+d x))+3 \left (-7 a^2+40 b^2+a^2 \cos (4 (c+d x))\right )\right ) \tan (c+d x)}{96 d} \]
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Time = 1.80 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(133\) |
default | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(133\) |
parts | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(138\) |
parallelrisch | \(\frac {72 a^{2} x d \cos \left (d x +c \right )-288 b^{2} d x \cos \left (d x +c \right )+384 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-384 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-24 a^{2} \sin \left (d x +c \right )+216 b^{2} \sin \left (d x +c \right )-21 a^{2} \sin \left (3 d x +3 c \right )+24 b^{2} \sin \left (3 d x +3 c \right )+3 a^{2} \sin \left (5 d x +5 c \right )-224 a b \sin \left (2 d x +2 c \right )+16 a b \sin \left (4 d x +4 c \right )}{192 d \cos \left (d x +c \right )}\) | \(175\) |
risch | \(\frac {3 a^{2} x}{8}-\frac {3 b^{2} x}{2}+\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {5 i a b \,{\mathrm e}^{i \left (d x +c \right )}}{4 d}-\frac {5 i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{4 d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {a b \sin \left (3 d x +3 c \right )}{6 d}\) | \(215\) |
norman | \(\frac {\left (-\frac {3 a^{2}}{8}+\frac {3 b^{2}}{2}\right ) x +\left (-\frac {9 a^{2}}{8}+\frac {9 b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {3 a^{2}}{4}+3 b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3 a^{2}}{4}-3 b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 a^{2}}{8}-\frac {3 b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {9 a^{2}}{8}-\frac {9 b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {\left (11 a^{2}+20 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {2 \left (3 a^{2}-20 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {\left (3 a^{2}-16 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {\left (3 a^{2}+16 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 \left (3 a^{2}+20 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(354\) |
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Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.80 \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\frac {9 \, {\left (a^{2} - 4 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 24 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 24 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a b \cos \left (d x + c\right )^{3} - 64 \, a b \cos \left (d x + c\right ) - 3 \, {\left (5 \, a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, b^{2}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]
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\[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{4}{\left (c + d x \right )}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.70 \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 32 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b - 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{2}}{96 \, d} \]
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Time = 0.37 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.60 \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\frac {48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, {\left (a^{2} - 4 \, b^{2}\right )} {\left (d x + c\right )} - \frac {48 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 208 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 33 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 208 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
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Time = 14.82 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.16 \[ \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx=\frac {3\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}-\frac {3\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {8\,a\,b\,\sin \left (c+d\,x\right )}{3\,d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {5\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]
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