Integrand size = 31, antiderivative size = 88 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a^2 B x+\frac {a^2 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d} \]
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Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3054, 3047, 3100, 2814, 3855} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^2 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+a^2 B x \]
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Rule 2814
Rule 3047
Rule 3054
Rule 3100
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+a \cos (c+d x)) (a (3 A+2 B)+2 a B \cos (c+d x)) \sec ^2(c+d x) \, dx \\ & = \frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (a^2 (3 A+2 B)+\left (2 a^2 B+a^2 (3 A+2 B)\right ) \cos (c+d x)+2 a^2 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \left (a^2 (3 A+4 B)+2 a^2 B \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = a^2 B x+\frac {a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^2 (3 A+4 B)\right ) \int \sec (c+d x) \, dx \\ & = a^2 B x+\frac {a^2 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (3 A+2 B) \tan (c+d x)}{2 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}
Time = 0.54 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a^2 (2 B d x+(3 A+4 B) \text {arctanh}(\sin (c+d x))+(4 A+2 B+A \sec (c+d x)) \tan (c+d x))}{2 d} \]
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Time = 3.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20
method | result | size |
parts | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (d x +c \right )}{d}\) | \(106\) |
derivativedivides | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )}{d}\) | \(114\) |
default | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )}{d}\) | \(114\) |
parallelrisch | \(-\frac {3 \left (\left (A +\frac {4 B}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +\frac {4 B}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 d x B \cos \left (2 d x +2 c \right )}{3}+\frac {2 \left (-B -2 A \right ) \sin \left (2 d x +2 c \right )}{3}-\frac {2 d x B}{3}-\frac {2 A \sin \left (d x +c \right )}{3}\right ) a^{2}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(128\) |
risch | \(a^{2} B x -\frac {i a^{2} \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-4 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\) | \(172\) |
norman | \(\frac {a^{2} B x +a^{2} B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} B x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} B x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (5 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {6 A \,a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-2 a^{2} B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{2} B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 a^{2} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (3 A +2 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2} \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(296\) |
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Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, B a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
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\[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
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Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.61 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} B a^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \tan \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right )}{4 \, d} \]
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Time = 0.31 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} B a^{2} + {\left (3 \, A a^{2} + 4 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} + 4 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
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Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.84 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]
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