Integrand size = 15, antiderivative size = 96 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {B \text {arctanh}(\sin (x))}{a^4}+\frac {(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac {2 (3 A-80 B) \sin (x)}{105 a^4 (1+\cos (x))}+\frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3} \]
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Time = 0.46 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2907, 3057, 12, 3855} \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {2 (3 A-80 B) \sin (x)}{105 a^4 (\cos (x)+1)}+\frac {(6 A-55 B) \sin (x)}{105 a^4 (\cos (x)+1)^2}+\frac {B \text {arctanh}(\sin (x))}{a^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a \cos (x)+a)^3}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4} \]
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Rule 12
Rule 2907
Rule 3057
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \int \frac {(B+A \cos (x)) \sec (x)}{(a+a \cos (x))^4} \, dx \\ & = \frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {\int \frac {(7 a B+3 a (A-B) \cos (x)) \sec (x)}{(a+a \cos (x))^3} \, dx}{7 a^2} \\ & = \frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac {\int \frac {\left (35 a^2 B+2 a^2 (3 A-10 B) \cos (x)\right ) \sec (x)}{(a+a \cos (x))^2} \, dx}{35 a^4} \\ & = \frac {(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac {\int \frac {\left (105 a^3 B+a^3 (6 A-55 B) \cos (x)\right ) \sec (x)}{a+a \cos (x)} \, dx}{105 a^6} \\ & = \frac {(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac {2 (3 A-80 B) \sin (x)}{105 \left (a^4+a^4 \cos (x)\right )}+\frac {\int 105 a^4 B \sec (x) \, dx}{105 a^8} \\ & = \frac {(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac {2 (3 A-80 B) \sin (x)}{105 \left (a^4+a^4 \cos (x)\right )}+\frac {B \int \sec (x) \, dx}{a^4} \\ & = \frac {B \text {arctanh}(\sin (x))}{a^4}+\frac {(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac {2 (3 A-80 B) \sin (x)}{105 \left (a^4+a^4 \cos (x)\right )} \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {-3360 B \cos ^8\left (\frac {x}{2}\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+(96 A-1055 B+(87 A-1480 B) \cos (x)+(24 A-535 B) \cos (2 x)+3 A \cos (3 x)-80 B \cos (3 x)) \sin (x)}{210 a^4 (1+\cos (x))^4} \]
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Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \(\frac {-56 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+56 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+\tan \left (\frac {x}{2}\right ) \left (\left (A -B \right ) \tan \left (\frac {x}{2}\right )^{6}+7 \left (\frac {3 A}{5}-B \right ) \tan \left (\frac {x}{2}\right )^{4}+7 \left (A -\frac {11 B}{3}\right ) \tan \left (\frac {x}{2}\right )^{2}+7 A -105 B \right )}{56 a^{4}}\) | \(79\) |
default | \(\frac {-8 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+A \tan \left (\frac {x}{2}\right )-15 B \tan \left (\frac {x}{2}\right )+8 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+A \tan \left (\frac {x}{2}\right )^{3}-\frac {11 B \tan \left (\frac {x}{2}\right )^{3}}{3}+\frac {3 A \tan \left (\frac {x}{2}\right )^{5}}{5}-B \tan \left (\frac {x}{2}\right )^{5}+\frac {\tan \left (\frac {x}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {x}{2}\right )^{7} B}{7}}{8 a^{4}}\) | \(93\) |
norman | \(\frac {\frac {\left (A -15 B \right ) \tan \left (\frac {x}{2}\right )}{8 a}+\frac {\left (A -B \right ) \tan \left (\frac {x}{2}\right )^{7}}{56 a}+\frac {\left (3 A -11 B \right ) \tan \left (\frac {x}{2}\right )^{3}}{24 a}+\frac {\left (3 A -5 B \right ) \tan \left (\frac {x}{2}\right )^{5}}{40 a}}{a^{3}}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a^{4}}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a^{4}}\) | \(98\) |
risch | \(-\frac {2 i \left (105 B \,{\mathrm e}^{6 i x}+735 B \,{\mathrm e}^{5 i x}+2170 B \,{\mathrm e}^{4 i x}-210 A \,{\mathrm e}^{3 i x}+3430 B \,{\mathrm e}^{3 i x}-126 A \,{\mathrm e}^{2 i x}+2625 B \,{\mathrm e}^{2 i x}-42 A \,{\mathrm e}^{i x}+1015 B \,{\mathrm e}^{i x}-6 A +160 B \right )}{105 \left ({\mathrm e}^{i x}+1\right )^{7} a^{4}}+\frac {B \ln \left (i+{\mathrm e}^{i x}\right )}{a^{4}}-\frac {B \ln \left ({\mathrm e}^{i x}-i\right )}{a^{4}}\) | \(125\) |
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Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {105 \, {\left (B \cos \left (x\right )^{4} + 4 \, B \cos \left (x\right )^{3} + 6 \, B \cos \left (x\right )^{2} + 4 \, B \cos \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - 105 \, {\left (B \cos \left (x\right )^{4} + 4 \, B \cos \left (x\right )^{3} + 6 \, B \cos \left (x\right )^{2} + 4 \, B \cos \left (x\right ) + B\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B\right )} \cos \left (x\right )^{3} + {\left (24 \, A - 535 \, B\right )} \cos \left (x\right )^{2} + {\left (39 \, A - 620 \, B\right )} \cos \left (x\right ) + 36 \, A - 260 \, B\right )} \sin \left (x\right )}{210 \, {\left (a^{4} \cos \left (x\right )^{4} + 4 \, a^{4} \cos \left (x\right )^{3} + 6 \, a^{4} \cos \left (x\right )^{2} + 4 \, a^{4} \cos \left (x\right ) + a^{4}\right )}} \]
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\[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {\int \frac {A}{\cos ^{4}{\left (x \right )} + 4 \cos ^{3}{\left (x \right )} + 6 \cos ^{2}{\left (x \right )} + 4 \cos {\left (x \right )} + 1}\, dx + \int \frac {B \sec {\left (x \right )}}{\cos ^{4}{\left (x \right )} + 4 \cos ^{3}{\left (x \right )} + 6 \cos ^{2}{\left (x \right )} + 4 \cos {\left (x \right )} + 1}\, dx}{a^{4}} \]
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Time = 0.21 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=-\frac {1}{168} \, B {\left (\frac {\frac {315 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {77 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {A {\left (\frac {35 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {35 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}\right )}}{280 \, a^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.31 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a^{4}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, x\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )}{840 \, a^{28}} \]
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Time = 26.96 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.46 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {A-B}{8\,a^4}-\frac {3\,B}{4\,a^4}+\frac {2\,A-4\,B}{8\,a^4}-\frac {2\,A+4\,B}{8\,a^4}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (\frac {A-B}{40\,a^4}+\frac {2\,A-4\,B}{40\,a^4}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {A-B}{24\,a^4}-\frac {B}{4\,a^4}+\frac {2\,A-4\,B}{24\,a^4}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^4} \]
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