Integrand size = 33, antiderivative size = 207 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]
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Time = 1.20 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3057, 3063, 3064, 2728, 212, 2852} \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
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Rule 212
Rule 2728
Rule 2852
Rule 3057
Rule 3063
Rule 3064
Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\left (a (5 A-B)-\frac {5}{2} a (A-B) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (\frac {1}{2} a^2 (35 A-11 B)-\frac {3}{4} a^2 (15 A-7 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-4 a^3 (5 A-2 B)+\frac {1}{4} a^3 (35 A-11 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(115 A-43 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}-\frac {(5 A-2 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^3} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(115 A-43 B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{16 a^2 d}+\frac {(5 A-2 B) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^2 d} \\ & = -\frac {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}
Time = 2.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {8 (115 A-43 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )-128 \sqrt {2} (5 A-2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )+(67 A-11 B+10 (11 A-3 B) \cos (c+d x)+(35 A-11 B) \cos (2 (c+d x))) \tan (c+d x)}{32 d (a (1+\cos (c+d x)))^{5/2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(924\) vs. \(2(178)=356\).
Time = 6.61 (sec) , antiderivative size = 925, normalized size of antiderivative = 4.47
method | result | size |
parts | \(\text {Expression too large to display}\) | \(925\) |
default | \(\text {Expression too large to display}\) | \(1122\) |
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Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (178) = 356\).
Time = 0.37 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.95 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 16 \, {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (35 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (11 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
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\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
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Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.81 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (115 \, A \sqrt {a} - 43 \, B \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (115 \, A \sqrt {a} - 43 \, B \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, {\left (5 \, A - 2 \, B\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {32 \, {\left (5 \, A - 2 \, B\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {64 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (19 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]
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Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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