Integrand size = 25, antiderivative size = 142 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
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Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4002, 4000, 3859, 209, 3877} \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (5 A+8 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]
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Rule 209
Rule 3859
Rule 3877
Rule 4000
Rule 4002
Rubi steps \begin{align*} \text {integral}& = \frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2}{5} \int (a+a \sec (c+d x))^{3/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (5 A+8 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (5 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \sqrt {a+a \sec (c+d x)} \left (\frac {15 a^2 A}{4}+\frac {1}{4} a^2 (35 A+32 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (5 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\left (a^2 A\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{15} \left (a^2 (35 A+32 B)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^3 (35 A+32 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac {\left (2 a^3 A\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {2 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \\ \end{align*}
Time = 0.99 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.90 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \left (30 \sqrt {2} A \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (40 A+49 B+2 (5 A+14 B) \cos (c+d x)+(40 A+43 B) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{30 d} \]
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Time = 9.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.49
method | result | size |
default | \(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+15 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+40 A \sin \left (d x +c \right )+43 B \sin \left (d x +c \right )+5 A \tan \left (d x +c \right )+14 B \tan \left (d x +c \right )+3 B \tan \left (d x +c \right ) \sec \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(211\) |
parts | \(\frac {2 A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+8 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (43 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(236\) |
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Time = 0.28 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.66 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {15 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (40 \, A + 43 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (40 \, A + 43 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
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\[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 1396 vs. \(2 (124) = 248\).
Time = 0.41 (sec) , antiderivative size = 1396, normalized size of antiderivative = 9.83 \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]
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\[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]
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