Integrand size = 25, antiderivative size = 142 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\frac {2 a^{5/2} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f} \]
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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} \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\frac {2 a^{5/2} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}+\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 (5 c+8 d) \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{15 f}+\frac {2 a d \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{5 f} \]
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Rule 209
Rule 3859
Rule 3877
Rule 4000
Rule 4002
Rubi steps \begin{align*} \text {integral}& = \frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {2}{5} \int (a+a \sec (e+f x))^{3/2} \left (\frac {5 a c}{2}+\frac {1}{2} a (5 c+8 d) \sec (e+f x)\right ) \, dx \\ & = \frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {4}{15} \int \sqrt {a+a \sec (e+f x)} \left (\frac {15 a^2 c}{4}+\frac {1}{4} a^2 (35 c+32 d) \sec (e+f x)\right ) \, dx \\ & = \frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\left (a^2 c\right ) \int \sqrt {a+a \sec (e+f x)} \, dx+\frac {1}{15} \left (a^2 (35 c+32 d)\right ) \int \sec (e+f x) \sqrt {a+a \sec (e+f x)} \, dx \\ & = \frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}-\frac {\left (2 a^3 c\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f} \\ & = \frac {2 a^{5/2} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (5 c+8 d) \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f} \\ \end{align*}
Time = 0.91 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.90 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt {a (1+\sec (e+f x))} \left (30 \sqrt {2} c \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \cos ^{\frac {5}{2}}(e+f x)+2 (40 c+49 d+2 (5 c+14 d) \cos (e+f x)+(40 c+43 d) \cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{30 f} \]
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Time = 7.70 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.49
method | result | size |
default | \(\frac {2 a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (15 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) c \cos \left (f x +e \right )+15 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) c +40 \sin \left (f x +e \right ) c +43 \sin \left (f x +e \right ) d +5 c \tan \left (f x +e \right )+14 d \tan \left (f x +e \right )+3 d \tan \left (f x +e \right ) \sec \left (f x +e \right )\right )}{15 f \left (\cos \left (f x +e \right )+1\right )}\) | \(211\) |
parts | \(\frac {2 c \,a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+3 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+8 \sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{3 f \left (\cos \left (f x +e \right )+1\right )}+\frac {2 d \,a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (43 \sin \left (f x +e \right )+14 \tan \left (f x +e \right )+3 \sec \left (f x +e \right ) \tan \left (f x +e \right )\right )}{15 f \left (\cos \left (f x +e \right )+1\right )}\) | \(236\) |
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Time = 0.31 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.75 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\left [\frac {15 \, {\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (3 \, a^{2} d + {\left (40 \, a^{2} c + 43 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} + {\left (5 \, a^{2} c + 14 \, a^{2} d\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{15 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (3 \, a^{2} d + {\left (40 \, a^{2} c + 43 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} + {\left (5 \, a^{2} c + 14 \, a^{2} d\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{15 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \]
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\[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 1396 vs. \(2 (124) = 248\).
Time = 0.43 (sec) , antiderivative size = 1396, normalized size of antiderivative = 9.83 \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\text {Too large to display} \]
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\[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sec \left (f x + e\right ) + c\right )} \,d x } \]
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Timed out. \[ \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right ) \,d x \]
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