Integrand size = 35, antiderivative size = 175 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (6 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (6 A+7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a (6 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}} \]
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Time = 0.38 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {4100, 3890, 3889} \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a (6 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a (6 A+7 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a (6 A+7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]
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Rule 3889
Rule 3890
Rule 4100
Rubi steps \begin{align*} \text {integral}& = \frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1}{7} (6 A+7 B) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (6 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {1}{35} (4 (6 A+7 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (6 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (6 A+7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {1}{105} (8 (6 A+7 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (6 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (6 A+7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a (6 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \left (15 A+3 (6 A+7 B) \sec (c+d x)+4 (6 A+7 B) \sec ^2(c+d x)+8 (6 A+7 B) \sec ^3(c+d x)\right ) \sin (c+d x)}{105 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]
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Time = 4.33 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.57
method | result | size |
default | \(\frac {2 \left (15 A \cos \left (d x +c \right )^{3}+18 A \cos \left (d x +c \right )^{2}+21 B \cos \left (d x +c \right )^{2}+24 A \cos \left (d x +c \right )+28 B \cos \left (d x +c \right )+48 A +56 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(100\) |
parts | \(\frac {2 A \left (5 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )+16\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{35 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+8 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(141\) |
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Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]
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Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (151) = 302\).
Time = 0.53 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.85 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
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Time = 16.13 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (420\,A\,\sin \left (c+d\,x\right )+490\,B\,\sin \left (c+d\,x\right )+126\,A\,\sin \left (2\,c+2\,d\,x\right )+36\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+112\,B\,\sin \left (2\,c+2\,d\,x\right )+42\,B\,\sin \left (3\,c+3\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]
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