Integrand size = 33, antiderivative size = 283 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\frac {\left (2 a^2 A-3 A b^2+a b B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}+\frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
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Time = 0.69 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4115, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\frac {b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\left (2 a^2 A+a b B-3 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 d (a-b) (a+b)^2} \]
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4115
Rule 4191
Rubi steps \begin{align*} \text {integral}& = \frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (-2 a^2 A+3 A b^2-a b B\right )+a (A b-a B) \sec (c+d x)-\frac {1}{2} b (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\frac {1}{2} a \left (-2 a^2 A+3 A b^2-a b B\right )-\left (-a^2 (A b-a B)+\frac {1}{2} b \left (-2 a^2 A+3 A b^2-a b B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}+\frac {\left (b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )} \\ & = \frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (2 a^2 A-3 A b^2+a b B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )} \\ & = \frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (\left (2 a^2 A-3 A b^2+a b B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2 A-3 A b^2+a b B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}+\frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(652\) vs. \(2(283)=566\).
Time = 7.40 (sec) , antiderivative size = 652, normalized size of antiderivative = 2.30 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \left (-2 a^2 A+A b^2+a b B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (4 a A b-4 a^2 B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-2 a^2 A+3 A b^2-a b B\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{4 a (-a+b) (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {b (A b-a B) \sin (c+d x)}{a^2 \left (-a^2+b^2\right )}+\frac {-A b^3 \sin (c+d x)+a b^2 B \sin (c+d x)}{a^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))}\right )}{d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(842\) vs. \(2(347)=694\).
Time = 18.54 (sec) , antiderivative size = 843, normalized size of antiderivative = 2.98
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
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