Integrand size = 33, antiderivative size = 263 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {(A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-A b^2-a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
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Time = 0.59 (sec) , antiderivative size = 263, 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 = {4112, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\left (2 a^2 A-a b B-A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a-b) (a+b)^2} \]
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4112
Rule 4191
Rubi steps \begin{align*} \text {integral}& = -\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {\frac {1}{2} (-A b+a B)-(a A-b B) \sec (c+d x)+\frac {1}{2} (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{-a^2+b^2} \\ & = -\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\frac {1}{2} a (-A b+a B)-\left (\frac {1}{2} b (-A b+a B)-a (-a A+b B)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^2 \left (a^2-b^2\right )}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = -\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {(A b-a B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a \left (a^2-b^2\right )}+\frac {\left (2 a^2 A-A b^2-a b B\right ) \int \sqrt {\sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (\left (3 a^2 A b-A b^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^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a^2 A b-A b^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^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left ((A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a \left (a^2-b^2\right )}+\frac {\left (\left (2 a^2 A-A b^2-a b B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = \frac {(A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-A b^2-a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 a^2 A b-A b^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^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\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. \(722\) vs. \(2(263)=526\).
Time = 7.87 (sec) , antiderivative size = 722, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x))^2 \sec (c+d x) (A+B \sec (c+d x)) \left (\frac {2 (-A b+a B) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 (4 a A-4 b B) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {(A b-a B) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}\right )}{4 (a-b) (a+b) d (B+A \cos (c+d x)) (a+b \sec (c+d x))^2}+\frac {(b+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \left (\frac {(-A b+a B) \sin (c+d x)}{a \left (a^2-b^2\right )}+\frac {A b^2 \sin (c+d x)-a b B \sin (c+d x)}{a \left (a^2-b^2\right ) (b+a \cos (c+d x))}\right )}{d (B+A \cos (c+d x)) (a+b \sec (c+d x))^2} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(801\) vs. \(2(327)=654\).
Time = 15.37 (sec) , antiderivative size = 802, normalized size of antiderivative = 3.05
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Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
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\[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
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