Integrand size = 33, antiderivative size = 363 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^4 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \]
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Time = 1.24 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3039, 4115, 4189, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a \sec (c+d x)+b)}-\frac {\left (-5 a^2 B+3 a A b+2 b^2 B\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {\left (-5 a^3 B+3 a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac {a^2 \left (-5 a^3 B+3 a^2 A b+7 a b^2 B-5 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^4 d (a-b) (a+b)^2}-\frac {\left (-15 a^4 B+9 a^3 A b+16 a^2 b^2 B-12 a A b^3+2 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d \left (a^2-b^2\right )} \]
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Rule 2719
Rule 2720
Rule 2884
Rule 3039
Rule 3856
Rule 3872
Rule 3934
Rule 4115
Rule 4189
Rule 4191
Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (b+a \sec (c+d x))^2} \, dx \\ & = \frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (-3 a A b+5 a^2 B-2 b^2 B\right )-b (A b-a B) \sec (c+d x)+\frac {3}{2} a (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (b+a \sec (c+d x))} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}-\frac {2 \int \frac {-\frac {3}{4} \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\frac {1}{2} b \left (3 a A b-2 a^2 B-b^2 B\right ) \sec (c+d x)+\frac {1}{4} a \left (3 a A b-5 a^2 B+2 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}-\frac {2 \int \frac {-\frac {3}{4} b \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )-\left (\frac {1}{2} b^2 \left (3 a A b-2 a^2 B-b^2 B\right )-\frac {3}{4} a \left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 b^4 \left (a^2-b^2\right )}+\frac {\left (a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )} \\ & = -\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}+\frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac {\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}+\frac {\left (a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^4 \left (a^2-b^2\right )} \\ & = \frac {a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^4 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}+\frac {\left (\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac {\left (\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^4 \left (a^2-b^2\right )} \\ & = \frac {\left (3 a^2 A b-2 A b^3-5 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (9 a^3 A b-12 a A b^3-15 a^4 B+16 a^2 b^2 B+2 b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^4 \left (a^2-b^2\right ) d}+\frac {a^2 \left (3 a^2 A b-5 A b^3-5 a^3 B+7 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^4 (a+b)^2 d}-\frac {\left (3 a A b-5 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \\ \end{align*}
Time = 6.84 (sec) , antiderivative size = 701, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\frac {2 \left (-3 a^2 A b+6 A b^3+5 a^3 B-8 a b^2 B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-12 a A b^2+8 a^2 b B+4 b^3 B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-9 a^2 A b+6 A b^3+15 a^3 B-12 a b^2 B\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{12 (a-b) b^2 (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {a^2 (-A b+a B) \sin (c+d x)}{b^3 \left (a^2-b^2\right )}-\frac {a^3 A b \sin (c+d x)-a^4 B \sin (c+d x)}{b^3 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}+\frac {B \sin (2 (c+d x))}{3 b^2}\right )}{d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1065\) vs. \(2(421)=842\).
Time = 11.30 (sec) , antiderivative size = 1066, normalized size of antiderivative = 2.94
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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