Integrand size = 33, antiderivative size = 284 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \]
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Time = 0.72 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3039, 4115, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac {\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^3 d \left (a^2-b^2\right )}-\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 d (a-b) (a+b)^2} \]
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Rule 2719
Rule 2720
Rule 2884
Rule 3039
Rule 3856
Rule 3872
Rule 3934
Rule 4115
Rule 4191
Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \sec (c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2} \, dx \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (-a A b+3 a^2 B-2 b^2 B\right )-b (A b-a B) \sec (c+d x)+\frac {1}{2} a (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{b \left (a^2-b^2\right )} \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{2} b \left (-a A b+3 a^2 B-2 b^2 B\right )-\left (b^2 (A b-a B)+\frac {1}{2} a \left (-a A b+3 a^2 B-2 b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{b^3 \left (a^2-b^2\right )}-\frac {\left (a \left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )} \\ & = \frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac {\left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac {\left (a \left (a^2 A b-3 A b^3-3 a^3 B+5 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^3 \left (a^2-b^2\right )} \\ & = -\frac {a \left (a^2 A b-3 A b^3-3 a^3 B+5 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^3 (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac {\left (\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac {\left (\left (a^2 A b-2 A b^3-3 a^3 B+4 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 b^3 \left (a^2-b^2\right )} \\ & = -\frac {\left (a A b-3 a^2 B+2 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^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-2 A b^3-3 a^3 B+4 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)}}{b^3 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 A b-3 A b^3-3 a^3 B+5 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^3 (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \\ \end{align*}
Time = 6.52 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {4 a b^2 (-A b+a B) \sin (c+d x)}{-a^2+b^2}+\frac {\cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (-2 b^2 \left (-a A b-a^2 B+2 b^2 B\right ) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+8 a b^2 (-A b+a B) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+\left (a A b-3 a^2 B+2 b^2 B\right ) \left (4 a b-4 a b \sec ^2(c+d x)+4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{a (a-b) (a+b)}}{4 b^3 d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(848\) vs. \(2(348)=696\).
Time = 9.53 (sec) , antiderivative size = 849, normalized size of antiderivative = 2.99
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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