Optimal. Leaf size=260 \[ \frac{b (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b+a^3 (-B)-a b^2 B-A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.612991, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2960, 4029, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{b (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b+a^3 (-B)-a b^2 B-A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 2960
Rule 4029
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sqrt{\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx &=\int \frac{\sec ^{\frac{3}{2}}(c+d x) (B+A \sec (c+d x))}{(b+a \sec (c+d x))^2} \, dx\\ &=\frac{b (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}-\frac{\int \frac{\frac{1}{2} b (A b-a B)+a (A b-a B) \sec (c+d x)-\frac{1}{2} \left (2 a^2 A-A b^2-a b B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \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 (b+a \sec (c+d x))}-\frac{\int \frac{\frac{1}{2} b^2 (A b-a B)+\frac{1}{2} a b (A b-a B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a b^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)}{b+a \sec (c+d x)} \, dx}{2 a b \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 (b+a \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{(A b-a B) \int \sqrt{\sec (c+d x)} \, dx}{2 b \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)} (a+b \cos (c+d x))} \, dx}{2 a b \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)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a (a-b) 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 (b+a \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 ((A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b \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{(A b-a B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b \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)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a (a-b) 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 (b+a \sec (c+d x))}\\ \end{align*}
Mathematica [B] time = 6.81852, size = 645, normalized size = 2.48 \[ \frac{\sqrt{\sec (c+d x)} \left (\frac{a B \sin (c+d x)-A b \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{(a B-A b) \sin (c+d x)}{a \left (a^2-b^2\right )}\right )}{d}+\frac{\frac{2 \left (-4 a^2 A+a b B+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \left (\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{\left (A b^2-a b B\right ) \sin (c+d x) \cos (2 (c+d x)) (a \sec (c+d x)+b) \left (4 a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}-\frac{2 \left (4 a A b-4 a^2 B\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}}{4 a d (b-a) (a+b)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 8.523, size = 721, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \cos{\left (c + d x \right )}\right ) \sqrt{\sec{\left (c + d x \right )}}}{\left (a + b \cos{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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