Optimal. Leaf size=245 \[ -\frac{a^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac{a \left (3 a^2-4 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{a^2 \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.473508, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3238, 3847, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{a^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac{a \left (3 a^2-4 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{a^2 \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 3238
Rule 3847
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{1}{(a+b \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{1}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))^2} \, dx\\ &=-\frac{a^2 \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{3 a^2}{2}-b^2+a b \sec (c+d x)-\frac{1}{2} a^2 \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^2 \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{b \left (\frac{3 a^2}{2}-b^2\right )-\left (-a b^2+a \left (\frac{3 a^2}{2}-b^2\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{b^3 \left (a^2-b^2\right )}+\frac{\left (a^2 \left (3 a^2-5 b^2\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^2 \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 \left (3 a^2-4 b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac{\left (a^2 \left (3 a^2-5 b^2\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^2 \left (3 a^2-5 b^2\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-b) b^3 (a+b)^2 d}-\frac{a^2 \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 \left (3 a^2-4 b^2\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 (\left (3 a^2-2 b^2\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 (3 a^2-2 b^2\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{a \left (3 a^2-4 b^2\right ) \sqrt{\cos (c+d x)} F\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{a^2 \left (3 a^2-5 b^2\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-b) b^3 (a+b)^2 d}-\frac{a^2 \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*}
Mathematica [A] time = 6.63468, size = 323, normalized size = 1.32 \[ \frac{\frac{4 a^2 \sin (c+d x)}{b \left (b^2-a^2\right ) \sqrt{\sec (c+d x)} (a+b \cos (c+d x))}-\frac{2 \cot (c+d x) \left (2 b \left (-3 a^2+a b+2 b^2\right ) \sqrt{-\tan ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 b \left (3 a^2-2 b^2\right ) \sqrt{-\tan ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-3 a^2 b \sec ^{\frac{3}{2}}(c+d x)+3 a^2 b \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)-6 a^3 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+10 a b^2 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 b^3 \sec ^{\frac{3}{2}}(c+d x)-2 b^3 \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)\right )}{b^3 (a-b) (a+b)}}{4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 9.097, size = 815, normalized size = 3.3 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \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(-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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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