Optimal. Leaf size=227 \[ \frac{\left (2 a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a^2 d \left (a^2-b^2\right )}-\frac{b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac{b \left (3 a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.369199, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3843, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{b \sin (c+d x) \sqrt{\sec (c+d x)}}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\left (2 a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac{b \left (3 a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 3843
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{\sqrt{\sec (c+d x)}}{(a+b \sec (c+d x))^2} \, dx &=-\frac{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{b}{2}-a \sec (c+d x)+\frac{1}{2} b \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{-a^2+b^2}\\ &=-\frac{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{a b}{2}-\left (a^2-\frac{b^2}{2}\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^2 \left (a^2-b^2\right )}-\frac{\left (b \left (3-\frac{b^2}{a^2}\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{b \sqrt{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{b \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a \left (a^2-b^2\right )}+\frac{\left (2 a^2-b^2\right ) \int \sqrt{\sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac{\left (b \left (3-\frac{b^2}{a^2}\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 \left (a^2-b^2\right )}\\ &=-\frac{b \left (3 a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}-\frac{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 (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-b^2\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{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-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)}}{a^2 \left (a^2-b^2\right ) d}-\frac{b \left (3 a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}-\frac{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*}
Mathematica [A] time = 4.38511, size = 255, normalized size = 1.12 \[ -\frac{\cos (2 (c+d x)) \csc (c+d x) \sqrt{\sec (c+d x)} \left (-a (a-b) \sqrt{-\tan ^2(c+d x)} \sqrt{\sec (c+d x)} (a \cos (c+d x)+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )-\left (3 a^2-b^2\right ) \sqrt{-\tan ^2(c+d x)} \sqrt{\sec (c+d x)} (a \cos (c+d x)+b) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+a b \left (b \tan ^2(c+d x)-\sqrt{-\tan ^2(c+d x)} \sqrt{\sec (c+d x)} (a \cos (c+d x)+b) E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )\right )}{a^2 d (a-b) (a+b) \left (\sec ^2(c+d x)-2\right ) (a \cos (c+d x)+b)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 3.836, size = 788, normalized size = 3.5 \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{\sqrt{\sec{\left (c + d x \right )}}}{\left (a + b \sec{\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{\sqrt{\sec \left (d x + c\right )}}{{\left (b \sec \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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