Optimal. Leaf size=184 \[ -\frac{b \left (4 a^2-3 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))} \]
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Rubi [A] time = 0.478678, antiderivative size = 184, 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 = {4264, 3847, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{b \left (4 a^2-3 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))} \]
Antiderivative was successfully verified.
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Rule 4264
Rule 3847
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{(a+b \sec (c+d x))^2} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2} \, dx\\ &=\frac{b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-a^2+\frac{3 b^2}{2}+a b \sec (c+d x)-\frac{1}{2} b^2 \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a \left (-a^2+\frac{3 b^2}{2}\right )-\left (-a^2 b+b \left (-a^2+\frac{3 b^2}{2}\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^3 \left (a^2-b^2\right )}+\frac{\left (b^2 \left (5-\frac{3 b^2}{a^2}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (b^2 \left (5 a^2-3 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac{\left (\left (2 a^2-3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac{\left (b \left (4 a^2-3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac{b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}+\frac{\left (2 a^2-3 b^2\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac{\left (b \left (4 a^2-3 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2-3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac{b \left (4 a^2-3 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac{b^2 \left (5 a^2-3 b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac{b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.76664, size = 254, normalized size = 1.38 \[ \frac{\frac{-\frac{2 \left (2 a^2-3 b^2\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+8 b \left (\frac{b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}-\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )\right )+\frac{2 \left (2 a^2-b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b) (a+b)}+\frac{4 b^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a d} \]
Antiderivative was successfully verified.
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Maple [B] time = 4.926, size = 809, normalized size = 4.4 \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{\sqrt{\cos \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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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{\cos{\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{\cos \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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