Optimal. Leaf size=352 \[ \frac{\sqrt{a+b} (2 a C+A b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{a b d}+\frac{A b \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^2 d}+\frac{A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{a d}+\frac{A (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a b d} \]
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Rubi [A] time = 0.395225, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4105, 4058, 3921, 3784, 3832, 4004} \[ \frac{A b \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^2 d}+\frac{\sqrt{a+b} (2 a C+A b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a b d}+\frac{A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{a d}+\frac{A (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a b d} \]
Antiderivative was successfully verified.
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Rule 4105
Rule 4058
Rule 3921
Rule 3784
Rule 3832
Rule 4004
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx &=\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}-\frac{\int \frac{\frac{A b}{2}-a C \sec (c+d x)+\frac{1}{2} A b \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{a}\\ &=\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}-\frac{\int \frac{\frac{A b}{2}+\left (-\frac{A b}{2}-a C\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{a}-\frac{(A b) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 a}\\ &=\frac{A (a-b) \sqrt{a+b} \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{a b d}+\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}-\frac{(A b) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 a}-\frac{(-A b-2 a C) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 a}\\ &=\frac{A (a-b) \sqrt{a+b} \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{a b d}+\frac{\sqrt{a+b} (A b+2 a C) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{a b d}+\frac{A b \sqrt{a+b} \cot (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}\\ \end{align*}
Mathematica [A] time = 15.8516, size = 386, normalized size = 1.1 \[ \frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (A \cos (c+d x)+C \sec (c+d x)) \left (4 a C \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+A \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 A (a+b) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )+4 A b \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \Pi \left (-1;-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{a d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} (A \cos (2 c+2 d x)+A+2 C)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.462, size = 841, normalized size = 2.4 \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{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{\sqrt{b \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )}{\sqrt{b \sec \left (d x + c\right ) + a}}, x\right ) \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 + C \sec ^{2}{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )}}{\sqrt{a + b \sec{\left (c + d x \right )}}}\, 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 (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{\sqrt{b \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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