Optimal. Leaf size=91 \[ \frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.165683, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2985, 2649, 206, 2773} \[ \frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 2985
Rule 2649
Rule 206
Rule 2773
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx &=\frac{A \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{a}-(A-B) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=-\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}+\frac{(2 (A-B)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{\sqrt{a} d}\\ \end{align*}
Mathematica [A] time = 0.0768419, size = 72, normalized size = 0.79 \[ -\frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left ((A-B) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\sqrt{2} A \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{d \sqrt{a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 4.029, size = 268, normalized size = 3. \begin{align*} -{\frac{1}{d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( \sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) A-\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) B-A\ln \left ( -4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,a}{-2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) -A\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,a}{2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) \right ){\frac{1}{\sqrt{a}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.02556, size = 123, normalized size = 1.35 \begin{align*} \frac{{\left (\sqrt{2} \log \left (\cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) - \sqrt{2} \log \left (\cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )\right )} B}{2 \, \sqrt{a} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.80238, size = 464, normalized size = 5.1 \begin{align*} -\frac{\sqrt{2}{\left (A - B\right )} \sqrt{a} \log \left (-\frac{\cos \left (d x + c\right )^{2} - \frac{2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - A \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \, a d} \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 ) \sec{\left (c + d x \right )}}{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 2.7415, size = 227, normalized size = 2.49 \begin{align*} \frac{\frac{\sqrt{2}{\left (A \sqrt{a} - B \sqrt{a}\right )} \log \left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a} + \frac{2 \, A \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{\sqrt{a}} - \frac{2 \, A \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{\sqrt{a}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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