Optimal. Leaf size=192 \[ \frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}} \]
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Rubi [A] time = 0.524277, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {4022, 3920, 3774, 203, 3795} \[ \frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 4022
Rule 3920
Rule 3774
Rule 203
Rule 3795
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) (A+A \sec (c+d x))}{\sqrt{a-a \sec (c+d x)}} \, dx &=\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{\cos ^2(c+d x) \left (-\frac{7 a A}{2}-\frac{5}{2} a A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{3 a}\\ &=\frac{7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (\frac{27 a^2 A}{4}+\frac{21}{4} a^2 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{-\frac{69 a^3 A}{8}-\frac{27}{8} a^3 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{6 a^3}\\ &=\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+(2 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx+\frac{(23 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{16 a}\\ &=\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}+\frac{(23 A) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 d}-\frac{(4 A) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}\\ &=\frac{23 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{9 A \sin (c+d x)}{8 d \sqrt{a-a \sec (c+d x)}}+\frac{7 A \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a-a \sec (c+d x)}}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.88793, size = 330, normalized size = 1.72 \[ \frac{A e^{-4 i (c+d x)} \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right )+i \sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (9 e^{i (c+d x)}+40 e^{2 i (c+d x)}+47 e^{3 i (c+d x)}+47 e^{4 i (c+d x)}+40 e^{5 i (c+d x)}+9 e^{6 i (c+d x)}+2 e^{7 i (c+d x)}+69 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-96 \sqrt{2} e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+69 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+2\right )}{48 d \sqrt{a-a \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.404, size = 625, 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{{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt{-a \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 [A] time = 0.548307, size = 1258, normalized size = 6.55 \begin{align*} \left [\frac{48 \, \sqrt{2} A a \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} -{\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 69 \, A \sqrt{-a} \log \left (\frac{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} -{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \,{\left (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{48 \, a d \sin \left (d x + c\right )}, \frac{48 \, \sqrt{2} A \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 69 \, A \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) -{\left (8 \, A \cos \left (d x + c\right )^{4} + 22 \, A \cos \left (d x + c\right )^{3} + 41 \, A \cos \left (d x + c\right )^{2} + 27 \, A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{24 \, a d \sin \left (d x + c\right )}\right ] \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 [C] time = 2.21682, size = 412, normalized size = 2.15 \begin{align*} -\frac{A a{\left (\frac{48 \, \sqrt{2} \arctan \left (\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{\sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{69 \, \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\sqrt{2}{\left (21 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{5}{2}} + 80 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} a + 108 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{3} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}\right )} + \frac{{\left (48 i \, \sqrt{2} A \sqrt{-a} \arctan \left (-i\right ) - 69 i \, A \sqrt{-a} \arctan \left (-\frac{1}{2} i \, \sqrt{2}\right ) - 49 \, \sqrt{2} A \sqrt{-a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{a}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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