Optimal. Leaf size=90 \[ \frac{5 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}+\frac{3 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]
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Rubi [A] time = 0.0412816, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2664, 12, 2659, 206} \[ \frac{5 \sin (c+d x)}{16 d (5 \cos (c+d x)+3)}+\frac{3 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]
Antiderivative was successfully verified.
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Rule 2664
Rule 12
Rule 2659
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{(3+5 \cos (c+d x))^2} \, dx &=\frac{5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}+\frac{1}{16} \int -\frac{3}{3+5 \cos (c+d x)} \, dx\\ &=\frac{5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}-\frac{3}{16} \int \frac{1}{3+5 \cos (c+d x)} \, dx\\ &=\frac{5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{8-2 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}\\ &=\frac{3 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{3 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{5 \sin (c+d x)}{16 d (3+5 \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.0780956, size = 143, normalized size = 1.59 \[ \frac{20 \sin (c+d x)+9 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 \cos (c+d x) \left (\log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-9 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d (5 \cos (c+d x)+3)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 72, normalized size = 0.8 \begin{align*} -{\frac{5}{32\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) ^{-1}}-{\frac{3}{64\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) }-{\frac{5}{32\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) ^{-1}}+{\frac{3}{64\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.4163, size = 123, normalized size = 1.37 \begin{align*} -\frac{\frac{20 \, \sin \left (d x + c\right )}{{\left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 4\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + 3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.66956, size = 258, normalized size = 2.87 \begin{align*} -\frac{3 \,{\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac{3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac{5}{2}\right ) - 3 \,{\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac{3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac{5}{2}\right ) - 40 \, \sin \left (d x + c\right )}{128 \,{\left (5 \, d \cos \left (d x + c\right ) + 3 \, d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.27173, size = 228, normalized size = 2.53 \begin{align*} \begin{cases} \frac{x}{\left (5 \cos{\left (2 \operatorname{atan}{\left (2 \right )} \right )} + 3\right )^{2}} & \text{for}\: c = - d x - 2 \operatorname{atan}{\left (2 \right )} \vee c = - d x + 2 \operatorname{atan}{\left (2 \right )} \\\frac{x}{\left (5 \cos{\left (c \right )} + 3\right )^{2}} & \text{for}\: d = 0 \\\frac{3 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 \right )} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{64 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 256 d} - \frac{12 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 \right )}}{64 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 256 d} - \frac{3 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 \right )} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{64 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 256 d} + \frac{12 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 \right )}}{64 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 256 d} - \frac{20 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{64 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 256 d} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.16509, size = 84, normalized size = 0.93 \begin{align*} -\frac{\frac{20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4} + 3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \right |}\right ) - 3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \right |}\right )}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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