Optimal. Leaf size=117 \[ -\frac{(2 A-2 B+C) \tan (c+d x)}{a d}+\frac{(3 A-2 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(3 A-2 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)} \]
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Rubi [A] time = 0.201625, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3041, 2748, 3768, 3770, 3767, 8} \[ -\frac{(2 A-2 B+C) \tan (c+d x)}{a d}+\frac{(3 A-2 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac{(3 A-2 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)} \]
Antiderivative was successfully verified.
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Rule 3041
Rule 2748
Rule 3768
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac{\int (a (3 A-2 B+2 C)-a (2 A-2 B+C) \cos (c+d x)) \sec ^3(c+d x) \, dx}{a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}-\frac{(2 A-2 B+C) \int \sec ^2(c+d x) \, dx}{a}+\frac{(3 A-2 B+2 C) \int \sec ^3(c+d x) \, dx}{a}\\ &=\frac{(3 A-2 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac{(3 A-2 B+2 C) \int \sec (c+d x) \, dx}{2 a}+\frac{(2 A-2 B+C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=\frac{(3 A-2 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(2 A-2 B+C) \tan (c+d x)}{a d}+\frac{(3 A-2 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}\\ \end{align*}
Mathematica [B] time = 1.44026, size = 256, normalized size = 2.19 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \left (-4 (A-B+C) \tan \left (\frac{1}{2} (c+d x)\right )-2 (3 A-2 B+2 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (3 A-2 B+2 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 (B-A) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 (B-A) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{2 a d (\cos (c+d x)+1)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.066, size = 311, normalized size = 2.7 \begin{align*} -{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3\,A}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{B}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{C}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{3\,A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,A}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{B}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{C}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.01616, size = 481, normalized size = 4.11 \begin{align*} -\frac{A{\left (\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (d x + c\right )}{{\left (a - \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 2 \, C{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.96998, size = 428, normalized size = 3.66 \begin{align*} \frac{{\left ({\left (3 \, A - 2 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, A - 2 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (3 \, A - 2 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, A - 2 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (2 \, A - 2 \, B + C\right )} \cos \left (d x + c\right )^{2} +{\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - A\right )} \sin \left (d x + c\right )}{4 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\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 [A] time = 1.22098, size = 235, normalized size = 2.01 \begin{align*} \frac{\frac{{\left (3 \, A - 2 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{{\left (3 \, A - 2 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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