Optimal. Leaf size=165 \[ -\frac{2 (8 A-5 B+2 C) \tan (c+d x)}{3 a^2 d}+\frac{(7 A-4 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(7 A-4 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.360768, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3041, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{2 (8 A-5 B+2 C) \tan (c+d x)}{3 a^2 d}+\frac{(7 A-4 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac{(7 A-4 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{(8 A-5 B+2 C) \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3041
Rule 2978
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))^2} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(a (5 A-2 B+2 C)-3 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(8 A-5 B+2 C) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \left (3 a^2 (7 A-4 B+2 C)-2 a^2 (8 A-5 B+2 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{3 a^4}\\ &=-\frac{(8 A-5 B+2 C) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(2 (8 A-5 B+2 C)) \int \sec ^2(c+d x) \, dx}{3 a^2}+\frac{(7 A-4 B+2 C) \int \sec ^3(c+d x) \, dx}{a^2}\\ &=\frac{(7 A-4 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{(8 A-5 B+2 C) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 A-4 B+2 C) \int \sec (c+d x) \, dx}{2 a^2}+\frac{(2 (8 A-5 B+2 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=\frac{(7 A-4 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{2 (8 A-5 B+2 C) \tan (c+d x)}{3 a^2 d}+\frac{(7 A-4 B+2 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{(8 A-5 B+2 C) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}
Mathematica [B] time = 6.17683, size = 578, normalized size = 3.5 \[ -\frac{2 (7 A-4 B+2 C) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+a)^2}+\frac{2 (7 A-4 B+2 C) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+a)^2}-\frac{2 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (A \sin \left (\frac{1}{2} (c+d x)\right )-B \sin \left (\frac{1}{2} (c+d x)\right )+C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d (a \cos (c+d x)+a)^2}-\frac{4 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (10 A \sin \left (\frac{1}{2} (c+d x)\right )-7 B \sin \left (\frac{1}{2} (c+d x)\right )+4 C \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d (a \cos (c+d x)+a)^2}-\frac{4 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (2 A \sin \left (\frac{1}{2} (c+d x)\right )-B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+a)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{4 \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (2 A \sin \left (\frac{1}{2} (c+d x)\right )-B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+a)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{A \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a \cos (c+d x)+a)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a \cos (c+d x)+a)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.075, size = 373, normalized size = 2.3 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{7\,A}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{B\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{2}}}-{\frac{C}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{5\,A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{7\,A}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{B\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{2}}}+{\frac{C}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.01204, size = 582, normalized size = 3.53 \begin{align*} -\frac{A{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{21 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac{a^{2} \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 )}}\right )} + C{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.9805, size = 630, normalized size = 3.82 \begin{align*} \frac{3 \,{\left ({\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A - 4 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (8 \, A - 5 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (43 \, A - 28 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (A - B\right )} \cos \left (d x + c\right ) - 3 \, A\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} 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 = 1.24511, size = 317, normalized size = 1.92 \begin{align*} \frac{\frac{3 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \,{\left (7 \, A - 4 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{6 \,{\left (5 \, 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} - 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}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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