Optimal. Leaf size=172 \[ \frac{(12 A+5 C) \tan ^3(c+d x)}{3 a^2 d}+\frac{(12 A+5 C) \tan (c+d x)}{a^2 d}-\frac{(5 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(5 A+2 C) \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac{2 (5 A+2 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.334533, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3042, 2978, 2748, 3767, 3768, 3770} \[ \frac{(12 A+5 C) \tan ^3(c+d x)}{3 a^2 d}+\frac{(12 A+5 C) \tan (c+d x)}{a^2 d}-\frac{(5 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(5 A+2 C) \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac{2 (5 A+2 C) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3042
Rule 2978
Rule 2748
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(3 a (2 A+C)-a (4 A+C) \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \left (3 a^2 (12 A+5 C)-6 a^2 (5 A+2 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{3 a^4}\\ &=-\frac{2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(2 (5 A+2 C)) \int \sec ^3(c+d x) \, dx}{a^2}+\frac{(12 A+5 C) \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac{(5 A+2 C) \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac{2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(5 A+2 C) \int \sec (c+d x) \, dx}{a^2}-\frac{(12 A+5 C) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac{(5 A+2 C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{(12 A+5 C) \tan (c+d x)}{a^2 d}-\frac{(5 A+2 C) \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac{2 (5 A+2 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(12 A+5 C) \tan ^3(c+d x)}{3 a^2 d}\\ \end{align*}
Mathematica [B] time = 5.06395, size = 594, normalized size = 3.45 \[ \frac{192 (5 A+2 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac{c}{2}\right ) \sec (c) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-153 A \sin \left (c-\frac{d x}{2}\right )+21 A \sin \left (c+\frac{d x}{2}\right )-135 A \sin \left (2 c+\frac{d x}{2}\right )+25 A \sin \left (c+\frac{3 d x}{2}\right )+45 A \sin \left (2 c+\frac{3 d x}{2}\right )-85 A \sin \left (3 c+\frac{3 d x}{2}\right )+99 A \sin \left (c+\frac{5 d x}{2}\right )+21 A \sin \left (2 c+\frac{5 d x}{2}\right )+33 A \sin \left (3 c+\frac{5 d x}{2}\right )-45 A \sin \left (4 c+\frac{5 d x}{2}\right )+57 A \sin \left (2 c+\frac{7 d x}{2}\right )+18 A \sin \left (3 c+\frac{7 d x}{2}\right )+24 A \sin \left (4 c+\frac{7 d x}{2}\right )-15 A \sin \left (5 c+\frac{7 d x}{2}\right )+24 A \sin \left (3 c+\frac{9 d x}{2}\right )+11 A \sin \left (4 c+\frac{9 d x}{2}\right )+13 A \sin \left (5 c+\frac{9 d x}{2}\right )-3 (A+8 C) \sin \left (\frac{d x}{2}\right )+(155 A+66 C) \sin \left (\frac{3 d x}{2}\right )-60 C \sin \left (c-\frac{d x}{2}\right )+24 C \sin \left (c+\frac{d x}{2}\right )-60 C \sin \left (2 c+\frac{d x}{2}\right )-4 C \sin \left (c+\frac{3 d x}{2}\right )+36 C \sin \left (2 c+\frac{3 d x}{2}\right )-34 C \sin \left (3 c+\frac{3 d x}{2}\right )+42 C \sin \left (c+\frac{5 d x}{2}\right )+24 C \sin \left (3 c+\frac{5 d x}{2}\right )-18 C \sin \left (4 c+\frac{5 d x}{2}\right )+24 C \sin \left (2 c+\frac{7 d x}{2}\right )+3 C \sin \left (3 c+\frac{7 d x}{2}\right )+15 C \sin \left (4 c+\frac{7 d x}{2}\right )-6 C \sin \left (5 c+\frac{7 d x}{2}\right )+10 C \sin \left (3 c+\frac{9 d x}{2}\right )+3 C \sin \left (4 c+\frac{9 d x}{2}\right )+7 C \sin \left (5 c+\frac{9 d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.065, size = 338, normalized size = 2. \begin{align*}{\frac{A}{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{9\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{A}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{C}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+5\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{2}}}+2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{2}}}-{\frac{A}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{3\,A}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-5\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{2}}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{2}}}-5\,{\frac{A}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{C}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{3\,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 = 1.07824, size = 512, normalized size = 2.98 \begin{align*} \frac{A{\left (\frac{4 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \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{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} + C{\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 )}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.70117, size = 591, normalized size = 3.44 \begin{align*} -\frac{3 \,{\left ({\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (5 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (12 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (33 \, A + 14 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + A\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\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 = 1.29958, size = 304, normalized size = 1.77 \begin{align*} -\frac{\frac{6 \,{\left (5 \, A + 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 + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{4 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C \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 )}^{3} a^{2}} - \frac{A 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} + 27 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, 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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