Optimal. Leaf size=117 \[ -\frac {2 (11 B-C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 B-2 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rubi [A] time = 0.40, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3029, 2978, 12, 3770} \[ -\frac {2 (11 B-C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 B-2 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2978
Rule 3029
Rule 3770
Rubi steps
\begin {align*} \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=\int \frac {(B+C \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(5 a B-2 a (B-C) \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 B-2 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (15 a^2 B-a^2 (7 B-2 C) \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 B-2 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {2 (11 B-C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int 15 a^3 B \sec (c+d x) \, dx}{15 a^6}\\ &=-\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 B-2 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {2 (11 B-C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {B \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {B \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(B-C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 B-2 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {2 (11 B-C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.92, size = 197, normalized size = 1.68 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (75 B \sin \left (c+\frac {d x}{2}\right )-95 B \sin \left (c+\frac {3 d x}{2}\right )+15 B \sin \left (2 c+\frac {3 d x}{2}\right )-22 B \sin \left (2 c+\frac {5 d x}{2}\right )-5 (29 B-4 C) \sin \left (\frac {d x}{2}\right )+10 C \sin \left (c+\frac {3 d x}{2}\right )+2 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )-240 B \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.42, size = 185, normalized size = 1.58 \[ \frac {15 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (11 \, B - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, B - 2 \, C\right )} \cos \left (d x + c\right ) + 32 \, B - 7 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.09, size = 148, normalized size = 1.26 \[ \frac {\frac {60 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 159, normalized size = 1.36 \[ \frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.70, size = 187, normalized size = 1.60 \[ -\frac {B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.11, size = 130, normalized size = 1.11 \[ \frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{4\,a^3}+\frac {3\,B+C}{4\,a^3}+\frac {3\,B-C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{12\,a^3}+\frac {3\,B-C}{12\,a^3}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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