Optimal. Leaf size=136 \[ \frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}+\frac {3 C \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {3 C x}{a^3}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(A-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.46, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2977, 2968, 3023, 12, 2735, 2648} \[ \frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}+\frac {3 C \sin (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {3 C x}{a^3}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(A-9 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2648
Rule 2735
Rule 2968
Rule 2977
Rule 3023
Rule 3042
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) (a (2 A-3 C)+a (A+6 C) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos (c+d x) \left (2 a^2 (A-9 C)+a^2 (2 A+27 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {2 a^2 (A-9 C) \cos (c+d x)+a^2 (2 A+27 C) \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int -\frac {45 a^3 C \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^5}\\ &=\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 C) \int \frac {\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=-\frac {3 C x}{a^3}+\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 C) \int \frac {1}{a+a \cos (c+d x)} \, dx}{a^2}\\ &=-\frac {3 C x}{a^3}+\frac {(2 A+27 C) \sin (c+d x)}{15 a^3 d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-9 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {3 C \sin (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 0.94, size = 283, normalized size = 2.08 \[ -\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (120 A \sin \left (c+\frac {d x}{2}\right )-80 A \sin \left (c+\frac {3 d x}{2}\right )+60 A \sin \left (2 c+\frac {3 d x}{2}\right )-28 A \sin \left (2 c+\frac {5 d x}{2}\right )-160 A \sin \left (\frac {d x}{2}\right )+1125 C \sin \left (c+\frac {d x}{2}\right )-1215 C \sin \left (c+\frac {3 d x}{2}\right )+225 C \sin \left (2 c+\frac {3 d x}{2}\right )-363 C \sin \left (2 c+\frac {5 d x}{2}\right )-75 C \sin \left (3 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {7 d x}{2}\right )-15 C \sin \left (4 c+\frac {7 d x}{2}\right )+900 C d x \cos \left (c+\frac {d x}{2}\right )+450 C d x \cos \left (c+\frac {3 d x}{2}\right )+450 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+90 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+90 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-1755 C \sin \left (\frac {d x}{2}\right )+900 C d x \cos \left (\frac {d x}{2}\right )\right )}{960 a^3 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.19, size = 149, normalized size = 1.10 \[ -\frac {45 \, C d x \cos \left (d x + c\right )^{3} + 135 \, C d x \cos \left (d x + c\right )^{2} + 135 \, C d x \cos \left (d x + c\right ) + 45 \, C d x - {\left (15 \, C \cos \left (d x + c\right )^{3} + {\left (7 \, A + 117 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + 72 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.39, size = 151, normalized size = 1.11 \[ -\frac {\frac {180 \, {\left (d x + c\right )} C}{a^{3}} - \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 170, normalized size = 1.25 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{3}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.14, size = 205, normalized size = 1.51 \[ \frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.00, size = 153, normalized size = 1.12 \[ \frac {\left (\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {24\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {4\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-\frac {3\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}-\frac {3\,C\,x}{a^3}+\frac {2\,C\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 9.50, size = 422, normalized size = 3.10 \[ \begin {cases} \frac {3 A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {7 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {5 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {15 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {180 C d x}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {3 C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {27 C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {225 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {375 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\relax (c )}\right ) \cos ^{2}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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