Optimal. Leaf size=156 \[ -\frac {2 (8 A-5 B+2 C) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(8 A-5 B+2 C) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {x (7 A-4 B+2 C)}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.33, antiderivative size = 156, normalized size of antiderivative = 1.00, 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 = {4084, 4020, 3787, 2635, 8, 2637} \[ -\frac {2 (8 A-5 B+2 C) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(8 A-5 B+2 C) \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {x (7 A-4 B+2 C)}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2635
Rule 2637
Rule 3787
Rule 4020
Rule 4084
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) (a (5 A-2 B+2 C)-3 a (A-B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(8 A-5 B+2 C) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \cos ^2(c+d x) \left (3 a^2 (7 A-4 B+2 C)-2 a^2 (8 A-5 B+2 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(8 A-5 B+2 C) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 (8 A-5 B+2 C)) \int \cos (c+d x) \, dx}{3 a^2}+\frac {(7 A-4 B+2 C) \int \cos ^2(c+d x) \, dx}{a^2}\\ &=-\frac {2 (8 A-5 B+2 C) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(8 A-5 B+2 C) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(7 A-4 B+2 C) \int 1 \, dx}{2 a^2}\\ &=\frac {(7 A-4 B+2 C) x}{2 a^2}-\frac {2 (8 A-5 B+2 C) \sin (c+d x)}{3 a^2 d}+\frac {(7 A-4 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(8 A-5 B+2 C) \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 1.53, size = 377, normalized size = 2.42 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (36 d x (7 A-4 B+2 C) \cos \left (c+\frac {d x}{2}\right )+36 d x (7 A-4 B+2 C) \cos \left (\frac {d x}{2}\right )+147 A \sin \left (c+\frac {d x}{2}\right )-239 A \sin \left (c+\frac {3 d x}{2}\right )-63 A \sin \left (2 c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {5 d x}{2}\right )-15 A \sin \left (3 c+\frac {5 d x}{2}\right )+3 A \sin \left (3 c+\frac {7 d x}{2}\right )+3 A \sin \left (4 c+\frac {7 d x}{2}\right )+84 A d x \cos \left (c+\frac {3 d x}{2}\right )+84 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 A \sin \left (\frac {d x}{2}\right )-120 B \sin \left (c+\frac {d x}{2}\right )+164 B \sin \left (c+\frac {3 d x}{2}\right )+36 B \sin \left (2 c+\frac {3 d x}{2}\right )+12 B \sin \left (2 c+\frac {5 d x}{2}\right )+12 B \sin \left (3 c+\frac {5 d x}{2}\right )-48 B d x \cos \left (c+\frac {3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 B \sin \left (\frac {d x}{2}\right )+96 C \sin \left (c+\frac {d x}{2}\right )-80 C \sin \left (c+\frac {3 d x}{2}\right )+24 C d x \cos \left (c+\frac {3 d x}{2}\right )+24 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-144 C \sin \left (\frac {d x}{2}\right )\right )}{192 a^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 153, normalized size = 0.98 \[ \frac {3 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (7 \, A - 4 \, B + 2 \, C\right )} d x + {\left (3 \, A \cos \left (d x + c\right )^{3} - 6 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 28 \, B + 10 \, C\right )} \cos \left (d x + c\right ) - 32 \, A + 20 \, B - 8 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 198, normalized size = 1.27 \[ \frac {\frac {3 \, {\left (d x + c\right )} {\left (7 \, A - 4 \, B + 2 \, C\right )}}{a^{2}} - \frac {6 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.25, size = 309, normalized size = 1.98 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 352, normalized size = 2.26 \[ -\frac {A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.28, size = 159, normalized size = 1.02 \[ \frac {x\,\left (7\,A-4\,B+2\,C\right )}{2\,a^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{2\,a^2}+\frac {4\,A-2\,B}{2\,a^2}\right )}{d}-\frac {\left (5\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \]
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 \[ \frac {\int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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