Optimal. Leaf size=187 \[ -\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {4 (19 B-9 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 B-6 C)}{2 a^3}-\frac {(11 B-6 C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.55, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4072, 4020, 3787, 2635, 8, 2637} \[ -\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {4 (19 B-9 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 B-6 C)}{2 a^3}-\frac {(11 B-6 C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2635
Rule 2637
Rule 3787
Rule 4020
Rule 4072
Rubi steps
\begin {align*} \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac {\cos ^2(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) (a (7 B-2 C)-4 a (B-C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (43 B-18 C)-3 a^2 (11 B-6 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \cos ^2(c+d x) \left (15 a^3 (13 B-6 C)-8 a^3 (19 B-9 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(8 (19 B-9 C)) \int \cos (c+d x) \, dx}{15 a^3}+\frac {(13 B-6 C) \int \cos ^2(c+d x) \, dx}{a^3}\\ &=-\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(13 B-6 C) \int 1 \, dx}{2 a^3}\\ &=\frac {(13 B-6 C) x}{2 a^3}-\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 0.75, size = 435, normalized size = 2.33 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (600 d x (13 B-6 C) \cos \left (c+\frac {d x}{2}\right )+7560 B \sin \left (c+\frac {d x}{2}\right )-9230 B \sin \left (c+\frac {3 d x}{2}\right )+930 B \sin \left (2 c+\frac {3 d x}{2}\right )-2782 B \sin \left (2 c+\frac {5 d x}{2}\right )-750 B \sin \left (3 c+\frac {5 d x}{2}\right )-105 B \sin \left (3 c+\frac {7 d x}{2}\right )-105 B \sin \left (4 c+\frac {7 d x}{2}\right )+15 B \sin \left (4 c+\frac {9 d x}{2}\right )+15 B \sin \left (5 c+\frac {9 d x}{2}\right )+3900 B d x \cos \left (c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+600 d x (13 B-6 C) \cos \left (\frac {d x}{2}\right )-12760 B \sin \left (\frac {d x}{2}\right )-4500 C \sin \left (c+\frac {d x}{2}\right )+4860 C \sin \left (c+\frac {3 d x}{2}\right )-900 C \sin \left (2 c+\frac {3 d x}{2}\right )+1452 C \sin \left (2 c+\frac {5 d x}{2}\right )+300 C \sin \left (3 c+\frac {5 d x}{2}\right )+60 C \sin \left (3 c+\frac {7 d x}{2}\right )+60 C \sin \left (4 c+\frac {7 d x}{2}\right )-1800 C d x \cos \left (c+\frac {3 d x}{2}\right )-1800 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-360 C d x \cos \left (2 c+\frac {5 d x}{2}\right )-360 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+7020 C \sin \left (\frac {d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 190, normalized size = 1.02 \[ \frac {15 \, {\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (13 \, B - 6 \, C\right )} d x + {\left (15 \, B \cos \left (d x + c\right )^{4} - 15 \, {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (479 \, B - 234 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (239 \, B - 114 \, C\right )} \cos \left (d x + c\right ) - 304 \, B + 144 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 200, normalized size = 1.07 \[ \frac {\frac {30 \, {\left (d x + c\right )} {\left (13 \, B - 6 \, C\right )}}{a^{3}} - \frac {60 \, {\left (7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.48, size = 292, normalized size = 1.56 \[ -\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{2 d \,a^{3}}-\frac {31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3}}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.56, size = 322, normalized size = 1.72 \[ -\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.05, size = 204, normalized size = 1.09 \[ \frac {x\,\left (13\,B-6\,C\right )}{2\,a^3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^3}+\frac {3\,\left (5\,B-3\,C\right )}{4\,a^3}+\frac {10\,B-2\,C}{4\,a^3}\right )}{d}-\frac {\left (7\,B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,B-2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{4\,a^3}+\frac {5\,B-3\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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