Optimal. Leaf size=239 \[ \frac {(21 A-8 B+2 C) x}{2 a^4}-\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3} \]
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Rubi [A]
time = 0.49, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4169, 4105,
3872, 2715, 8, 2717} \begin {gather*} -\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B+20 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A-52 B+10 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A-8 B+2 C)}{2 a^4}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 2715
Rule 2717
Rule 3872
Rule 4105
Rule 4169
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))^4} \, dx &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\cos ^2(c+d x) (a (9 A-2 B+2 C)-a (5 A-5 B-2 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (73 A-24 B+10 C)-28 a^2 (2 A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) \left (a^3 (477 A-176 B+50 C)-3 a^3 (129 A-52 B+10 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {\int \cos ^2(c+d x) \left (105 a^4 (21 A-8 B+2 C)-8 a^4 (216 A-83 B+20 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(21 A-8 B+2 C) \int \cos ^2(c+d x) \, dx}{a^4}-\frac {(8 (216 A-83 B+20 C)) \int \cos (c+d x) \, dx}{105 a^4}\\ &=-\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(21 A-8 B+2 C) \int 1 \, dx}{2 a^4}\\ &=\frac {(21 A-8 B+2 C) x}{2 a^4}-\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 5.56, size = 345, normalized size = 1.44 \begin {gather*} \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \left (15 (A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-6 (39 A-32 B+25 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+4 (447 A-286 B+160 C) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-8 (1653 A-764 B+260 C) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+210 \cos ^7\left (\frac {1}{2} (c+d x)\right ) (2 (21 A-8 B+2 C) d x+4 (-4 A+B) \sin (c+d x)+A \sin (2 (c+d x)))+15 (A-B+C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )-6 (39 A-32 B+25 C) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+4 (447 A-286 B+160 C) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{105 a^4 d (1+\cos (c+d x))^4 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.86, size = 244, normalized size = 1.02 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 474 vs.
\(2 (225) = 450\).
time = 0.51, size = 474, normalized size = 1.98 \begin {gather*} -\frac {3 \, A {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {5880 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \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 {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + 5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.19, size = 267, normalized size = 1.12 \begin {gather*} \frac {105 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x + {\left (105 \, A \cos \left (d x + c\right )^{5} - 210 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (1509 \, A - 592 \, B + 130 \, C\right )} \cos \left (d x + c\right )^{3} - 4 \, {\left (3411 \, A - 1318 \, B + 310 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (11619 \, A - 4472 \, B + 1070 \, C\right )} \cos \left (d x + c\right ) - 3456 \, A + 1328 \, B - 320 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end {gather*}
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 \begin {gather*} \frac {\int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \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 ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \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 ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.50, size = 302, normalized size = 1.26 \begin {gather*} \frac {\frac {420 \, {\left (d x + c\right )} {\left (21 \, A - 8 \, B + 2 \, C\right )}}{a^{4}} - \frac {840 \, {\left (9 \, 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} + 7 \, 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^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 189 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11655 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.27, size = 285, normalized size = 1.19 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {6\,A-4\,B+2\,C}{8\,a^4}-\frac {5\,B-15\,A+C}{24\,a^4}+\frac {A-B+C}{4\,a^4}\right )}{d}-\frac {\left (9\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (7\,A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {6\,A-4\,B+2\,C}{40\,a^4}+\frac {3\,\left (A-B+C\right )}{40\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (6\,A-4\,B+2\,C\right )}{4\,a^4}-\frac {3\,\left (5\,B-15\,A+C\right )}{8\,a^4}+\frac {5\,\left (A-B+C\right )}{4\,a^4}+\frac {20\,A-4\,C}{8\,a^4}\right )}{d}+\frac {x\,\left (21\,A-8\,B+2\,C\right )}{2\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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