Optimal. Leaf size=115 \[ -\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {A x}{a^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.20, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4052, 3922, 3919, 3794} \[ -\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {A x}{a^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 3794
Rule 3919
Rule 3922
Rule 4052
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-5 a A+a (2 A-2 B-3 C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B-3 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 0.94, size = 289, normalized size = 2.51 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (270 A \sin \left (c+\frac {d x}{2}\right )-230 A \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-64 A \sin \left (2 c+\frac {5 d x}{2}\right )+150 A d x \cos \left (c+\frac {d x}{2}\right )+75 A d x \cos \left (c+\frac {3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-370 A \sin \left (\frac {d x}{2}\right )+150 A d x \cos \left (\frac {d x}{2}\right )-60 B \sin \left (c+\frac {d x}{2}\right )+40 B \sin \left (c+\frac {3 d x}{2}\right )-30 B \sin \left (2 c+\frac {3 d x}{2}\right )+14 B \sin \left (2 c+\frac {5 d x}{2}\right )+80 B \sin \left (\frac {d x}{2}\right )-30 C \sin \left (c+\frac {d x}{2}\right )+30 C \sin \left (c+\frac {3 d x}{2}\right )+6 C \sin \left (2 c+\frac {5 d x}{2}\right )+30 C \sin \left (\frac {d x}{2}\right )\right )}{480 a^3 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 147, normalized size = 1.28 \[ \frac {15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x - {\left ({\left (32 \, A - 7 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 22 \, A - 2 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 153, normalized size = 1.33 \[ \frac {\frac {60 \, {\left (d x + c\right )} A}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.84, size = 175, normalized size = 1.52 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{3}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 205, normalized size = 1.78 \[ -\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.44, size = 158, normalized size = 1.37 \[ \frac {A\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]
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 \[ \frac {\int \frac {A}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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