Optimal. Leaf size=110 \[ -\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-C) \tan (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B+7 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Rubi [A]
time = 0.15, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {4163, 4085,
3879} \begin {gather*} \frac {(2 A+3 B+7 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(A-C) \tan (c+d x)}{3 a d (a \sec (c+d x)+a)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 3879
Rule 4085
Rule 4163
Rubi steps
\begin {align*} \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) (a (4 A+B-C)-a (A-B-4 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-C) \tan (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B+7 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-C) \tan (c+d x)}{3 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B+7 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.62, size = 156, normalized size = 1.42 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (5 (8 A+3 B+4 C) \sin \left (\frac {d x}{2}\right )-15 (2 A+B) \sin \left (c+\frac {d x}{2}\right )+20 A \sin \left (c+\frac {3 d x}{2}\right )+15 B \sin \left (c+\frac {3 d x}{2}\right )+10 C \sin \left (c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {3 d x}{2}\right )+7 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+2 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{240 a^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.60, size = 113, normalized size = 1.03
method | result | size |
derivativedivides | \(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(113\) |
default | \(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(113\) |
risch | \(\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}+15 B \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{2 i \left (d x +c \right )}+20 C \,{\mathrm e}^{2 i \left (d x +c \right )}+20 \,{\mathrm e}^{i \left (d x +c \right )} A +15 B \,{\mathrm e}^{i \left (d x +c \right )}+10 C \,{\mathrm e}^{i \left (d x +c \right )}+7 A +3 B +2 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(141\) |
norman | \(\frac {\frac {\left (A -B +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (4 A +3 B +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (8 A -3 B -2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}+\frac {\left (19 A +6 B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{2}}\) | \(154\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 179, normalized size = 1.63 \begin {gather*} \frac {\frac {C {\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 {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}} + \frac {3 \, B {\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.44, size = 102, normalized size = 0.93 \begin {gather*} \frac {{\left ({\left (7 \, A + 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + 3 \, B + 7 \, 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 )}} \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 \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 {B \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 + \int \frac {C \sec ^{3}{\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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 115, normalized size = 1.05 \begin {gather*} \frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.19, size = 73, normalized size = 0.66 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-2\,C\right )}{12\,a^3\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+C\right )}{4\,a^3\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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