Optimal. Leaf size=148 \[ \frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(8 A+13 B+36 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]
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Rubi [A]
time = 0.28, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4169, 4093,
4085, 3879} \begin {gather*} \frac {(8 A+13 B+36 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 3879
Rule 4085
Rule 4093
Rule 4169
Rubi steps
\begin {align*} \int \frac {\sec ^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) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^2(c+d x) (a (5 A+2 B-2 C)-a (A-B-6 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (-3 a^2 (6 A+B-8 C)+5 a^2 (A-B-6 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B+36 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=\frac {(23 A-2 B-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(6 A+B-8 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B+36 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.76, size = 200, normalized size = 1.35 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (70 (4 A+2 B+3 C) \sin \left (\frac {d x}{2}\right )-35 (5 A+4 B) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )+126 C \sin \left (c+\frac {3 d x}{2}\right )-105 A \sin \left (2 c+\frac {3 d x}{2}\right )+91 A \sin \left (2 c+\frac {5 d x}{2}\right )+56 B \sin \left (2 c+\frac {5 d x}{2}\right )+42 C \sin \left (2 c+\frac {5 d x}{2}\right )+13 A \sin \left (3 c+\frac {7 d x}{2}\right )+8 B \sin \left (3 c+\frac {7 d x}{2}\right )+6 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.65, size = 106, normalized size = 0.72
method | result | size |
derivativedivides | \(\frac {\frac {\left (A -B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A +3 C -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-A +B +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(106\) |
default | \(\frac {\frac {\left (A -B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (-A +3 C -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {\left (-A +B +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(106\) |
risch | \(\frac {2 i \left (105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+175 A \,{\mathrm e}^{4 i \left (d x +c \right )}+140 B \,{\mathrm e}^{4 i \left (d x +c \right )}+280 A \,{\mathrm e}^{3 i \left (d x +c \right )}+140 B \,{\mathrm e}^{3 i \left (d x +c \right )}+210 C \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+126 C \,{\mathrm e}^{2 i \left (d x +c \right )}+91 \,{\mathrm e}^{i \left (d x +c \right )} A +56 B \,{\mathrm e}^{i \left (d x +c \right )}+42 C \,{\mathrm e}^{i \left (d x +c \right )}+13 A +8 B +6 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(177\) |
norman | \(\frac {\frac {\left (A -B +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}-\frac {\left (A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (5 A +4 B +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (11 A -4 B -3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{70 a d}-\frac {\left (11 A -4 B -3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{140 a d}-\frac {\left (19 A +9 B +3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 a d}+\frac {\left (73 A +53 B -39 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a^{3}}\) | \(212\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 259, normalized size = 1.75 \begin {gather*} \frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \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 {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \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 {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \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 {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.01, size = 135, normalized size = 0.91 \begin {gather*} \frac {{\left ({\left (13 \, A + 8 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A + 8 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A + 52 \, B + 39 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 13 \, B + 36 \, C\right )} \sin \left (d x + c\right )}{105 \, {\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 \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 + \int \frac {B \sec ^{3}{\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 \sec ^{4}{\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 = 171, normalized size = 1.16 \begin {gather*} \frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.30, size = 99, normalized size = 0.67 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+B-3\,C\right )}{40\,a^4\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+C\right )}{8\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-A+3\,C\right )}{24\,a^4\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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