Optimal. Leaf size=161 \[ \frac {(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.45, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4084, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac {(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}+\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(A+4 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 4008
Rule 4019
Rule 4084
Rubi steps
\begin {align*} \int \frac {\sec ^3(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 ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) (a (2 A+3 B-3 C)+a (A-B+6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A+4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2 (A+4 B-9 C)+a^2 (2 A-7 B+27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A+4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec (c+d x) \left (-15 a^3 (B-3 C)-a^3 (2 A-7 B+27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A+4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(B-3 C) \int \sec (c+d x) \, dx}{a^3}+\frac {(2 A-7 B+27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A+4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(2 A-7 B+27 C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(B-3 C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(2 A-7 B+27 C) \tan (c+d x)}{15 a^3 d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A+4 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(B-3 C) \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 6.38, size = 839, normalized size = 5.21 \[ \frac {16 (3 C-B) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^3}-\frac {16 (3 C-B) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^3}+\frac {\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-20 A \sin \left (\frac {d x}{2}\right )+160 B \sin \left (\frac {d x}{2}\right )-255 C \sin \left (\frac {d x}{2}\right )+22 A \sin \left (\frac {3 d x}{2}\right )-167 B \sin \left (\frac {3 d x}{2}\right )+567 C \sin \left (\frac {3 d x}{2}\right )-10 A \sin \left (c-\frac {d x}{2}\right )+170 B \sin \left (c-\frac {d x}{2}\right )-600 C \sin \left (c-\frac {d x}{2}\right )+10 A \sin \left (c+\frac {d x}{2}\right )-170 B \sin \left (c+\frac {d x}{2}\right )+375 C \sin \left (c+\frac {d x}{2}\right )-20 A \sin \left (2 c+\frac {d x}{2}\right )+160 B \sin \left (2 c+\frac {d x}{2}\right )-480 C \sin \left (2 c+\frac {d x}{2}\right )+75 B \sin \left (c+\frac {3 d x}{2}\right )-60 C \sin \left (c+\frac {3 d x}{2}\right )+22 A \sin \left (2 c+\frac {3 d x}{2}\right )-167 B \sin \left (2 c+\frac {3 d x}{2}\right )+402 C \sin \left (2 c+\frac {3 d x}{2}\right )+75 B \sin \left (3 c+\frac {3 d x}{2}\right )-225 C \sin \left (3 c+\frac {3 d x}{2}\right )+10 A \sin \left (c+\frac {5 d x}{2}\right )-95 B \sin \left (c+\frac {5 d x}{2}\right )+315 C \sin \left (c+\frac {5 d x}{2}\right )+15 B \sin \left (2 c+\frac {5 d x}{2}\right )+30 C \sin \left (2 c+\frac {5 d x}{2}\right )+10 A \sin \left (3 c+\frac {5 d x}{2}\right )-95 B \sin \left (3 c+\frac {5 d x}{2}\right )+240 C \sin \left (3 c+\frac {5 d x}{2}\right )+15 B \sin \left (4 c+\frac {5 d x}{2}\right )-45 C \sin \left (4 c+\frac {5 d x}{2}\right )+2 A \sin \left (2 c+\frac {7 d x}{2}\right )-22 B \sin \left (2 c+\frac {7 d x}{2}\right )+72 C \sin \left (2 c+\frac {7 d x}{2}\right )+15 C \sin \left (3 c+\frac {7 d x}{2}\right )+2 A \sin \left (4 c+\frac {7 d x}{2}\right )-22 B \sin \left (4 c+\frac {7 d x}{2}\right )+57 C \sin \left (4 c+\frac {7 d x}{2}\right )\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{60 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (\sec (c+d x) a+a)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 263, normalized size = 1.63 \[ \frac {15 \, {\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 3 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (A - 11 \, B + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A - 17 \, B + 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 32 \, B + 117 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 234, normalized size = 1.45 \[ \frac {\frac {60 \, {\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (B - 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} + \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.66, size = 303, normalized size = 1.88 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{2 d \,a^{3}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{3}}-\frac {C}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{3}}-\frac {C}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 350, normalized size = 2.17 \[ \frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.32, size = 176, normalized size = 1.09 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B+C}{6\,a^3}-\frac {2\,B-4\,C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,A-6\,C}{4\,a^3}-\frac {3\,\left (A-B+C\right )}{4\,a^3}+\frac {2\,B-4\,C}{2\,a^3}\right )}{d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-3\,C\right )}{a^3\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \]
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 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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