Optimal. Leaf size=145 \[ \frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {3 C \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(A-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.43, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4085, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {3 C \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(A-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 4008
Rule 4019
Rule 4085
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A+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 C)-a (A+6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-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-9 C)-a^2 (2 A+27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sec (c+d x) \left (-45 a^3 C+a^3 (2 A+27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(3 C) \int \sec (c+d x) \, dx}{a^3}+\frac {(2 A+27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(2 A+27 C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac {3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {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 = 3.06, size = 457, normalized size = 3.15 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-10 A \sin \left (c-\frac {d x}{2}\right )+10 A \sin \left (c+\frac {d x}{2}\right )-20 A \sin \left (2 c+\frac {d x}{2}\right )+22 A \sin \left (2 c+\frac {3 d x}{2}\right )+10 A \sin \left (c+\frac {5 d x}{2}\right )+10 A \sin \left (3 c+\frac {5 d x}{2}\right )+2 A \sin \left (2 c+\frac {7 d x}{2}\right )+2 A \sin \left (4 c+\frac {7 d x}{2}\right )-5 (4 A+51 C) \sin \left (\frac {d x}{2}\right )+(22 A+567 C) \sin \left (\frac {3 d x}{2}\right )-600 C \sin \left (c-\frac {d x}{2}\right )+375 C \sin \left (c+\frac {d x}{2}\right )-480 C \sin \left (2 c+\frac {d x}{2}\right )-60 C \sin \left (c+\frac {3 d x}{2}\right )+402 C \sin \left (2 c+\frac {3 d x}{2}\right )-225 C \sin \left (3 c+\frac {3 d x}{2}\right )+315 C \sin \left (c+\frac {5 d x}{2}\right )+30 C \sin \left (2 c+\frac {5 d x}{2}\right )+240 C \sin \left (3 c+\frac {5 d x}{2}\right )-45 C \sin \left (4 c+\frac {5 d x}{2}\right )+72 C \sin \left (2 c+\frac {7 d x}{2}\right )+15 C \sin \left (3 c+\frac {7 d x}{2}\right )+57 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )+2880 C \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{60 a^3 d (\sec (c+d x)+1)^3 (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 222, normalized size = 1.53 \[ -\frac {45 \, {\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (A + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A + 117 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 178, normalized size = 1.23 \[ -\frac {\frac {180 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {180 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.65, size = 204, normalized size = 1.41 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{3}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {C}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{3}}-\frac {C}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 233, normalized size = 1.61 \[ \frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.64, size = 150, normalized size = 1.03 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{6\,a^3}+\frac {C}{3\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{4\,a^3}+\frac {2\,C}{a^3}-\frac {2\,A-6\,C}{4\,a^3}\right )}{d}-\frac {6\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]
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 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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