Optimal. Leaf size=129 \[ \frac {2 (36 A+C) \tan (c+d x)}{15 a^3 d}-\frac {3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {3 A \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rubi [A] time = 0.44, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 2978, 2748, 3767, 8, 3770} \[ \frac {2 (36 A+C) \tan (c+d x)}{15 a^3 d}-\frac {3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {3 A \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2978
Rule 3042
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (6 A+C)-a (3 A-2 C) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (27 A+2 C)-2 a^2 (9 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (2 a^3 (36 A+C)-45 a^3 A \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6}\\ &=-\frac {(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(3 A) \int \sec (c+d x) \, dx}{a^3}+\frac {(2 (36 A+C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=-\frac {3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(2 (36 A+C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac {3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A+C) \tan (c+d x)}{15 a^3 d}-\frac {(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 6.32, size = 596, normalized size = 4.62 \[ \frac {\frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos (c+d x) \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-600 A \sin \left (c-\frac {d x}{2}\right )+375 A \sin \left (c+\frac {d x}{2}\right )-480 A \sin \left (2 c+\frac {d x}{2}\right )-60 A \sin \left (c+\frac {3 d x}{2}\right )+402 A \sin \left (2 c+\frac {3 d x}{2}\right )-225 A \sin \left (3 c+\frac {3 d x}{2}\right )+315 A \sin \left (c+\frac {5 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )+240 A \sin \left (3 c+\frac {5 d x}{2}\right )-45 A \sin \left (4 c+\frac {5 d x}{2}\right )+72 A \sin \left (2 c+\frac {7 d x}{2}\right )+15 A \sin \left (3 c+\frac {7 d x}{2}\right )+57 A \sin \left (4 c+\frac {7 d x}{2}\right )-255 A \sin \left (\frac {d x}{2}\right )+567 A \sin \left (\frac {3 d x}{2}\right )-10 C \sin \left (c-\frac {d x}{2}\right )+10 C \sin \left (c+\frac {d x}{2}\right )-20 C \sin \left (2 c+\frac {d x}{2}\right )+22 C \sin \left (2 c+\frac {3 d x}{2}\right )+10 C \sin \left (c+\frac {5 d x}{2}\right )+10 C \sin \left (3 c+\frac {5 d x}{2}\right )+2 C \sin \left (2 c+\frac {7 d x}{2}\right )+2 C \sin \left (4 c+\frac {7 d x}{2}\right )-20 C \sin \left (\frac {d x}{2}\right )+22 C \sin \left (\frac {3 d x}{2}\right )\right ) \left (A \sec ^2(c+d x)+C\right )}{60 d (\cos (c+d x)+1)^3 (2 A+C \cos (2 c+2 d x)+C)}+\frac {48 A \cos ^2(c+d x) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sec ^2(c+d x)+C\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (\cos (c+d x)+1)^3 (2 A+C \cos (2 c+2 d x)+C)}-\frac {48 A \cos ^2(c+d x) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sec ^2(c+d x)+C\right ) \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (\cos (c+d x)+1)^3 (2 A+C \cos (2 c+2 d x)+C)}}{a^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 222, normalized size = 1.72 \[ -\frac {45 \, {\left (A \cos \left (d x + c\right )^{4} + 3 \, A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (A \cos \left (d x + c\right )^{4} + 3 \, A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (36 \, A + C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (57 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (117 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.48, size = 178, normalized size = 1.38 \[ -\frac {\frac {180 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {180 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.21, size = 204, normalized size = 1.58 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 233, normalized size = 1.81 \[ \frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {C {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.89, size = 150, normalized size = 1.16 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{6\,a^3}+\frac {A}{3\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{4\,a^3}+\frac {2\,A}{a^3}+\frac {6\,A-2\,C}{4\,a^3}\right )}{d}-\frac {6\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]
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 \[ \frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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