Optimal. Leaf size=91 \[ \frac {(10 A+C) \tan (c+d x)}{3 a^2 d}-\frac {2 A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {2 A \tan (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.30, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {(10 A+C) \tan (c+d x)}{3 a^2 d}-\frac {2 A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {2 A \tan (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
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))^2} \, dx &=-\frac {(A+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(a (4 A+C)-a (2 A-C) \cos (c+d x)) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 A \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \left (a^2 (10 A+C)-6 a^2 A \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{3 a^4}\\ &=-\frac {2 A \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(2 A) \int \sec (c+d x) \, dx}{a^2}+\frac {(10 A+C) \int \sec ^2(c+d x) \, dx}{3 a^2}\\ &=-\frac {2 A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {2 A \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(10 A+C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=-\frac {2 A \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {(10 A+C) \tan (c+d x)}{3 a^2 d}-\frac {2 A \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 1.55, size = 288, normalized size = 3.16 \[ \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left ((A+C) \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+(A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (7 A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+6 A \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2 (2 A+C \cos (2 (c+d x))+C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 167, normalized size = 1.84 \[ -\frac {3 \, {\left (A \cos \left (d x + c\right )^{3} + 2 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A \cos \left (d x + c\right )^{3} + 2 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (10 \, A + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A + C\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.47, size = 142, normalized size = 1.56 \[ -\frac {\frac {12 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {12 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 164, normalized size = 1.80 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 191, normalized size = 2.10 \[ \frac {A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.95, size = 113, normalized size = 1.24 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}+\frac {3\,A-C}{2\,a^2}\right )}{d}-\frac {4\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]
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 \[ \frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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