Optimal. Leaf size=82 \[ \frac {(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac {2 A \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {2 A x}{a^2}-\frac {(A+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.22, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4085, 4020, 3787, 2637, 8} \[ \frac {(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac {2 A \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {2 A x}{a^2}-\frac {(A+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2637
Rule 3787
Rule 4020
Rule 4085
Rubi steps
\begin {align*} \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos (c+d x) (-a (4 A+C)+a (2 A-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \cos (c+d x) \left (-a^2 (10 A+C)+6 a^2 A \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 A) \int 1 \, dx}{a^2}+\frac {(10 A+C) \int \cos (c+d x) \, dx}{3 a^2}\\ &=-\frac {2 A x}{a^2}+\frac {(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 0.99, size = 195, normalized size = 2.38 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-30 A \sin \left (c+\frac {d x}{2}\right )+41 A \sin \left (c+\frac {3 d x}{2}\right )+9 A \sin \left (2 c+\frac {3 d x}{2}\right )+3 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 A \sin \left (3 c+\frac {5 d x}{2}\right )-36 A d x \cos \left (c+\frac {d x}{2}\right )-12 A d x \cos \left (c+\frac {3 d x}{2}\right )-12 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+66 A \sin \left (\frac {d x}{2}\right )-36 A d x \cos \left (\frac {d x}{2}\right )-12 C \sin \left (c+\frac {d x}{2}\right )+8 C \sin \left (c+\frac {3 d x}{2}\right )+12 C \sin \left (\frac {d x}{2}\right )\right )}{48 a^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 102, normalized size = 1.24 \[ -\frac {6 \, A d x \cos \left (d x + c\right )^{2} + 12 \, A d x \cos \left (d x + c\right ) + 6 \, A d x - {\left (3 \, A \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A + C\right )} \cos \left (d x + c\right ) + 10 \, A + C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.54, size = 114, normalized size = 1.39 \[ -\frac {\frac {12 \, {\left (d x + c\right )} A}{a^{2}} - \frac {12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.26, size = 130, normalized size = 1.59 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 165, normalized size = 2.01 \[ \frac {A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.65, size = 99, normalized size = 1.21 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}+\frac {3\,A-C}{2\,a^2}\right )}{d}-\frac {2\,A\,x}{a^2}+\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]
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 \[ \frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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