Optimal. Leaf size=167 \[ -\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}+\frac {a \left (2 a^2 C+6 A b^2+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+3 a^2 A b x-\frac {a b^2 (6 A-5 C) \tan (c+d x) \sec (c+d x)}{6 d}-\frac {b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]
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Rubi [A] time = 0.31, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4095, 4056, 4048, 3770, 3767, 8} \[ -\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}+\frac {a \left (2 a^2 C+6 A b^2+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+3 a^2 A b x-\frac {a b^2 (6 A-5 C) \tan (c+d x) \sec (c+d x)}{6 d}-\frac {b (3 A-C) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 4048
Rule 4056
Rule 4095
Rubi steps
\begin {align*} \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^2 \left (3 A b+a C \sec (c+d x)-b (3 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \sec (c+d x)) \left (9 a A b+\left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sec (c+d x)-a b (6 A-5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (18 a^2 A b+3 a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \sec (c+d x)-2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=3 a^2 A b x+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{2} \left (a \left (6 A b^2+2 a^2 C+3 b^2 C\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{3} \left (b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=3 a^2 A b x+\frac {a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {\left (b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=3 a^2 A b x+\frac {a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 1.63, size = 325, normalized size = 1.95 \[ \frac {\sec ^3(c+d x) \left (9 a \cos (c+d x) \left (-\left (2 a^2 C+6 A b^2+3 b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (2 a^2 C+6 A b^2+3 b^2 C\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 a A b (c+d x)\right )+3 a \cos (3 (c+d x)) \left (-\left (2 a^2 C+6 A b^2+3 b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (2 a^2 C+6 A b^2+3 b^2 C\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 a A b (c+d x)\right )+2 \sin (c+d x) \left (3 a^3 A \cos (3 (c+d x))+2 \left (9 a^2 b C+3 A b^3+2 b^3 C\right ) \cos (2 (c+d x))+9 a \left (a^2 A+2 b^2 C\right ) \cos (c+d x)+18 a^2 b C+6 A b^3+8 b^3 C\right )\right )}{24 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 178, normalized size = 1.07 \[ \frac {36 \, A a^{2} b d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, C a^{3} + 3 \, {\left (2 \, A + C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, C a^{3} + 3 \, {\left (2 \, A + C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, C a b^{2} \cos \left (d x + c\right ) + 2 \, C b^{3} + 2 \, {\left (9 \, C a^{2} b + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.32, size = 322, normalized size = 1.93 \[ \frac {18 \, {\left (d x + c\right )} A a^{2} b + \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 3 \, {\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.26, size = 195, normalized size = 1.17 \[ \frac {a^{3} A \sin \left (d x +c \right )}{d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 a^{2} A b x +\frac {3 A \,a^{2} b c}{d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A \,b^{3} \tan \left (d x +c \right )}{d}+\frac {2 b^{3} C \tan \left (d x +c \right )}{3 d}+\frac {b^{3} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.37, size = 181, normalized size = 1.08 \[ \frac {36 \, {\left (d x + c\right )} A a^{2} b + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{3} - 9 \, C a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{2} b \tan \left (d x + c\right ) + 12 \, A b^{3} \tan \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.37, size = 464, normalized size = 2.78 \[ \frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {A\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{6}+\frac {A\,b^3\,\sin \left (c+d\,x\right )}{4}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,C\,a^2\,b\,\sin \left (c+d\,x\right )}{4}-\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{2}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,C\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4}-\frac {C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}+\frac {3\,A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{2}-\frac {C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{4}+\frac {9\,A\,a^2\,b\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-\frac {A\,a\,b^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,9{}\mathrm {i}}{2}-\frac {C\,a\,b^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,9{}\mathrm {i}}{4}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]
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 \[ \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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