Optimal. Leaf size=119 \[ \frac {b \left (6 a^2 B+6 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 (2 a A-b B) \sin (c+d x)}{2 d}+a^2 x (a B+3 A b)+\frac {b^2 (2 a B+A b) \tan (c+d x)}{d}+\frac {b B \sin (c+d x) (a+b \sec (c+d x))^2}{2 d} \]
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Rubi [A] time = 0.22, antiderivative size = 131, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4025, 4048, 3770, 3767, 8} \[ -\frac {b \left (2 a^2 A-3 a b B-A b^2\right ) \tan (c+d x)}{d}+\frac {b \left (6 a^2 B+6 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (a B+3 A b)-\frac {b^2 (2 a A-b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a A \sin (c+d x) (a+b \sec (c+d x))^2}{d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 4025
Rule 4048
Rubi steps
\begin {align*} \int \cos (c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\int (a+b \sec (c+d x)) \left (-a (3 A b+a B)-b (A b+2 a B) \sec (c+d x)+b (2 a A-b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b^2 (2 a A-b B) \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 a^2 (3 A b+a B)-b \left (6 a A b+6 a^2 B+b^2 B\right ) \sec (c+d x)+2 b \left (2 a^2 A-A b^2-3 a b B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 (3 A b+a B) x+\frac {a A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b^2 (2 a A-b B) \sec (c+d x) \tan (c+d x)}{2 d}-\left (b \left (2 a^2 A-A b^2-3 a b B\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (b \left (6 a A b+6 a^2 B+b^2 B\right )\right ) \int \sec (c+d x) \, dx\\ &=a^2 (3 A b+a B) x+\frac {b \left (6 a A b+6 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b^2 (2 a A-b B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (b \left (2 a^2 A-A b^2-3 a b B\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=a^2 (3 A b+a B) x+\frac {b \left (6 a A b+6 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b \left (2 a^2 A-A b^2-3 a b B\right ) \tan (c+d x)}{d}-\frac {b^2 (2 a A-b B) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [B] time = 0.98, size = 399, normalized size = 3.35 \[ \frac {\sec ^2(c+d x) \left (\left (a^3 A+2 b^3 B\right ) \sin (c+d x)+a^3 A \sin (3 (c+d x))+2 a^3 B c+2 a^3 B d x+\cos (2 (c+d x)) \left (-b \left (6 a^2 B+6 a A b+b^2 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b \left (6 a^2 B+6 a A b+b^2 B\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 (c+d x) (a B+3 A b)\right )+6 a^2 A b c+6 a^2 A b d x-6 a^2 b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a^2 b B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-6 a A b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a A b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 a b^2 B \sin (2 (c+d x))+2 A b^3 \sin (2 (c+d x))-b^3 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^3 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 167, normalized size = 1.40 \[ \frac {4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d x \cos \left (d x + c\right )^{2} + {\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{3} \cos \left (d x + c\right )^{2} + B b^{3} + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.38, size = 241, normalized size = 2.03 \[ \frac {\frac {4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} {\left (d x + c\right )} + {\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.17, size = 172, normalized size = 1.45 \[ \frac {a^{3} A \sin \left (d x +c \right )}{d}+a^{3} B x +\frac {a^{3} B c}{d}+3 A x \,a^{2} b +\frac {3 A \,a^{2} b c}{d}+\frac {3 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 B a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {A \,b^{3} \tan \left (d x +c \right )}{d}+\frac {b^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.77, size = 169, normalized size = 1.42 \[ \frac {4 \, {\left (d x + c\right )} B a^{3} + 12 \, {\left (d x + c\right )} A a^{2} b - B b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{3} \sin \left (d x + c\right ) + 12 \, B a b^{2} \tan \left (d x + c\right ) + 4 \, A b^{3} \tan \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.60, size = 249, normalized size = 2.09 \[ \frac {\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {B\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{2}-3\,A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}+B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}\right )}{d} \]
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 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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