Optimal. Leaf size=147 \[ \frac {5 a^3 (4 A+3 C) \tan (c+d x)}{8 d}+\frac {a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A+5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+a^3 A x+\frac {C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.22, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4055, 3917, 3914, 3767, 8, 3770} \[ \frac {5 a^3 (4 A+3 C) \tan (c+d x)}{8 d}+\frac {a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A+5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+a^3 A x+\frac {C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3914
Rule 3917
Rule 4055
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\int (a+a \sec (c+d x))^3 (4 a A+3 a C \sec (c+d x)) \, dx}{4 a}\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac {\int (a+a \sec (c+d x))^2 \left (12 a^2 A+3 a^2 (4 A+5 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac {(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}+\frac {\int (a+a \sec (c+d x)) \left (24 a^3 A+15 a^3 (4 A+3 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=a^3 A x+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac {(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}+\frac {1}{8} \left (5 a^3 (4 A+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (a^3 (28 A+15 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac {a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac {(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}-\frac {\left (5 a^3 (4 A+3 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=a^3 A x+\frac {a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^3 (4 A+3 C) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac {(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}\\ \end {align*}
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Mathematica [B] time = 2.06, size = 363, normalized size = 2.47 \[ \frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (\sec (c) (4 A \sin (2 c+d x)+72 A \sin (c+2 d x)-24 A \sin (3 c+2 d x)+4 A \sin (2 c+3 d x)+4 A \sin (4 c+3 d x)+24 A \sin (3 c+4 d x)+24 A d x \cos (c)+16 A d x \cos (c+2 d x)+16 A d x \cos (3 c+2 d x)+4 A d x \cos (3 c+4 d x)+4 A d x \cos (5 c+4 d x)-72 A \sin (c)+4 A \sin (d x)+23 C \sin (2 c+d x)+88 C \sin (c+2 d x)-8 C \sin (3 c+2 d x)+15 C \sin (2 c+3 d x)+15 C \sin (4 c+3 d x)+24 C \sin (3 c+4 d x)-72 C \sin (c)+23 C \sin (d x))-8 (28 A+15 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{256 d (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.87, size = 151, normalized size = 1.03 \[ \frac {16 \, A a^{3} d x \cos \left (d x + c\right )^{4} + {\left (28 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (28 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, {\left (A + C\right )} a^{3} \cos \left (d x + c\right )^{3} + {\left (4 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, C a^{3} \cos \left (d x + c\right ) + 2 \, C a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.35, size = 222, normalized size = 1.51 \[ \frac {8 \, {\left (d x + c\right )} A a^{3} + {\left (28 \, A a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (28 \, A a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (20 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 68 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 55 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 73 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 49 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.66, size = 180, normalized size = 1.22 \[ a^{3} A x +\frac {A \,a^{3} c}{d}+\frac {3 a^{3} C \tan \left (d x +c \right )}{d}+\frac {7 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {15 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 250, normalized size = 1.70 \[ \frac {16 \, {\left (d x + c\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, A a^{3} \tan \left (d x + c\right ) + 16 \, C a^{3} \tan \left (d x + c\right )}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.65, size = 231, normalized size = 1.57 \[ \frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {15\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {15\,C\,a^3\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4} \]
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 \[ a^{3} \left (\int A\, dx + \int 3 A \sec {\left (c + d x \right )}\, dx + \int 3 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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