Optimal. Leaf size=125 \[ \frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.14, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {4054, 12, 3791, 3770, 3767, 8, 3768} \[ \frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 12
Rule 3767
Rule 3768
Rule 3770
Rule 3791
Rule 4054
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+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 (4 B+3 C) \sec (c+d x) (a+a \sec (c+d x))^3 \, dx}{4 a}\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} (4 B+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} (4 B+3 C) \int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \left (a^3 (4 B+3 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{4} \left (a^3 (4 B+3 C)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{8} \left (3 a^3 (4 B+3 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (4 B+3 C)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{4 d}-\frac {\left (3 a^3 (4 B+3 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{4 d}\\ &=\frac {5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}\\ \end {align*}
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Mathematica [A] time = 0.50, size = 81, normalized size = 0.65 \[ \frac {a^3 \left (15 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 (B+3 C) \tan ^2(c+d x)+9 (4 B+5 C) \sec (c+d x)+96 (B+C)+6 C \sec ^3(c+d x)\right )\right )}{24 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 145, normalized size = 1.16 \[ \frac {15 \, {\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (4 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.86, size = 212, normalized size = 1.70 \[ \frac {15 \, {\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.47, size = 188, normalized size = 1.50 \[ \frac {5 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} C \tan \left (d x +c \right )}{d}+\frac {11 a^{3} B \tan \left (d x +c \right )}{3 d}+\frac {15 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 262, normalized size = 2.10 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.42, size = 185, normalized size = 1.48 \[ \frac {\left (-5\,B\,a^3-\frac {15\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,B\,a^3}{3}+\frac {55\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,B\,a^3}{3}-\frac {73\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,B\,a^3+\frac {49\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,B+3\,C\right )}{4\,d} \]
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 \[ a^{3} \left (\int B \sec {\left (c + d x \right )}\, dx + \int 3 B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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