Optimal. Leaf size=140 \[ \frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
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Rubi [A] time = 0.19, antiderivative size = 140, 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 = {4077, 4047, 3767, 4046, 3768, 3770} \[ \frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 3767
Rule 3768
Rule 3770
Rule 4046
Rule 4047
Rule 4077
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+a (5 A+4 C) \sec (c+d x)+5 a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+5 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (a (5 A+4 C)) \int \sec ^4(c+d x) \, dx\\ &=\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (a (4 A+3 C)) \int \sec ^3(c+d x) \, dx-\frac {(a (5 A+4 C)) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (a (4 A+3 C)) \int \sec (c+d x) \, dx\\ &=\frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 0.75, size = 93, normalized size = 0.66 \[ \frac {a \left (15 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 (A+2 C) \tan ^2(c+d x)+15 (A+C)+3 C \tan ^4(c+d x)\right )+15 (4 A+3 C) \sec (c+d x)+30 C \sec ^3(c+d x)\right )\right )}{120 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 147, normalized size = 1.05 \[ \frac {15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 30 \, C a \cos \left (d x + c\right ) + 24 \, C a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 218, normalized size = 1.56 \[ \frac {15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 200 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 130 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 440 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.76, size = 192, normalized size = 1.37 \[ \frac {a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a C \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a A \tan \left (d x +c \right )}{3 d}+\frac {a A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {8 a C \tan \left (d x +c \right )}{15 d}+\frac {a C \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a C \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 175, normalized size = 1.25 \[ \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.19, size = 201, normalized size = 1.44 \[ \frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {\left (A\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {10\,A\,a}{3}-\frac {13\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,C\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {22\,A\,a}{3}-\frac {19\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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 \[ a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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