Optimal. Leaf size=122 \[ \frac {3 a^2 \tan ^3(c+d x)}{5 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{4 d} \]
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Rubi [A] time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3788, 3768, 3770, 4046, 3767} \[ \frac {3 a^2 \tan ^3(c+d x)}{5 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 3767
Rule 3768
Rule 3770
Rule 3788
Rule 4046
Rubi steps
\begin {align*} \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int \sec ^5(c+d x) \, dx+\int \sec ^4(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} \left (3 a^2\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (9 a^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (3 a^2\right ) \int \sec (c+d x) \, dx-\frac {\left (9 a^2\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {3 a^2 \tan ^3(c+d x)}{5 d}\\ \end {align*}
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Mathematica [B] time = 1.72, size = 487, normalized size = 3.99 \[ -\frac {a^2 \sec (c) \sec ^5(c+d x) \left (80 \sin (2 c+d x)-140 \sin (c+2 d x)-140 \sin (3 c+2 d x)-240 \sin (2 c+3 d x)-30 \sin (3 c+4 d x)-30 \sin (5 c+4 d x)-48 \sin (4 c+5 d x)+75 \cos (2 c+3 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+75 \cos (4 c+3 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+15 \cos (4 c+5 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+15 \cos (6 c+5 d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+150 \cos (d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+150 \cos (2 c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-75 \cos (2 c+3 d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-75 \cos (4 c+3 d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (4 c+5 d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (6 c+5 d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-400 \sin (d x)\right )}{640 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 124, normalized size = 1.02 \[ \frac {15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, a^{2} \cos \left (d x + c\right )^{4} + 15 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{40 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.47, size = 138, normalized size = 1.13 \[ \frac {15 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 65 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.91, size = 124, normalized size = 1.02 \[ \frac {6 a^{2} \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {a^{2} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 133, normalized size = 1.09 \[ \frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 15 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.70, size = 170, normalized size = 1.39 \[ \frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}-9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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 \[ a^{2} \left (\int \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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