Optimal. Leaf size=158 \[ \frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {7 a^2 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {25 a^2 x}{16} \]
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Rubi [A] time = 0.23, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 2638, 2633, 2635} \[ \frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {7 a^2 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {25 a^2 x}{16} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 2638
Rule 2872
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (-2 a^8+2 a^8 \csc (c+d x)+a^8 \csc ^2(c+d x)-6 a^8 \sin (c+d x)+6 a^8 \sin ^3(c+d x)+2 a^8 \sin ^4(c+d x)-2 a^8 \sin ^5(c+d x)-a^8 \sin ^6(c+d x)\right ) \, dx}{a^6}\\ &=-2 a^2 x+a^2 \int \csc ^2(c+d x) \, dx-a^2 \int \sin ^6(c+d x) \, dx+\left (2 a^2\right ) \int \csc (c+d x) \, dx+\left (2 a^2\right ) \int \sin ^4(c+d x) \, dx-\left (2 a^2\right ) \int \sin ^5(c+d x) \, dx-\left (6 a^2\right ) \int \sin (c+d x) \, dx+\left (6 a^2\right ) \int \sin ^3(c+d x) \, dx\\ &=-2 a^2 x-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {6 a^2 \cos (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{6} \left (5 a^2\right ) \int \sin ^4(c+d x) \, dx+\frac {1}{2} \left (3 a^2\right ) \int \sin ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-2 a^2 x-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{8} \left (5 a^2\right ) \int \sin ^2(c+d x) \, dx+\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac {5 a^2 x}{4}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{16} \left (5 a^2\right ) \int 1 \, dx\\ &=-\frac {25 a^2 x}{16}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 110, normalized size = 0.70 \[ \frac {a^2 \left (-255 \sin (2 (c+d x))+15 \sin (4 (c+d x))+5 \sin (6 (c+d x))+2640 \cos (c+d x)+280 \cos (3 (c+d x))+24 \cos (5 (c+d x))-960 \cot (c+d x)+1920 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1920 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-1500 c-1500 d x\right )}{960 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.80, size = 161, normalized size = 1.02 \[ -\frac {40 \, a^{2} \cos \left (d x + c\right )^{7} - 50 \, a^{2} \cos \left (d x + c\right )^{5} - 125 \, a^{2} \cos \left (d x + c\right )^{3} + 240 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 240 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 375 \, a^{2} \cos \left (d x + c\right ) - {\left (96 \, a^{2} \cos \left (d x + c\right )^{5} + 160 \, a^{2} \cos \left (d x + c\right )^{3} - 375 \, a^{2} d x + 480 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 274, normalized size = 1.73 \[ -\frac {375 \, {\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {120 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6720 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2976 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 736 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.42, size = 175, normalized size = 1.11 \[ -\frac {5 a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6 d}-\frac {25 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{24 d}-\frac {25 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}-\frac {25 a^{2} x}{16}-\frac {25 a^{2} c}{16 d}+\frac {2 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {2 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.78, size = 173, normalized size = 1.09 \[ \frac {64 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 120 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{960 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.97, size = 401, normalized size = 2.54 \[ \frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {25\,a^2\,\mathrm {atan}\left (\frac {625\,a^4}{64\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{4}+24\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {47\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{12}+72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {368\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+112\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {299\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+\frac {248\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}-\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {184\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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