Optimal. Leaf size=119 \[ -\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^2 x}{4} \]
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Rubi [A] time = 0.14, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2873, 2635, 8, 2592, 302, 206, 2565, 30} \[ -\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^2 x}{4} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 206
Rule 302
Rule 2565
Rule 2592
Rule 2635
Rule 2873
Rubi steps
\begin {align*} \int \cos ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (2 a^2 \cos ^4(c+d x)+a^2 \cos ^3(c+d x) \cot (c+d x)+a^2 \cos ^4(c+d x) \sin (c+d x)\right ) \, dx\\ &=a^2 \int \cos ^3(c+d x) \cot (c+d x) \, dx+a^2 \int \cos ^4(c+d x) \sin (c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \, dx\\ &=\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^2 \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}+\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx-\frac {a^2 \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {3 a^2 x}{4}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {3 a^2 x}{4}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.85, size = 96, normalized size = 0.81 \[ \frac {a^2 \left (270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \cos (5 (c+d x))+15 \left (8 \sin (2 (c+d x))+\sin (4 (c+d x))+4 \left (4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 c+3 d x\right )\right )\right )}{240 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 115, normalized size = 0.97 \[ -\frac {12 \, a^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a^{2} d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 181, normalized size = 1.52 \[ \frac {45 \, {\left (d x + c\right )} a^{2} + 60 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 68 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.46, size = 127, normalized size = 1.07 \[ -\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{4 d}+\frac {3 a^{2} x}{4}+\frac {3 a^{2} c}{4 d}+\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 98, normalized size = 0.82 \[ -\frac {48 \, a^{2} \cos \left (d x + c\right )^{5} - 40 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.22, size = 293, normalized size = 2.46 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {3\,a^2\,\mathrm {atan}\left (\frac {9\,a^4}{4\,\left (3\,a^4-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}+\frac {3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^4-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}\right )}{2\,d}+\frac {-\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {34\,a^2}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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