Optimal. Leaf size=103 \[ \frac {a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 x}{4} \]
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Rubi [A] time = 0.10, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 2633, 2635, 8} \[ \frac {a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 x}{4} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 2757
Rubi steps
\begin {align*} \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx &=\int \left (a^2 \cos ^3(c+d x)+2 a^2 \cos ^4(c+d x)+a^2 \cos ^5(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^3(c+d x) \, dx+a^2 \int \cos ^5(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 \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {a^2 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {2 a^2 \sin (c+d x)}{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 \sin ^3(c+d x)}{d}+\frac {a^2 \sin ^5(c+d x)}{5 d}+\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac {3 a^2 x}{4}+\frac {2 a^2 \sin (c+d x)}{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 \sin ^3(c+d x)}{d}+\frac {a^2 \sin ^5(c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 61, normalized size = 0.59 \[ \frac {a^2 (110 \sin (c+d x)+40 \sin (2 (c+d x))+15 \sin (3 (c+d x))+5 \sin (4 (c+d x))+\sin (5 (c+d x))+60 d x)}{80 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.46, size = 76, normalized size = 0.74 \[ \frac {15 \, a^{2} d x + {\left (4 \, a^{2} \cos \left (d x + c\right )^{4} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 15 \, a^{2} \cos \left (d x + c\right ) + 24 \, a^{2}\right )} \sin \left (d x + c\right )}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.54, size = 89, normalized size = 0.86 \[ \frac {3}{4} \, a^{2} x + \frac {a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {3 \, a^{2} \sin \left (3 \, d x + 3 \, c\right )}{16 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {11 \, a^{2} \sin \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 96, normalized size = 0.93 \[ \frac {\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.13, size = 95, normalized size = 0.92 \[ \frac {16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.63, size = 105, normalized size = 1.02 \[ \frac {3\,a^2\,x}{4}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.98, size = 221, normalized size = 2.15 \[ \begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{2} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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