Optimal. Leaf size=78 \[ -\frac {5 a^2 \cos ^3(c+d x)}{12 d}-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^2 x}{8} \]
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Rubi [A] time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2678, 2669, 2635, 8} \[ -\frac {5 a^2 \cos ^3(c+d x)}{12 d}-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^2 x}{8} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2635
Rule 2669
Rule 2678
Rubi steps
\begin {align*} \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}+\frac {1}{4} (5 a) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {5 a^2 \cos ^3(c+d x)}{12 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}+\frac {1}{4} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {5 a^2 \cos ^3(c+d x)}{12 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}+\frac {1}{8} \left (5 a^2\right ) \int 1 \, dx\\ &=\frac {5 a^2 x}{8}-\frac {5 a^2 \cos ^3(c+d x)}{12 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.30, size = 131, normalized size = 1.68 \[ -\frac {a^2 \left (30 \sqrt {1-\sin (c+d x)} \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\sin (c+d x)+1} \left (6 \sin ^4(c+d x)+10 \sin ^3(c+d x)-7 \sin ^2(c+d x)-25 \sin (c+d x)+16\right )\right ) \cos ^3(c+d x)}{24 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{3/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 59, normalized size = 0.76 \[ -\frac {16 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} d x + 3 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.44, size = 72, normalized size = 0.92 \[ \frac {5}{8} \, a^{2} x - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{2 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 87, normalized size = 1.12 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right ) a^{2}}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 65, normalized size = 0.83 \[ -\frac {64 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.71, size = 237, normalized size = 3.04 \[ \frac {5\,a^2\,x}{8}-\frac {\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {a^2\,\left (15\,c+15\,d\,x\right )}{24}-\frac {a^2\,\left (15\,c+15\,d\,x-32\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{6}-\frac {a^2\,\left (60\,c+60\,d\,x-32\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{6}-\frac {a^2\,\left (60\,c+60\,d\,x-96\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{4}-\frac {a^2\,\left (90\,c+90\,d\,x-96\right )}{24}\right )-\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.00, size = 180, normalized size = 2.31 \[ \begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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