Optimal. Leaf size=132 \[ -\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16} \]
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Rubi [A] time = 0.30, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2568, 2635, 8, 2565, 14, 270} \[ -\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16} \]
Antiderivative was successfully verified.
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Rule 8
Rule 14
Rule 270
Rule 2565
Rule 2568
Rule 2635
Rule 2873
Rubi steps
\begin {align*} \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (a^3 \cos ^2(c+d x) \sin ^2(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^3(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^4(c+d x)+a^3 \cos ^2(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+a^3 \int \cos ^2(c+d x) \sin ^5(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{4} a^3 \int \cos ^2(c+d x) \, dx+\frac {1}{2} \left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{8} a^3 \int 1 \, dx+\frac {1}{8} \left (3 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 x}{8}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{16} \left (3 a^3\right ) \int 1 \, dx\\ &=\frac {5 a^3 x}{16}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.67, size = 86, normalized size = 0.65 \[ \frac {a^3 (-63 \sin (2 (c+d x))-105 \sin (4 (c+d x))+21 \sin (6 (c+d x))-609 \cos (c+d x)-91 \cos (3 (c+d x))+63 \cos (5 (c+d x))-3 \cos (7 (c+d x))+420 c+420 d x)}{1344 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 98, normalized size = 0.74 \[ -\frac {48 \, a^{3} \cos \left (d x + c\right )^{7} - 336 \, a^{3} \cos \left (d x + c\right )^{5} + 448 \, a^{3} \cos \left (d x + c\right )^{3} - 105 \, a^{3} d x - 21 \, {\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 18 \, a^{3} \cos \left (d x + c\right )^{3} + 5 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{336 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 123, normalized size = 0.93 \[ \frac {5}{16} \, a^{3} x - \frac {a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {3 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {13 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {29 \, a^{3} \cos \left (d x + c\right )}{64 \, d} + \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac {5 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {3 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 194, normalized size = 1.47 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{7}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (d x +c \right )\right )}{105}\right )+3 a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+3 a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+a^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 129, normalized size = 0.98 \[ -\frac {64 \, {\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 1344 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} + 105 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 210 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{6720 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.16, size = 331, normalized size = 2.51 \[ \frac {5\,a^3\,x}{16}-\frac {\frac {5\,a^3\,\left (c+d\,x\right )}{16}+\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {119\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}+\frac {119\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{2}-\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}-\frac {a^3\,\left (105\,c+105\,d\,x-320\right )}{336}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {35\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (735\,c+735\,d\,x-2240\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {105\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (2205\,c+2205\,d\,x-2688\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {175\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (3675\,c+3675\,d\,x-896\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {105\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (2205\,c+2205\,d\,x-4032\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {175\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (3675\,c+3675\,d\,x-10304\right )}{336}\right )+\frac {5\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 7.84, size = 379, normalized size = 2.87 \[ \begin {cases} \frac {3 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {4 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {8 a^{3} \cos ^{7}{\left (c + d x \right )}}{105 d} - \frac {2 a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \sin ^{2}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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