Optimal. Leaf size=216 \[ \frac {a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16}-\frac {a^2 b \cos ^6(c+d x)}{2 d}-\frac {a b^2 \sin (c+d x) \cos ^5(c+d x)}{2 d}+\frac {a b^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {3}{16} a b^2 x-\frac {b^3 \sin ^6(c+d x)}{6 d}+\frac {b^3 \sin ^4(c+d x)}{4 d} \]
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Rubi [A] time = 0.21, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3090, 2635, 8, 2565, 30, 2568, 2564, 14} \[ -\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16}-\frac {a b^2 \sin (c+d x) \cos ^5(c+d x)}{2 d}+\frac {a b^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {3}{16} a b^2 x-\frac {b^3 \sin ^6(c+d x)}{6 d}+\frac {b^3 \sin ^4(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 14
Rule 30
Rule 2564
Rule 2565
Rule 2568
Rule 2635
Rule 3090
Rubi steps
\begin {align*} \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=\int \left (a^3 \cos ^6(c+d x)+3 a^2 b \cos ^5(c+d x) \sin (c+d x)+3 a b^2 \cos ^4(c+d x) \sin ^2(c+d x)+b^3 \cos ^3(c+d x) \sin ^3(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^6(c+d x) \, dx+\left (3 a^2 b\right ) \int \cos ^5(c+d x) \sin (c+d x) \, dx+\left (3 a b^2\right ) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+b^3 \int \cos ^3(c+d x) \sin ^3(c+d x) \, dx\\ &=\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {1}{6} \left (5 a^3\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{2} \left (a b^2\right ) \int \cos ^4(c+d x) \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int x^5 \, dx,x,\cos (c+d x)\right )}{d}+\frac {b^3 \operatorname {Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {1}{8} \left (5 a^3\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{8} \left (3 a b^2\right ) \int \cos ^2(c+d x) \, dx+\frac {b^3 \operatorname {Subst}\left (\int \left (x^3-x^5\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {b^3 \sin ^4(c+d x)}{4 d}-\frac {b^3 \sin ^6(c+d x)}{6 d}+\frac {1}{16} \left (5 a^3\right ) \int 1 \, dx+\frac {1}{16} \left (3 a b^2\right ) \int 1 \, dx\\ &=\frac {5 a^3 x}{16}+\frac {3}{16} a b^2 x-\frac {a^2 b \cos ^6(c+d x)}{2 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a b^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a b^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {b^3 \sin ^4(c+d x)}{4 d}-\frac {b^3 \sin ^6(c+d x)}{6 d}\\ \end {align*}
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Mathematica [A] time = 0.30, size = 171, normalized size = 0.79 \[ \frac {a \left (5 a^2+3 b^2\right ) (c+d x)}{16 d}+\frac {3 a \left (5 a^2+b^2\right ) \sin (2 (c+d x))}{64 d}+\frac {3 a \left (a^2-b^2\right ) \sin (4 (c+d x))}{64 d}+\frac {a \left (a^2-3 b^2\right ) \sin (6 (c+d x))}{192 d}-\frac {3 b \left (5 a^2+b^2\right ) \cos (2 (c+d x))}{64 d}-\frac {b \left (3 a^2-b^2\right ) \cos (6 (c+d x))}{192 d}-\frac {3 a^2 b \cos (4 (c+d x))}{32 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 128, normalized size = 0.59 \[ -\frac {12 \, b^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} d x - {\left (8 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 157, normalized size = 0.73 \[ -\frac {3 \, a^{2} b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{16} \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} x - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {3 \, {\left (5 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {3 \, {\left (a^{3} - a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {3 \, {\left (5 \, a^{3} + a b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 9.72, size = 155, normalized size = 0.72 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+3 b^{2} a \left (-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a^{2} b \left (\cos ^{6}\left (d x +c \right )\right )}{2}+a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 131, normalized size = 0.61 \[ -\frac {96 \, a^{2} b \cos \left (d x + c\right )^{6} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 3 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} + 16 \, {\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} b^{3}}{192 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.03, size = 407, normalized size = 1.88 \[ \frac {4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b^2}{8}-\frac {11\,a^3}{8}\right )+4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {3\,a\,b^2}{8}-\frac {11\,a^3}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {39\,a\,b^2}{4}-\frac {15\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {39\,a\,b^2}{4}-\frac {15\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {47\,a\,b^2}{8}-\frac {5\,a^3}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {47\,a\,b^2}{8}-\frac {5\,a^3}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (20\,a^2\,b-\frac {8\,b^3}{3}\right )+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^2+3\,b^2\right )}{8\,\left (\frac {5\,a^3}{8}+\frac {3\,a\,b^2}{8}\right )}\right )\,\left (5\,a^2+3\,b^2\right )}{8\,d}-\frac {a\,\left (5\,a^2+3\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.69, size = 400, normalized size = 1.85 \[ \begin {cases} \frac {5 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a^{2} b \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {3 a b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {b^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{3} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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