Optimal. Leaf size=126 \[ \frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a^2 x}{8}-\frac {a b \cos ^4(c+d x)}{2 d}-\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {b^2 x}{8} \]
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Rubi [A] time = 0.13, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3090, 2635, 8, 2565, 30, 2568} \[ \frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a^2 x}{8}-\frac {a b \cos ^4(c+d x)}{2 d}-\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {b^2 x}{8} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2565
Rule 2568
Rule 2635
Rule 3090
Rubi steps
\begin {align*} \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^4(c+d x)+2 a b \cos ^3(c+d x) \sin (c+d x)+b^2 \cos ^2(c+d x) \sin ^2(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^4(c+d x) \, dx+(2 a b) \int \cos ^3(c+d x) \sin (c+d x) \, dx+b^2 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx\\ &=\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{4} b^2 \int \cos ^2(c+d x) \, dx-\frac {(2 a b) \operatorname {Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} \left (3 a^2\right ) \int 1 \, dx+\frac {1}{8} b^2 \int 1 \, dx\\ &=\frac {3 a^2 x}{8}+\frac {b^2 x}{8}-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 98, normalized size = 0.78 \[ \frac {\left (3 a^2+b^2\right ) (c+d x)}{8 d}+\frac {\left (a^2-b^2\right ) \sin (4 (c+d x))}{32 d}+\frac {a^2 \sin (2 (c+d x))}{4 d}-\frac {a b \cos (2 (c+d x))}{4 d}-\frac {a b \cos (4 (c+d x))}{16 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 75, normalized size = 0.60 \[ -\frac {4 \, a b \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} + b^{2}\right )} d x - {\left (2 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 4.97, size = 85, normalized size = 0.67 \[ \frac {1}{8} \, {\left (3 \, a^{2} + b^{2}\right )} x - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {a b \cos \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.42, size = 97, normalized size = 0.77 \[ \frac {b^{2} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a b \left (\cos ^{4}\left (d x +c \right )\right )}{2}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 75, normalized size = 0.60 \[ -\frac {16 \, a b \cos \left (d x + c\right )^{4} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{32 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.58, size = 89, normalized size = 0.71 \[ \frac {4\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {a^2\,\sin \left (4\,c+4\,d\,x\right )}{2}-\frac {b^2\,\sin \left (4\,c+4\,d\,x\right )}{2}+2\,a\,b\,{\sin \left (2\,c+2\,d\,x\right )}^2+8\,a\,b\,{\sin \left (c+d\,x\right )}^2+6\,a^2\,d\,x+2\,b^2\,d\,x}{16\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.05, size = 238, normalized size = 1.89 \[ \begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {a b \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{2} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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