Optimal. Leaf size=88 \[ \frac {\sin (2 (a-c)+2 x (b-d))}{16 (b-d)}+\frac {\sin (2 (a+c)+2 x (b+d))}{16 (b+d)}+\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 c+2 d x)}{8 d}+\frac {x}{4} \]
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Rubi [A] time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4570, 2637} \[ \frac {\sin (2 (a-c)+2 x (b-d))}{16 (b-d)}+\frac {\sin (2 (a+c)+2 x (b+d))}{16 (b+d)}+\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 c+2 d x)}{8 d}+\frac {x}{4} \]
Antiderivative was successfully verified.
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Rule 2637
Rule 4570
Rubi steps
\begin {align*} \int \cos ^2(a+b x) \cos ^2(c+d x) \, dx &=\int \left (\frac {1}{4}+\frac {1}{4} \cos (2 a+2 b x)+\frac {1}{8} \cos (2 (a-c)+2 (b-d) x)+\frac {1}{4} \cos (2 c+2 d x)+\frac {1}{8} \cos (2 (a+c)+2 (b+d) x)\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{8} \int \cos (2 (a-c)+2 (b-d) x) \, dx+\frac {1}{8} \int \cos (2 (a+c)+2 (b+d) x) \, dx+\frac {1}{4} \int \cos (2 a+2 b x) \, dx+\frac {1}{4} \int \cos (2 c+2 d x) \, dx\\ &=\frac {x}{4}+\frac {\sin (2 a+2 b x)}{8 b}+\frac {\sin (2 (a-c)+2 (b-d) x)}{16 (b-d)}+\frac {\sin (2 c+2 d x)}{8 d}+\frac {\sin (2 (a+c)+2 (b+d) x)}{16 (b+d)}\\ \end {align*}
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Mathematica [A] time = 0.72, size = 105, normalized size = 1.19 \[ \frac {2 d \left (b^2-d^2\right ) \sin (2 (a+b x))+b d (b+d) \sin (2 (a+x (b-d)-c))+b (b-d) (d (\sin (2 (a+x (b+d)+c))+4 x (b+d))+2 (b+d) \sin (2 (c+d x)))}{16 b d (b-d) (b+d)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 105, normalized size = 1.19 \[ -\frac {{\left (2 \, b d^{2} \cos \left (b x + a\right )^{2} - b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b^{3} d - b d^{3}\right )} x - {\left (2 \, b^{2} d \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} - d^{3} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{4 \, {\left (b^{3} d - b d^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.99, size = 80, normalized size = 0.91 \[ \frac {1}{4} \, x + \frac {\sin \left (2 \, b x + 2 \, d x + 2 \, a + 2 \, c\right )}{16 \, {\left (b + d\right )}} + \frac {\sin \left (2 \, b x - 2 \, d x + 2 \, a - 2 \, c\right )}{16 \, {\left (b - d\right )}} + \frac {\sin \left (2 \, b x + 2 \, a\right )}{8 \, b} + \frac {\sin \left (2 \, d x + 2 \, c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.69, size = 83, normalized size = 0.94 \[ \frac {x}{4}+\frac {\sin \left (2 b x +2 a \right )}{8 b}+\frac {\sin \left (2 d x +2 c \right )}{8 d}+\frac {\sin \left (\left (2 b -2 d \right ) x +2 a -2 c \right )}{16 b -16 d}+\frac {\sin \left (\left (2 b +2 d \right ) x +2 a +2 c \right )}{16 b +16 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 620, normalized size = 7.05 \[ \frac {8 \, {\left ({\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{3} - {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2}\right )} d\right )} x + {\left (b^{2} d \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, {\left (b + d\right )} x + 2 \, a + 4 \, c\right ) - {\left (b^{2} d \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, {\left (b + d\right )} x + 2 \, a\right ) - {\left (b^{2} d \sin \left (2 \, c\right ) + b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (-2 \, {\left (b - d\right )} x - 2 \, a + 4 \, c\right ) + {\left (b^{2} d \sin \left (2 \, c\right ) + b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (-2 \, {\left (b - d\right )} x - 2 \, a\right ) + 2 \, {\left (b^{2} d \sin \left (2 \, c\right ) - d^{3} \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) - 2 \, {\left (b^{2} d \sin \left (2 \, c\right ) - d^{3} \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + 2 \, a - 2 \, c\right ) - 2 \, {\left (b^{3} \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, d x\right ) + 2 \, {\left (b^{3} \sin \left (2 \, c\right ) - b d^{2} \sin \left (2 \, c\right )\right )} \cos \left (2 \, d x + 4 \, c\right ) - {\left (b^{2} d \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, {\left (b + d\right )} x + 2 \, a + 4 \, c\right ) - {\left (b^{2} d \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, {\left (b + d\right )} x + 2 \, a\right ) + {\left (b^{2} d \cos \left (2 \, c\right ) + b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (-2 \, {\left (b - d\right )} x - 2 \, a + 4 \, c\right ) + {\left (b^{2} d \cos \left (2 \, c\right ) + b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (-2 \, {\left (b - d\right )} x - 2 \, a\right ) - 2 \, {\left (b^{2} d \cos \left (2 \, c\right ) - d^{3} \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) - 2 \, {\left (b^{2} d \cos \left (2 \, c\right ) - d^{3} \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + 2 \, a - 2 \, c\right ) - 2 \, {\left (b^{3} \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, d x\right ) - 2 \, {\left (b^{3} \cos \left (2 \, c\right ) - b d^{2} \cos \left (2 \, c\right )\right )} \sin \left (2 \, d x + 4 \, c\right )}{32 \, {\left ({\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{3} - {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2}\right )} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.06, size = 177, normalized size = 2.01 \[ \frac {2\,b^3\,\sin \left (2\,c+2\,d\,x\right )-2\,d^3\,\sin \left (2\,a+2\,b\,x\right )+b\,d^2\,\sin \left (2\,a-2\,c+2\,b\,x-2\,d\,x\right )-b\,d^2\,\sin \left (2\,a+2\,c+2\,b\,x+2\,d\,x\right )+b^2\,d\,\sin \left (2\,a-2\,c+2\,b\,x-2\,d\,x\right )+b^2\,d\,\sin \left (2\,a+2\,c+2\,b\,x+2\,d\,x\right )+2\,b^2\,d\,\sin \left (2\,a+2\,b\,x\right )-2\,b\,d^2\,\sin \left (2\,c+2\,d\,x\right )-4\,b\,d^3\,x+4\,b^3\,d\,x}{16\,b\,d\,\left (b^2-d^2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 23.83, size = 1027, normalized size = 11.67 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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