Optimal. Leaf size=62 \[ -\frac {\cos (a+b x)}{2 b}-\frac {\cos (a-2 c+(b-2 d) x)}{4 (b-2 d)}-\frac {\cos (a+2 c+(b+2 d) x)}{4 (b+2 d)} \]
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Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4670, 2718}
\begin {gather*} -\frac {\cos (a+x (b-2 d)-2 c)}{4 (b-2 d)}-\frac {\cos (a+x (b+2 d)+2 c)}{4 (b+2 d)}-\frac {\cos (a+b x)}{2 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 2718
Rule 4670
Rubi steps
\begin {align*} \int \cos ^2(c+d x) \sin (a+b x) \, dx &=\int \left (\frac {1}{2} \sin (a+b x)+\frac {1}{4} \sin (a-2 c+(b-2 d) x)+\frac {1}{4} \sin (a+2 c+(b+2 d) x)\right ) \, dx\\ &=\frac {1}{4} \int \sin (a-2 c+(b-2 d) x) \, dx+\frac {1}{4} \int \sin (a+2 c+(b+2 d) x) \, dx+\frac {1}{2} \int \sin (a+b x) \, dx\\ &=-\frac {\cos (a+b x)}{2 b}-\frac {\cos (a-2 c+(b-2 d) x)}{4 (b-2 d)}-\frac {\cos (a+2 c+(b+2 d) x)}{4 (b+2 d)}\\ \end {align*}
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Mathematica [A]
time = 0.83, size = 71, normalized size = 1.15 \begin {gather*} \frac {1}{4} \left (-\frac {2 \cos (a) \cos (b x)}{b}-\frac {\cos (a-2 c+b x-2 d x)}{b-2 d}-\frac {\cos (a+2 c+b x+2 d x)}{b+2 d}+\frac {2 \sin (a) \sin (b x)}{b}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.21, size = 57, normalized size = 0.92
method | result | size |
default | \(-\frac {\cos \left (b x +a \right )}{2 b}-\frac {\cos \left (a -2 c +\left (b -2 d \right ) x \right )}{4 \left (b -2 d \right )}-\frac {\cos \left (a +2 c +\left (b +2 d \right ) x \right )}{4 \left (b +2 d \right )}\) | \(57\) |
risch | \(-\frac {\cos \left (b x +a \right )}{2 b}-\frac {\cos \left (b x -2 d x +a -2 c \right )}{4 \left (b -2 d \right )}-\frac {\cos \left (b x +2 d x +a +2 c \right )}{4 \left (b +2 d \right )}\) | \(57\) |
norman | \(\frac {\frac {-2 b^{2}+4 d^{2}}{b \left (b^{2}-4 d^{2}\right )}+\frac {\left (-2 b^{2}+4 d^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b \left (b^{2}-4 d^{2}\right )}-\frac {8 d \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}-4 d^{2}}+\frac {8 d \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}-4 d^{2}}-\frac {4 b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}-4 d^{2}}+\frac {8 d^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b \left (b^{2}-4 d^{2}\right )}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\) | \(228\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 414 vs.
\(2 (56) = 112\).
time = 0.30, size = 414, normalized size = 6.68 \begin {gather*} -\frac {{\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left ({\left (b + 2 \, d\right )} x + a + 4 \, c\right ) + {\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left ({\left (b + 2 \, d\right )} x + a\right ) + {\left (b^{2} \cos \left (2 \, c\right ) + 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left (-{\left (b - 2 \, d\right )} x - a + 4 \, c\right ) + {\left (b^{2} \cos \left (2 \, c\right ) + 2 \, b d \cos \left (2 \, c\right )\right )} \cos \left (-{\left (b - 2 \, d\right )} x - a\right ) + 2 \, {\left (b^{2} \cos \left (2 \, c\right ) - 4 \, d^{2} \cos \left (2 \, c\right )\right )} \cos \left (b x + a + 2 \, c\right ) + 2 \, {\left (b^{2} \cos \left (2 \, c\right ) - 4 \, d^{2} \cos \left (2 \, c\right )\right )} \cos \left (b x + a - 2 \, c\right ) + {\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left ({\left (b + 2 \, d\right )} x + a + 4 \, c\right ) - {\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left ({\left (b + 2 \, d\right )} x + a\right ) + {\left (b^{2} \sin \left (2 \, c\right ) + 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left (-{\left (b - 2 \, d\right )} x - a + 4 \, c\right ) - {\left (b^{2} \sin \left (2 \, c\right ) + 2 \, b d \sin \left (2 \, c\right )\right )} \sin \left (-{\left (b - 2 \, d\right )} x - a\right ) + 2 \, {\left (b^{2} \sin \left (2 \, c\right ) - 4 \, d^{2} \sin \left (2 \, c\right )\right )} \sin \left (b x + a + 2 \, c\right ) - 2 \, {\left (b^{2} \sin \left (2 \, c\right ) - 4 \, d^{2} \sin \left (2 \, c\right )\right )} \sin \left (b x + a - 2 \, c\right )}{8 \, {\left (b^{3} \cos \left (2 \, c\right )^{2} + b^{3} \sin \left (2 \, c\right )^{2} - 4 \, {\left (b \cos \left (2 \, c\right )^{2} + b \sin \left (2 \, c\right )^{2}\right )} d^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.96, size = 66, normalized size = 1.06 \begin {gather*} -\frac {b^{2} \cos \left (b x + a\right ) \cos \left (d x + c\right )^{2} + 2 \, b d \cos \left (d x + c\right ) \sin \left (b x + a\right ) \sin \left (d x + c\right ) - 2 \, d^{2} \cos \left (b x + a\right )}{b^{3} - 4 \, b d^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 408 vs.
\(2 (51) = 102\).
time = 0.80, size = 408, normalized size = 6.58 \begin {gather*} \begin {cases} x \sin {\left (a \right )} \cos ^{2}{\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\left (\frac {x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {\sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d}\right ) \sin {\left (a \right )} & \text {for}\: b = 0 \\- \frac {x \sin {\left (a - 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{4} + \frac {x \sin {\left (a - 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {x \sin {\left (c + d x \right )} \cos {\left (a - 2 d x \right )} \cos {\left (c + d x \right )}}{2} + \frac {3 \sin {\left (a - 2 d x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {\sin ^{2}{\left (c + d x \right )} \cos {\left (a - 2 d x \right )}}{2 d} & \text {for}\: b = - 2 d \\- \frac {x \sin {\left (a + 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{4} + \frac {x \sin {\left (a + 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {x \sin {\left (c + d x \right )} \cos {\left (a + 2 d x \right )} \cos {\left (c + d x \right )}}{2} + \frac {3 \sin {\left (a + 2 d x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {\sin ^{2}{\left (c + d x \right )} \cos {\left (a + 2 d x \right )}}{2 d} & \text {for}\: b = 2 d \\- \frac {b^{2} \cos {\left (a + b x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 b d \sin {\left (a + b x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 d^{2} \sin ^{2}{\left (c + d x \right )} \cos {\left (a + b x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 d^{2} \cos {\left (a + b x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 56, normalized size = 0.90 \begin {gather*} -\frac {\cos \left (b x + 2 \, d x + a + 2 \, c\right )}{4 \, {\left (b + 2 \, d\right )}} - \frac {\cos \left (b x - 2 \, d x + a - 2 \, c\right )}{4 \, {\left (b - 2 \, d\right )}} - \frac {\cos \left (b x + a\right )}{2 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.77, size = 97, normalized size = 1.56 \begin {gather*} \frac {d\,\left (2\,b\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )-2\,b\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )\right )+b^2\,\cos \left (a-2\,c+b\,x-2\,d\,x\right )+b^2\,\cos \left (a+2\,c+b\,x+2\,d\,x\right )}{16\,b\,d^2-4\,b^3}-\frac {\cos \left (a+b\,x\right )}{2\,b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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