3.3.39 \(\int \cos (a+b x) \cos ^2(c+d x) \, dx\) [239]

Optimal. Leaf size=62 \[ \frac {\sin (a+b x)}{2 b}+\frac {\sin (a-2 c+(b-2 d) x)}{4 (b-2 d)}+\frac {\sin (a+2 c+(b+2 d) x)}{4 (b+2 d)} \]

[Out]

1/2*sin(b*x+a)/b+1/4*sin(a-2*c+(b-2*d)*x)/(b-2*d)+1/4*sin(a+2*c+(b+2*d)*x)/(b+2*d)

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Rubi [A]
time = 0.04, 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 = {4666, 2717} \begin {gather*} \frac {\sin (a+x (b-2 d)-2 c)}{4 (b-2 d)}+\frac {\sin (a+x (b+2 d)+2 c)}{4 (b+2 d)}+\frac {\sin (a+b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cos[c + d*x]^2,x]

[Out]

Sin[a + b*x]/(2*b) + Sin[a - 2*c + (b - 2*d)*x]/(4*(b - 2*d)) + Sin[a + 2*c + (b + 2*d)*x]/(4*(b + 2*d))

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 4666

Int[Cos[v_]^(p_.)*Cos[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Cos[v]^p*Cos[w]^q, x], x] /; ((PolynomialQ[
v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x])) && IGtQ[p, 0] && IGtQ[q
, 0]

Rubi steps

\begin {align*} \int \cos (a+b x) \cos ^2(c+d x) \, dx &=\int \left (\frac {1}{2} \cos (a+b x)+\frac {1}{4} \cos (a-2 c+(b-2 d) x)+\frac {1}{4} \cos (a+2 c+(b+2 d) x)\right ) \, dx\\ &=\frac {1}{4} \int \cos (a-2 c+(b-2 d) x) \, dx+\frac {1}{4} \int \cos (a+2 c+(b+2 d) x) \, dx+\frac {1}{2} \int \cos (a+b x) \, dx\\ &=\frac {\sin (a+b x)}{2 b}+\frac {\sin (a-2 c+(b-2 d) x)}{4 (b-2 d)}+\frac {\sin (a+2 c+(b+2 d) x)}{4 (b+2 d)}\\ \end {align*}

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Mathematica [A]
time = 0.81, size = 69, normalized size = 1.11 \begin {gather*} \frac {1}{4} \left (\frac {2 \cos (b x) \sin (a)}{b}+\frac {2 \cos (a) \sin (b x)}{b}+\frac {\sin (a-2 c+b x-2 d x)}{b-2 d}+\frac {\sin (a+2 c+b x+2 d x)}{b+2 d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Cos[c + d*x]^2,x]

[Out]

((2*Cos[b*x]*Sin[a])/b + (2*Cos[a]*Sin[b*x])/b + Sin[a - 2*c + b*x - 2*d*x]/(b - 2*d) + Sin[a + 2*c + b*x + 2*
d*x]/(b + 2*d))/4

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Maple [A]
time = 0.23, size = 57, normalized size = 0.92

method	result	size
default	\(\frac {\sin \left (b x +a \right )}{2 b}+\frac {\sin \left (a -2 c +\left (b -2 d \right ) x \right )}{4 b -8 d}+\frac {\sin \left (a +2 c +\left (b +2 d \right ) x \right )}{4 b +8 d}\)	\(57\)
risch	\(\frac {\sin \left (b x +a \right )}{2 b}+\frac {\sin \left (b x -2 d x +a -2 c \right ) b}{4 \left (b -2 d \right ) \left (b +2 d \right )}+\frac {\sin \left (b x -2 d x +a -2 c \right ) d}{2 \left (b -2 d \right ) \left (b +2 d \right )}+\frac {\sin \left (b x +2 d x +a +2 c \right ) b}{4 \left (b -2 d \right ) \left (b +2 d \right )}-\frac {\sin \left (b x +2 d x +a +2 c \right ) d}{2 \left (b -2 d \right ) \left (b +2 d \right )}\)	\(133\)
norman	\(\frac {-\frac {4 d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}-4 d^{2}}+\frac {4 d \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}-4 d^{2}}+\frac {4 d \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}-4 d^{2}}-\frac {4 d \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}-4 d^{2}}+\frac {2 \left (b^{2}-2 d^{2}\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\left (b^{2}-4 d^{2}\right ) b}+\frac {2 \left (b^{2}-2 d^{2}\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (b^{2}-4 d^{2}\right ) b}-\frac {4 \left (b^{2}+2 d^{2}\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b \left (b^{2}-4 d^{2}\right )}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*cos(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*sin(b*x+a)/b+1/4*sin(a-2*c+(b-2*d)*x)/(b-2*d)+1/4*sin(a+2*c+(b+2*d)*x)/(b+2*d)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (56) = 112\).
time = 0.34, size = 416, normalized size = 6.71 \begin {gather*} -\frac {{\left (b^{2} \sin \left (2 \, c\right ) - 2 \, b d \sin \left (2 \, c\right )\right )} \cos \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 )} \cos \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 )} \cos \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 )} \cos \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 )} \cos \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 )} \cos \left (b x + a - 2 \, c\right ) - {\left (b^{2} \cos \left (2 \, c\right ) - 2 \, b d \cos \left (2 \, c\right )\right )} \sin \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 )} \sin \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 )} \sin \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 )} \sin \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 )} \sin \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 )} \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.

[In]

integrate(cos(b*x+a)*cos(d*x+c)^2,x, algorithm="maxima")

[Out]

-1/8*((b^2*sin(2*c) - 2*b*d*sin(2*c))*cos((b + 2*d)*x + a + 4*c) - (b^2*sin(2*c) - 2*b*d*sin(2*c))*cos((b + 2*
d)*x + a) - (b^2*sin(2*c) + 2*b*d*sin(2*c))*cos(-(b - 2*d)*x - a + 4*c) + (b^2*sin(2*c) + 2*b*d*sin(2*c))*cos(
-(b - 2*d)*x - a) + 2*(b^2*sin(2*c) - 4*d^2*sin(2*c))*cos(b*x + a + 2*c) - 2*(b^2*sin(2*c) - 4*d^2*sin(2*c))*c
os(b*x + a - 2*c) - (b^2*cos(2*c) - 2*b*d*cos(2*c))*sin((b + 2*d)*x + a + 4*c) - (b^2*cos(2*c) - 2*b*d*cos(2*c
))*sin((b + 2*d)*x + a) + (b^2*cos(2*c) + 2*b*d*cos(2*c))*sin(-(b - 2*d)*x - a + 4*c) + (b^2*cos(2*c) + 2*b*d*
cos(2*c))*sin(-(b - 2*d)*x - a) - 2*(b^2*cos(2*c) - 4*d^2*cos(2*c))*sin(b*x + a + 2*c) - 2*(b^2*cos(2*c) - 4*d
^2*cos(2*c))*sin(b*x + a - 2*c))/(b^3*cos(2*c)^2 + b^3*sin(2*c)^2 - 4*(b*cos(2*c)^2 + b*sin(2*c)^2)*d^2)

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Fricas [A]
time = 2.39, size = 63, normalized size = 1.02 \begin {gather*} -\frac {2 \, b d \cos \left (b x + a\right ) \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{b^{3} - 4 \, b d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cos(d*x+c)^2,x, algorithm="fricas")

[Out]

-(2*b*d*cos(b*x + a)*cos(d*x + c)*sin(d*x + c) - (b^2*cos(d*x + c)^2 - 2*d^2)*sin(b*x + a))/(b^3 - 4*b*d^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (49) = 98\).
time = 0.80, size = 408, normalized size = 6.58 \begin {gather*} \begin {cases} x \cos {\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 ) \cos {\left (a \right )} & \text {for}\: b = 0 \\- \frac {x \sin {\left (a - 2 d x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2} - \frac {x \sin ^{2}{\left (c + d x \right )} \cos {\left (a - 2 d x \right )}}{4} + \frac {x \cos {\left (a - 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {\sin {\left (a - 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 \sin {\left (c + d x \right )} \cos {\left (a - 2 d x \right )} \cos {\left (c + d x \right )}}{4 d} & \text {for}\: b = - 2 d \\\frac {x \sin {\left (a + 2 d x \right )} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2} - \frac {x \sin ^{2}{\left (c + d x \right )} \cos {\left (a + 2 d x \right )}}{4} + \frac {x \cos {\left (a + 2 d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {\sin {\left (a + 2 d x \right )} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 \sin {\left (c + d x \right )} \cos {\left (a + 2 d x \right )} \cos {\left (c + d x \right )}}{4 d} & \text {for}\: b = 2 d \\\frac {b^{2} \sin {\left (a + b x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 b d \sin {\left (c + d x \right )} \cos {\left (a + b x \right )} \cos {\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 d^{2} \sin {\left (a + b x \right )} \sin ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 d^{2} \sin {\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.

[In]

integrate(cos(b*x+a)*cos(d*x+c)**2,x)

[Out]

Piecewise((x*cos(a)*cos(c)**2, Eq(b, 0) & Eq(d, 0)), ((x*sin(c + d*x)**2/2 + x*cos(c + d*x)**2/2 + sin(c + d*x
)*cos(c + d*x)/(2*d))*cos(a), Eq(b, 0)), (-x*sin(a - 2*d*x)*sin(c + d*x)*cos(c + d*x)/2 - x*sin(c + d*x)**2*co
s(a - 2*d*x)/4 + x*cos(a - 2*d*x)*cos(c + d*x)**2/4 - sin(a - 2*d*x)*sin(c + d*x)**2/(2*d) + 3*sin(c + d*x)*co
s(a - 2*d*x)*cos(c + d*x)/(4*d), Eq(b, -2*d)), (x*sin(a + 2*d*x)*sin(c + d*x)*cos(c + d*x)/2 - x*sin(c + d*x)*
*2*cos(a + 2*d*x)/4 + x*cos(a + 2*d*x)*cos(c + d*x)**2/4 + sin(a + 2*d*x)*sin(c + d*x)**2/(2*d) + 3*sin(c + d*
x)*cos(a + 2*d*x)*cos(c + d*x)/(4*d), Eq(b, 2*d)), (b**2*sin(a + b*x)*cos(c + d*x)**2/(b**3 - 4*b*d**2) - 2*b*
d*sin(c + d*x)*cos(a + b*x)*cos(c + d*x)/(b**3 - 4*b*d**2) - 2*d**2*sin(a + b*x)*sin(c + d*x)**2/(b**3 - 4*b*d
**2) - 2*d**2*sin(a + b*x)*cos(c + d*x)**2/(b**3 - 4*b*d**2), True))

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Giac [A]
time = 0.39, size = 56, normalized size = 0.90 \begin {gather*} \frac {\sin \left (b x + 2 \, d x + a + 2 \, c\right )}{4 \, {\left (b + 2 \, d\right )}} + \frac {\sin \left (b x - 2 \, d x + a - 2 \, c\right )}{4 \, {\left (b - 2 \, d\right )}} + \frac {\sin \left (b x + a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cos(d*x+c)^2,x, algorithm="giac")

[Out]

1/4*sin(b*x + 2*d*x + a + 2*c)/(b + 2*d) + 1/4*sin(b*x - 2*d*x + a - 2*c)/(b - 2*d) + 1/2*sin(b*x + a)/b

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Mupad [B]
time = 1.02, size = 98, normalized size = 1.58 \begin {gather*} \frac {\sin \left (a+b\,x\right )}{2\,b}-\frac {d\,\left (2\,b\,\sin \left (a-2\,c+b\,x-2\,d\,x\right )-2\,b\,\sin \left (a+2\,c+b\,x+2\,d\,x\right )\right )+b^2\,\sin \left (a-2\,c+b\,x-2\,d\,x\right )+b^2\,\sin \left (a+2\,c+b\,x+2\,d\,x\right )}{16\,b\,d^2-4\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*cos(c + d*x)^2,x)

[Out]

sin(a + b*x)/(2*b) - (d*(2*b*sin(a - 2*c + b*x - 2*d*x) - 2*b*sin(a + 2*c + b*x + 2*d*x)) + b^2*sin(a - 2*c +
b*x - 2*d*x) + b^2*sin(a + 2*c + b*x + 2*d*x))/(16*b*d^2 - 4*b^3)

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