3.135 \(\int \cos (a+b x) \sin (2 a+2 b x) \, dx\)

Optimal. Leaf size=30 \[ -\frac {\cos (a+b x)}{2 b}-\frac {\cos (3 a+3 b x)}{6 b} \]

[Out]

-1/2*cos(b*x+a)/b-1/6*cos(3*b*x+3*a)/b

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4284} \[ -\frac {\cos (a+b x)}{2 b}-\frac {\cos (3 a+3 b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x],x]

[Out]

-Cos[a + b*x]/(2*b) - Cos[3*a + 3*b*x]/(6*b)

Rule 4284

Int[cos[(c_.) + (d_.)*(x_)]*sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[Cos[a - c + (b - d)*x]/(2*(b - d)), x]
- Simp[Cos[a + c + (b + d)*x]/(2*(b + d)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - d^2, 0]

Rubi steps

\begin {align*} \int \cos (a+b x) \sin (2 a+2 b x) \, dx &=-\frac {\cos (a+b x)}{2 b}-\frac {\cos (3 a+3 b x)}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.50 \[ -\frac {2 \cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x],x]

[Out]

(-2*Cos[a + b*x]^3)/(3*b)

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fricas [A]  time = 0.48, size = 13, normalized size = 0.43 \[ -\frac {2 \, \cos \left (b x + a\right )^{3}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a),x, algorithm="fricas")

[Out]

-2/3*cos(b*x + a)^3/b

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giac [A]  time = 0.28, size = 26, normalized size = 0.87 \[ -\frac {\cos \left (3 \, b x + 3 \, a\right )}{6 \, b} - \frac {\cos \left (b x + a\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a),x, algorithm="giac")

[Out]

-1/6*cos(3*b*x + 3*a)/b - 1/2*cos(b*x + a)/b

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maple [A]  time = 0.22, size = 27, normalized size = 0.90 \[ -\frac {\cos \left (b x +a \right )}{2 b}-\frac {\cos \left (3 b x +3 a \right )}{6 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*sin(2*b*x+2*a),x)

[Out]

-1/2*cos(b*x+a)/b-1/6*cos(3*b*x+3*a)/b

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maxima [A]  time = 0.32, size = 26, normalized size = 0.87 \[ -\frac {\cos \left (3 \, b x + 3 \, a\right )}{6 \, b} - \frac {\cos \left (b x + a\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a),x, algorithm="maxima")

[Out]

-1/6*cos(3*b*x + 3*a)/b - 1/2*cos(b*x + a)/b

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mupad [B]  time = 0.19, size = 43, normalized size = 1.43 \[ \left \{\begin {array}{cl} x\,\left (2\,\sin \relax (a)-2\,{\sin \relax (a)}^3\right ) & \text {\ if\ \ }b=0\\ -\frac {3\,\cos \left (a+b\,x\right )+\cos \left (3\,a+3\,b\,x\right )}{6\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*sin(2*a + 2*b*x),x)

[Out]

piecewise(b == 0, x*(2*sin(a) - 2*sin(a)^3), b ~= 0, -(3*cos(a + b*x) + cos(3*a + 3*b*x))/(6*b))

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sympy [A]  time = 0.76, size = 53, normalized size = 1.77 \[ \begin {cases} - \frac {\sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )}}{3 b} - \frac {2 \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sin {\left (2 a \right )} \cos {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a),x)

[Out]

Piecewise((-sin(a + b*x)*sin(2*a + 2*b*x)/(3*b) - 2*cos(a + b*x)*cos(2*a + 2*b*x)/(3*b), Ne(b, 0)), (x*sin(2*a
)*cos(a), True))

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