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]

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

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Rubi [A]  time = 0.01103, antiderivative size = 30, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0056336, size = 15, normalized size = 0.5 \[ -\frac{2 \cos ^3(a+b x)}{3 b} \]

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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Maple [A]  time = 0.014, size = 27, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( bx+a \right ) }{2\,b}}-{\frac{\cos \left ( 3\,bx+3\,a \right ) }{6\,b}} \end{align*}

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 = 1.06651, size = 35, normalized size = 1.17 \begin{align*} -\frac{\cos \left (3 \, b x + 3 \, a\right )}{6 \, b} - \frac{\cos \left (b x + a\right )}{2 \, b} \end{align*}

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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Fricas [A]  time = 0.483418, size = 31, normalized size = 1.03 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{3}}{3 \, b} \end{align*}

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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Sympy [A]  time = 1.3295, size = 53, normalized size = 1.77 \begin{align*} \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{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.25087, size = 35, normalized size = 1.17 \begin{align*} -\frac{\cos \left (3 \, b x + 3 \, a\right )}{6 \, b} - \frac{\cos \left (b x + a\right )}{2 \, b} \end{align*}

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