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

Optimal. Leaf size=49 \[ \frac{\sin ^3(2 a+2 b x)}{12 b}-\frac{\sin (2 a+2 b x) \cos (2 a+2 b x)}{8 b}+\frac{x}{4} \]

[Out]

x/4 - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(8*b) + Sin[2*a + 2*b*x]^3/(12*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0536167, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4285, 2635, 8, 2564, 30} \[ \frac{\sin ^3(2 a+2 b x)}{12 b}-\frac{\sin (2 a+2 b x) \cos (2 a+2 b x)}{8 b}+\frac{x}{4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[2*a + 2*b*x]^2,x]

[Out]

x/4 - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(8*b) + Sin[2*a + 2*b*x]^3/(12*b)

Rule 4285

Int[cos[(a_.) + (b_.)*(x_)]^2*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[1/2, Int[(g*Sin[c + d*x]
)^p, x], x] + Dist[1/2, Int[Cos[c + d*x]*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c -
a*d, 0] && EqQ[d/b, 2] && IGtQ[p/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx &=\frac{1}{2} \int \sin ^2(2 a+2 b x) \, dx+\frac{1}{2} \int \cos (2 a+2 b x) \sin ^2(2 a+2 b x) \, dx\\ &=-\frac{\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}+\frac{\int 1 \, dx}{4}+\frac{\operatorname{Subst}\left (\int x^2 \, dx,x,\sin (2 a+2 b x)\right )}{4 b}\\ &=\frac{x}{4}-\frac{\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}+\frac{\sin ^3(2 a+2 b x)}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.094312, size = 40, normalized size = 0.82 \[ -\frac{-3 \sin (2 (a+b x))+3 \sin (4 (a+b x))+\sin (6 (a+b x))-12 b x}{48 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-12*b*x - 3*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*x)] + Sin[6*(a + b*x)])/(48*b)

________________________________________________________________________________________

Maple [A]  time = 0.023, size = 47, normalized size = 1. \begin{align*}{\frac{x}{4}}+{\frac{\sin \left ( 2\,bx+2\,a \right ) }{16\,b}}-{\frac{\sin \left ( 4\,bx+4\,a \right ) }{16\,b}}-{\frac{\sin \left ( 6\,bx+6\,a \right ) }{48\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x+1/16*sin(2*b*x+2*a)/b-1/16/b*sin(4*b*x+4*a)-1/48/b*sin(6*b*x+6*a)

________________________________________________________________________________________

Maxima [A]  time = 1.20071, size = 58, normalized size = 1.18 \begin{align*} \frac{12 \, b x - \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*b*x - sin(6*b*x + 6*a) - 3*sin(4*b*x + 4*a) + 3*sin(2*b*x + 2*a))/b

________________________________________________________________________________________

Fricas [A]  time = 0.491132, size = 116, normalized size = 2.37 \begin{align*} \frac{3 \, b x -{\left (8 \, \cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b*x - (8*cos(b*x + a)^5 - 2*cos(b*x + a)^3 - 3*cos(b*x + a))*sin(b*x + a))/b

________________________________________________________________________________________

Sympy [A]  time = 28.5597, size = 231, normalized size = 4.71 \begin{align*} \begin{cases} \frac{x \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac{x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac{x \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{x \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac{\sin ^{2}{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos{\left (2 a + 2 b x \right )}}{24 b} + \frac{\sin{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )}}{6 b} + \frac{\sin{\left (a + b x \right )} \cos{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{3 b} - \frac{7 \sin{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{24 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\left (2 a \right )} \cos ^{2}{\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)**2*sin(2*b*x+2*a)**2,x)

[Out]

Piecewise((x*sin(a + b*x)**2*sin(2*a + 2*b*x)**2/4 + x*sin(a + b*x)**2*cos(2*a + 2*b*x)**2/4 + x*sin(2*a + 2*b
*x)**2*cos(a + b*x)**2/4 + x*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/4 + sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(2*a
+ 2*b*x)/(24*b) + sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(a + b*x)/(6*b) + sin(a + b*x)*cos(a + b*x)*cos(2*a + 2*
b*x)**2/(3*b) - 7*sin(2*a + 2*b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)/(24*b), Ne(b, 0)), (x*sin(2*a)**2*cos(a)**
2, True))

________________________________________________________________________________________

Giac [A]  time = 1.28467, size = 62, normalized size = 1.27 \begin{align*} \frac{1}{4} \, x - \frac{\sin \left (6 \, b x + 6 \, a\right )}{48 \, b} - \frac{\sin \left (4 \, b x + 4 \, a\right )}{16 \, b} + \frac{\sin \left (2 \, b x + 2 \, a\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x - 1/48*sin(6*b*x + 6*a)/b - 1/16*sin(4*b*x + 4*a)/b + 1/16*sin(2*b*x + 2*a)/b