3.2.44 \(\int \cos ^2(a+b x) \sin ^2(2 a+2 b x) \, dx\) [144]

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

[Out]

1/4*x-1/8*cos(2*b*x+2*a)*sin(2*b*x+2*a)/b+1/12*sin(2*b*x+2*a)^3/b

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Rubi [A]
time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4370, 2715, 8, 2644, 30} \begin {gather*} \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} \end {gather*}

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 8

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

Rule 30

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

Rule 2644

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 2715

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] && Integ
erQ[2*n]

Rule 4370

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]

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 {\text {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*}

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Mathematica [A]
time = 0.11, size = 40, normalized size = 0.82 \begin {gather*} -\frac {-12 b x-3 \sin (2 (a+b x))+3 \sin (4 (a+b x))+\sin (6 (a+b x))}{48 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 47, normalized size = 0.96

method	result	size
default	\(\frac {x}{4}+\frac {\sin \left (2 x b +2 a \right )}{16 b}-\frac {\sin \left (4 x b +4 a \right )}{16 b}-\frac {\sin \left (6 x b +6 a \right )}{48 b}\)	\(47\)
risch	\(\frac {x}{4}+\frac {\sin \left (2 x b +2 a \right )}{16 b}-\frac {\sin \left (4 x b +4 a \right )}{16 b}-\frac {\sin \left (6 x b +6 a \right )}{48 b}\)	\(47\)

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,method=_RETURNVERBOSE)

[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)

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Maxima [A]
time = 0.27, size = 43, normalized size = 0.88 \begin {gather*} \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 {gather*}

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

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Fricas [A]
time = 2.02, size = 47, normalized size = 0.96 \begin {gather*} \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 {gather*}

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (41) = 82\).
time = 1.13, size = 231, normalized size = 4.71 \begin {gather*} \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 {gather*}

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))

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Giac [A]
time = 0.45, size = 46, normalized size = 0.94 \begin {gather*} \frac {12 \, b x + 12 \, a - \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 {gather*}

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]

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

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Mupad [B]
time = 0.31, size = 43, normalized size = 0.88 \begin {gather*} \frac {x}{4}-\frac {\frac {\sin \left (4\,a+4\,b\,x\right )}{16}-\frac {\sin \left (2\,a+2\,b\,x\right )}{16}+\frac {\sin \left (6\,a+6\,b\,x\right )}{48}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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