\(\int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx\) [134]

Optimal result

Integrand size = 18, antiderivative size = 31 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4 \sin ^3(a+b x)}{3 b}-\frac {4 \sin ^5(a+b x)}{5 b} \]

[Out]

4/3*sin(b*x+a)^3/b-4/5*sin(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4372, 2644, 14} \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4 \sin ^3(a+b x)}{3 b}-\frac {4 \sin ^5(a+b x)}{5 b} \]

[In]

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

[Out]

(4*Sin[a + b*x]^3)/(3*b) - (4*Sin[a + b*x]^5)/(5*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 4372

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

Rubi steps \begin{align*} \text {integral}& = 4 \int \cos ^3(a+b x) \sin ^2(a+b x) \, dx \\ & = \frac {4 \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {4 \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {4 \sin ^3(a+b x)}{3 b}-\frac {4 \sin ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {2 (7+3 \cos (2 (a+b x))) \sin ^3(a+b x)}{15 b} \]

[In]

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

[Out]

(2*(7 + 3*Cos[2*(a + b*x)])*Sin[a + b*x]^3)/(15*b)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {\sin \left (x b +a \right )}{2 b}-\frac {\sin \left (3 x b +3 a \right )}{12 b}-\frac {\sin \left (5 x b +5 a \right )}{20 b}\)	\(41\)
risch	\(\frac {\sin \left (x b +a \right )}{2 b}-\frac {\sin \left (3 x b +3 a \right )}{12 b}-\frac {\sin \left (5 x b +5 a \right )}{20 b}\)	\(41\)
parallelrisch	\(\frac {\frac {8 \left (-\tan \left (x b +a \right )^{3}+\tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{15}+\frac {8 \left (2 \tan \left (x b +a \right )^{4}+3 \tan \left (x b +a \right )^{2}+2\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{15}+\frac {8 \tan \left (x b +a \right )^{3}}{15}-\frac {8 \tan \left (x b +a \right )}{15}}{b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{2}}\)	\(111\)
norman	\(\frac {\frac {16 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{15 b}-\frac {8 \tan \left (x b +a \right )}{15 b}+\frac {8 \tan \left (x b +a \right )^{3}}{15 b}+\frac {8 \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (x b +a \right )^{2}}{5 b}+\frac {16 \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \tan \left (x b +a \right )^{4}}{15 b}+\frac {8 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} \tan \left (x b +a \right )}{15 b}-\frac {8 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2} \tan \left (x b +a \right )^{3}}{15 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{2}}\)	\(158\)

[In]

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

[Out]

1/2*sin(b*x+a)/b-1/12*sin(3*b*x+3*a)/b-1/20/b*sin(5*b*x+5*a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \, {\left (3 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{15 \, b} \]

[In]

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

[Out]

-4/15*(3*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)*sin(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (26) = 52\).

Time = 0.36 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.90 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=\begin {cases} \frac {7 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{15 b} + \frac {8 \sin {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{15 b} - \frac {4 \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{15 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((7*sin(a + b*x)*sin(2*a + 2*b*x)**2/(15*b) + 8*sin(a + b*x)*cos(2*a + 2*b*x)**2/(15*b) - 4*sin(2*a +
2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)/(15*b), Ne(b, 0)), (x*sin(2*a)**2*cos(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {3 \, \sin \left (5 \, b x + 5 \, a\right ) + 5 \, \sin \left (3 \, b x + 3 \, a\right ) - 30 \, \sin \left (b x + a\right )}{60 \, b} \]

[In]

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

[Out]

-1/60*(3*sin(5*b*x + 5*a) + 5*sin(3*b*x + 3*a) - 30*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {3 \, \sin \left (5 \, b x + 5 \, a\right ) + 5 \, \sin \left (3 \, b x + 3 \, a\right ) - 30 \, \sin \left (b x + a\right )}{60 \, b} \]

[In]

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

[Out]

-1/60*(3*sin(5*b*x + 5*a) + 5*sin(3*b*x + 3*a) - 30*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos (a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4\,\left (5\,{\sin \left (a+b\,x\right )}^3-3\,{\sin \left (a+b\,x\right )}^5\right )}{15\,b} \]

[In]

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

[Out]

(4*(5*sin(a + b*x)^3 - 3*sin(a + b*x)^5))/(15*b)