\(\int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx\) [133]

Optimal result

Integrand size = 18, antiderivative size = 31 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8 \cos ^5(a+b x)}{5 b}+\frac {8 \cos ^7(a+b x)}{7 b} \]

[Out]

-8/5*cos(b*x+a)^5/b+8/7*cos(b*x+a)^7/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, 2645, 14} \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \cos ^7(a+b x)}{7 b}-\frac {8 \cos ^5(a+b x)}{5 b} \]

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^3,x]

[Out]

(-8*Cos[a + b*x]^5)/(5*b) + (8*Cos[a + b*x]^7)/(7*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 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[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}& = 8 \int \cos ^4(a+b x) \sin ^3(a+b x) \, dx \\ & = -\frac {8 \text {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {8 \text {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {8 \cos ^5(a+b x)}{5 b}+\frac {8 \cos ^7(a+b x)}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {4 \cos ^5(a+b x) (-9+5 \cos (2 (a+b x)))}{35 b} \]

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^3,x]

[Out]

(4*Cos[a + b*x]^5*(-9 + 5*Cos[2*(a + b*x)]))/(35*b)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77

method	result	size
default	\(-\frac {3 \cos \left (x b +a \right )}{8 b}-\frac {\cos \left (3 x b +3 a \right )}{8 b}+\frac {\cos \left (5 x b +5 a \right )}{40 b}+\frac {\cos \left (7 x b +7 a \right )}{56 b}\)	\(55\)
risch	\(-\frac {3 \cos \left (x b +a \right )}{8 b}-\frac {\cos \left (3 x b +3 a \right )}{8 b}+\frac {\cos \left (5 x b +5 a \right )}{40 b}+\frac {\cos \left (7 x b +7 a \right )}{56 b}\)	\(55\)
parallelrisch	\(\frac {\frac {8 \left (\tan \left (x b +a \right )^{4}+11 \tan \left (x b +a \right )^{2}+4\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{35}+\frac {16 \left (-2 \tan \left (x b +a \right )^{5}-5 \tan \left (x b +a \right )^{3}-2 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{35}+\frac {32 \tan \left (x b +a \right )^{6}}{35}+\frac {88 \tan \left (x b +a \right )^{4}}{35}+\frac {8 \tan \left (x b +a \right )^{2}}{35}}{b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{3}}\)	\(134\)

[In]

int(cos(b*x+a)*sin(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)

[Out]

-3/8*cos(b*x+a)/b-1/8*cos(3*b*x+3*a)/b+1/40*cos(5*b*x+5*a)/b+1/56*cos(7*b*x+7*a)/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \, {\left (5 \, \cos \left (b x + a\right )^{7} - 7 \, \cos \left (b x + a\right )^{5}\right )}}{35 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^3,x, algorithm="fricas")

[Out]

8/35*(5*cos(b*x + a)^7 - 7*cos(b*x + a)^5)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (26) = 52\).

Time = 0.77 (sec) , antiderivative size = 128, normalized size of antiderivative = 4.13 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=\begin {cases} - \frac {9 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{35 b} - \frac {8 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac {22 \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} - \frac {16 \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{35 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)**3,x)

[Out]

Piecewise((-9*sin(a + b*x)*sin(2*a + 2*b*x)**3/(35*b) - 8*sin(a + b*x)*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**2/(3
5*b) - 22*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/(35*b) - 16*cos(a + b*x)*cos(2*a + 2*b*x)**3/(35*b
), Ne(b, 0)), (x*sin(2*a)**3*cos(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {5 \, \cos \left (7 \, b x + 7 \, a\right ) + 7 \, \cos \left (5 \, b x + 5 \, a\right ) - 35 \, \cos \left (3 \, b x + 3 \, a\right ) - 105 \, \cos \left (b x + a\right )}{280 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^3,x, algorithm="maxima")

[Out]

1/280*(5*cos(7*b*x + 7*a) + 7*cos(5*b*x + 5*a) - 35*cos(3*b*x + 3*a) - 105*cos(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=\frac {8 \, {\left (5 \, \cos \left (b x + a\right )^{7} - 7 \, \cos \left (b x + a\right )^{5}\right )}}{35 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^3,x, algorithm="giac")

[Out]

8/35*(5*cos(b*x + a)^7 - 7*cos(b*x + a)^5)/b

Mupad [B] (verification not implemented)

Time = 19.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx=-\frac {8\,\left (7\,{\cos \left (a+b\,x\right )}^5-5\,{\cos \left (a+b\,x\right )}^7\right )}{35\,b} \]

[In]

int(cos(a + b*x)*sin(2*a + 2*b*x)^3,x)

[Out]

-(8*(7*cos(a + b*x)^5 - 5*cos(a + b*x)^7))/(35*b)