\(\int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx\) [132]

Optimal result

Integrand size = 18, antiderivative size = 46 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \sin ^5(a+b x)}{5 b}-\frac {32 \sin ^7(a+b x)}{7 b}+\frac {16 \sin ^9(a+b x)}{9 b} \]

[Out]

16/5*sin(b*x+a)^5/b-32/7*sin(b*x+a)^7/b+16/9*sin(b*x+a)^9/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, 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, 276} \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \sin ^9(a+b x)}{9 b}-\frac {32 \sin ^7(a+b x)}{7 b}+\frac {16 \sin ^5(a+b x)}{5 b} \]

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(16*Sin[a + b*x]^5)/(5*b) - (32*Sin[a + b*x]^7)/(7*b) + (16*Sin[a + b*x]^9)/(9*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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}& = 16 \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx \\ & = \frac {16 \text {Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {16 \text {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {16 \sin ^5(a+b x)}{5 b}-\frac {32 \sin ^7(a+b x)}{7 b}+\frac {16 \sin ^9(a+b x)}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {2 (249+220 \cos (2 (a+b x))+35 \cos (4 (a+b x))) \sin ^5(a+b x)}{315 b} \]

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(2*(249 + 220*Cos[2*(a + b*x)] + 35*Cos[4*(a + b*x)])*Sin[a + b*x]^5)/(315*b)

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {3 \sin \left (x b +a \right )}{8 b}-\frac {\sin \left (3 x b +3 a \right )}{12 b}-\frac {\sin \left (5 x b +5 a \right )}{20 b}+\frac {\sin \left (7 x b +7 a \right )}{112 b}+\frac {\sin \left (9 x b +9 a \right )}{144 b}\)	\(69\)
risch	\(\frac {3 \sin \left (x b +a \right )}{8 b}-\frac {\sin \left (3 x b +3 a \right )}{12 b}-\frac {\sin \left (5 x b +5 a \right )}{20 b}+\frac {\sin \left (7 x b +7 a \right )}{112 b}+\frac {\sin \left (9 x b +9 a \right )}{144 b}\)	\(69\)
parallelrisch	\(\frac {\left (-128 \tan \left (x b +a \right )^{7}-448 \tan \left (x b +a \right )^{5}+448 \tan \left (x b +a \right )^{3}+128 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\left (256 \tan \left (x b +a \right )^{8}+896 \tan \left (x b +a \right )^{6}+1120 \tan \left (x b +a \right )^{4}+896 \tan \left (x b +a \right )^{2}+256\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+128 \tan \left (x b +a \right )^{7}+448 \tan \left (x b +a \right )^{5}-448 \tan \left (x b +a \right )^{3}-128 \tan \left (x b +a \right )}{315 b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{4}}\)	\(175\)

[In]

int(cos(b*x+a)*sin(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)

[Out]

3/8*sin(b*x+a)/b-1/12*sin(3*b*x+3*a)/b-1/20/b*sin(5*b*x+5*a)+1/112/b*sin(7*b*x+7*a)+1/144/b*sin(9*b*x+9*a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \, {\left (35 \, \cos \left (b x + a\right )^{8} - 50 \, \cos \left (b x + a\right )^{6} + 3 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{315 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

16/315*(35*cos(b*x + a)^8 - 50*cos(b*x + a)^6 + 3*cos(b*x + a)^4 + 4*cos(b*x + a)^2 + 8)*sin(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (39) = 78\).

Time = 2.04 (sec) , antiderivative size = 162, normalized size of antiderivative = 3.52 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\begin {cases} \frac {107 \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )}}{315 b} + \frac {16 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{21 b} + \frac {128 \sin {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{315 b} - \frac {104 \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{315 b} - \frac {64 \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{315 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)**4,x)

[Out]

Piecewise((107*sin(a + b*x)*sin(2*a + 2*b*x)**4/(315*b) + 16*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*x)
**2/(21*b) + 128*sin(a + b*x)*cos(2*a + 2*b*x)**4/(315*b) - 104*sin(2*a + 2*b*x)**3*cos(a + b*x)*cos(2*a + 2*b
*x)/(315*b) - 64*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**3/(315*b), Ne(b, 0)), (x*sin(2*a)**4*cos(a),
True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {35 \, \sin \left (9 \, b x + 9 \, a\right ) + 45 \, \sin \left (7 \, b x + 7 \, a\right ) - 252 \, \sin \left (5 \, b x + 5 \, a\right ) - 420 \, \sin \left (3 \, b x + 3 \, a\right ) + 1890 \, \sin \left (b x + a\right )}{5040 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

1/5040*(35*sin(9*b*x + 9*a) + 45*sin(7*b*x + 7*a) - 252*sin(5*b*x + 5*a) - 420*sin(3*b*x + 3*a) + 1890*sin(b*x
+ a))/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {35 \, \sin \left (9 \, b x + 9 \, a\right ) + 45 \, \sin \left (7 \, b x + 7 \, a\right ) - 252 \, \sin \left (5 \, b x + 5 \, a\right ) - 420 \, \sin \left (3 \, b x + 3 \, a\right ) + 1890 \, \sin \left (b x + a\right )}{5040 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

1/5040*(35*sin(9*b*x + 9*a) + 45*sin(7*b*x + 7*a) - 252*sin(5*b*x + 5*a) - 420*sin(3*b*x + 3*a) + 1890*sin(b*x
+ a))/b

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16\,\left (35\,{\sin \left (a+b\,x\right )}^9-90\,{\sin \left (a+b\,x\right )}^7+63\,{\sin \left (a+b\,x\right )}^5\right )}{315\,b} \]

[In]

int(cos(a + b*x)*sin(2*a + 2*b*x)^4,x)

[Out]

(16*(63*sin(a + b*x)^5 - 90*sin(a + b*x)^7 + 35*sin(a + b*x)^9))/(315*b)