\(\int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx\) [152]

Optimal result

Integrand size = 20, antiderivative size = 61 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \sin ^5(a+b x)}{5 b}-\frac {48 \sin ^7(a+b x)}{7 b}+\frac {16 \sin ^9(a+b x)}{3 b}-\frac {16 \sin ^{11}(a+b x)}{11 b} \]

[Out]

16/5*sin(b*x+a)^5/b-48/7*sin(b*x+a)^7/b+16/3*sin(b*x+a)^9/b-16/11*sin(b*x+a)^11/b

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4372, 2644, 276} \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=-\frac {16 \sin ^{11}(a+b x)}{11 b}+\frac {16 \sin ^9(a+b x)}{3 b}-\frac {48 \sin ^7(a+b x)}{7 b}+\frac {16 \sin ^5(a+b x)}{5 b} \]

[In]

Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

(16*Sin[a + b*x]^5)/(5*b) - (48*Sin[a + b*x]^7)/(7*b) + (16*Sin[a + b*x]^9)/(3*b) - (16*Sin[a + b*x]^11)/(11*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 ^7(a+b x) \sin ^4(a+b x) \, dx \\ & = \frac {16 \text {Subst}\left (\int x^4 \left (1-x^2\right )^3 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {16 \text {Subst}\left (\int \left (x^4-3 x^6+3 x^8-x^{10}\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {16 \sin ^5(a+b x)}{5 b}-\frac {48 \sin ^7(a+b x)}{7 b}+\frac {16 \sin ^9(a+b x)}{3 b}-\frac {16 \sin ^{11}(a+b x)}{11 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {(3042+3335 \cos (2 (a+b x))+910 \cos (4 (a+b x))+105 \cos (6 (a+b x))) \sin ^5(a+b x)}{2310 b} \]

[In]

Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

((3042 + 3335*Cos[2*(a + b*x)] + 910*Cos[4*(a + b*x)] + 105*Cos[6*(a + b*x)])*Sin[a + b*x]^5)/(2310*b)

Maple [A] (verified)

Time = 3.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15

method	result	size
parallelrisch	\(\frac {16170 \sin \left (x b +a \right )-2310 \sin \left (3 x b +3 a \right )-2541 \sin \left (5 x b +5 a \right )-165 \sin \left (7 x b +7 a \right )+385 \sin \left (9 x b +9 a \right )+105 \sin \left (11 x b +11 a \right )}{73920 b}\)	\(70\)
default	\(\frac {7 \sin \left (x b +a \right )}{32 b}-\frac {\sin \left (3 x b +3 a \right )}{32 b}-\frac {11 \sin \left (5 x b +5 a \right )}{320 b}-\frac {\sin \left (7 x b +7 a \right )}{448 b}+\frac {\sin \left (9 x b +9 a \right )}{192 b}+\frac {\sin \left (11 x b +11 a \right )}{704 b}\)	\(83\)
risch	\(\frac {7 \sin \left (x b +a \right )}{32 b}-\frac {\sin \left (3 x b +3 a \right )}{32 b}-\frac {11 \sin \left (5 x b +5 a \right )}{320 b}-\frac {\sin \left (7 x b +7 a \right )}{448 b}+\frac {\sin \left (9 x b +9 a \right )}{192 b}+\frac {\sin \left (11 x b +11 a \right )}{704 b}\)	\(83\)

[In]

int(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x,method=_RETURNVERBOSE)

[Out]

1/73920*(16170*sin(b*x+a)-2310*sin(3*b*x+3*a)-2541*sin(5*b*x+5*a)-165*sin(7*b*x+7*a)+385*sin(9*b*x+9*a)+105*si
n(11*b*x+11*a))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {16 \, {\left (105 \, \cos \left (b x + a\right )^{10} - 140 \, \cos \left (b x + a\right )^{8} + 5 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4} + 8 \, \cos \left (b x + a\right )^{2} + 16\right )} \sin \left (b x + a\right )}{1155 \, b} \]

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

16/1155*(105*cos(b*x + a)^10 - 140*cos(b*x + a)^8 + 5*cos(b*x + a)^6 + 6*cos(b*x + a)^4 + 8*cos(b*x + a)^2 + 1
6)*sin(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (53) = 106\).

Time = 11.33 (sec) , antiderivative size = 366, normalized size of antiderivative = 6.00 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\begin {cases} \frac {46 \sin ^{3}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )}}{165 b} + \frac {192 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{385 b} + \frac {256 \sin ^{3}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{1155 b} + \frac {272 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{1155 b} + \frac {256 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{1155 b} + \frac {211 \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{1155 b} + \frac {304 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{385 b} + \frac {128 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{231 b} - \frac {472 \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{1155 b} - \frac {64 \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{231 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (2 a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**4,x)

[Out]

Piecewise((46*sin(a + b*x)**3*sin(2*a + 2*b*x)**4/(165*b) + 192*sin(a + b*x)**3*sin(2*a + 2*b*x)**2*cos(2*a +
2*b*x)**2/(385*b) + 256*sin(a + b*x)**3*cos(2*a + 2*b*x)**4/(1155*b) + 272*sin(a + b*x)**2*sin(2*a + 2*b*x)**3
*cos(a + b*x)*cos(2*a + 2*b*x)/(1155*b) + 256*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**
3/(1155*b) + 211*sin(a + b*x)*sin(2*a + 2*b*x)**4*cos(a + b*x)**2/(1155*b) + 304*sin(a + b*x)*sin(2*a + 2*b*x)
**2*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/(385*b) + 128*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**4/(231*b)
- 472*sin(2*a + 2*b*x)**3*cos(a + b*x)**3*cos(2*a + 2*b*x)/(1155*b) - 64*sin(2*a + 2*b*x)*cos(a + b*x)**3*cos
(2*a + 2*b*x)**3/(231*b), Ne(b, 0)), (x*sin(2*a)**4*cos(a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {105 \, \sin \left (11 \, b x + 11 \, a\right ) + 385 \, \sin \left (9 \, b x + 9 \, a\right ) - 165 \, \sin \left (7 \, b x + 7 \, a\right ) - 2541 \, \sin \left (5 \, b x + 5 \, a\right ) - 2310 \, \sin \left (3 \, b x + 3 \, a\right ) + 16170 \, \sin \left (b x + a\right )}{73920 \, b} \]

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

1/73920*(105*sin(11*b*x + 11*a) + 385*sin(9*b*x + 9*a) - 165*sin(7*b*x + 7*a) - 2541*sin(5*b*x + 5*a) - 2310*s
in(3*b*x + 3*a) + 16170*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {105 \, \sin \left (11 \, b x + 11 \, a\right ) + 385 \, \sin \left (9 \, b x + 9 \, a\right ) - 165 \, \sin \left (7 \, b x + 7 \, a\right ) - 2541 \, \sin \left (5 \, b x + 5 \, a\right ) - 2310 \, \sin \left (3 \, b x + 3 \, a\right ) + 16170 \, \sin \left (b x + a\right )}{73920 \, b} \]

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

1/73920*(105*sin(11*b*x + 11*a) + 385*sin(9*b*x + 9*a) - 165*sin(7*b*x + 7*a) - 2541*sin(5*b*x + 5*a) - 2310*s
in(3*b*x + 3*a) + 16170*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {-\frac {16\,{\sin \left (a+b\,x\right )}^{11}}{11}+\frac {16\,{\sin \left (a+b\,x\right )}^9}{3}-\frac {48\,{\sin \left (a+b\,x\right )}^7}{7}+\frac {16\,{\sin \left (a+b\,x\right )}^5}{5}}{b} \]

[In]

int(cos(a + b*x)^3*sin(2*a + 2*b*x)^4,x)

[Out]

((16*sin(a + b*x)^5)/5 - (48*sin(a + b*x)^7)/7 + (16*sin(a + b*x)^9)/3 - (16*sin(a + b*x)^11)/11)/b