\(\int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx\) [131]

Optimal result

Integrand size = 18, antiderivative size = 46 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {32 \cos ^7(a+b x)}{7 b}+\frac {64 \cos ^9(a+b x)}{9 b}-\frac {32 \cos ^{11}(a+b x)}{11 b} \]

[Out]

-32/7*cos(b*x+a)^7/b+64/9*cos(b*x+a)^9/b-32/11*cos(b*x+a)^11/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, 2645, 276} \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {32 \cos ^{11}(a+b x)}{11 b}+\frac {64 \cos ^9(a+b x)}{9 b}-\frac {32 \cos ^7(a+b x)}{7 b} \]

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^5,x]

[Out]

(-32*Cos[a + b*x]^7)/(7*b) + (64*Cos[a + b*x]^9)/(9*b) - (32*Cos[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 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}& = 32 \int \cos ^6(a+b x) \sin ^5(a+b x) \, dx \\ & = -\frac {32 \text {Subst}\left (\int x^6 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {32 \text {Subst}\left (\int \left (x^6-2 x^8+x^{10}\right ) \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {32 \cos ^7(a+b x)}{7 b}+\frac {64 \cos ^9(a+b x)}{9 b}-\frac {32 \cos ^{11}(a+b x)}{11 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=\frac {4 \cos ^7(a+b x) (-365+364 \cos (2 (a+b x))-63 \cos (4 (a+b x)))}{693 b} \]

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*Cos[a + b*x]^7*(-365 + 364*Cos[2*(a + b*x)] - 63*Cos[4*(a + b*x)]))/(693*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(82\) vs. \(2(40)=80\).

Time = 1.53 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.80

method	result	size
default	\(-\frac {5 \cos \left (x b +a \right )}{16 b}-\frac {5 \cos \left (3 x b +3 a \right )}{48 b}+\frac {\cos \left (5 x b +5 a \right )}{32 b}+\frac {5 \cos \left (7 x b +7 a \right )}{224 b}-\frac {\cos \left (9 x b +9 a \right )}{288 b}-\frac {\cos \left (11 x b +11 a \right )}{352 b}\)	\(83\)
risch	\(-\frac {5 \cos \left (x b +a \right )}{16 b}-\frac {5 \cos \left (3 x b +3 a \right )}{48 b}+\frac {\cos \left (5 x b +5 a \right )}{32 b}+\frac {5 \cos \left (7 x b +7 a \right )}{224 b}-\frac {\cos \left (9 x b +9 a \right )}{288 b}-\frac {\cos \left (11 x b +11 a \right )}{352 b}\)	\(83\)
parallelrisch	\(\frac {\left (128 \tan \left (x b +a \right )^{8}+544 \tan \left (x b +a \right )^{6}+4576 \tan \left (x b +a \right )^{4}+2432 \tan \left (x b +a \right )^{2}+512\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\left (-512 \tan \left (x b +a \right )^{9}-2304 \tan \left (x b +a \right )^{7}-4032 \tan \left (x b +a \right )^{5}-2304 \tan \left (x b +a \right )^{3}-512 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+512 \tan \left (x b +a \right )^{10}+2432 \tan \left (x b +a \right )^{8}+4576 \tan \left (x b +a \right )^{6}+544 \tan \left (x b +a \right )^{4}+128 \tan \left (x b +a \right )^{2}}{693 b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{5}}\)	\(197\)

[In]

int(cos(b*x+a)*sin(2*b*x+2*a)^5,x,method=_RETURNVERBOSE)

[Out]

-5/16*cos(b*x+a)/b-5/48*cos(3*b*x+3*a)/b+1/32*cos(5*b*x+5*a)/b+5/224*cos(7*b*x+7*a)/b-1/288*cos(9*b*x+9*a)/b-1
/352*cos(11*b*x+11*a)/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {32 \, {\left (63 \, \cos \left (b x + a\right )^{11} - 154 \, \cos \left (b x + a\right )^{9} + 99 \, \cos \left (b x + a\right )^{7}\right )}}{693 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="fricas")

[Out]

-32/693*(63*cos(b*x + a)^11 - 154*cos(b*x + a)^9 + 99*cos(b*x + a)^7)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (39) = 78\).

Time = 4.84 (sec) , antiderivative size = 199, normalized size of antiderivative = 4.33 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=\begin {cases} - \frac {151 \sin {\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )}}{693 b} - \frac {272 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{693 b} - \frac {128 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{693 b} - \frac {422 \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{693 b} - \frac {608 \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{693 b} - \frac {256 \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{693 b} & \text {for}\: b \neq 0 \\x \sin ^{5}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)**5,x)

[Out]

Piecewise((-151*sin(a + b*x)*sin(2*a + 2*b*x)**5/(693*b) - 272*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(2*a + 2*b*
x)**2/(693*b) - 128*sin(a + b*x)*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**4/(693*b) - 422*sin(2*a + 2*b*x)**4*cos(a
+ b*x)*cos(2*a + 2*b*x)/(693*b) - 608*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)**3/(693*b) - 256*cos(a
+ b*x)*cos(2*a + 2*b*x)**5/(693*b), Ne(b, 0)), (x*sin(2*a)**5*cos(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.50 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {63 \, \cos \left (11 \, b x + 11 \, a\right ) + 77 \, \cos \left (9 \, b x + 9 \, a\right ) - 495 \, \cos \left (7 \, b x + 7 \, a\right ) - 693 \, \cos \left (5 \, b x + 5 \, a\right ) + 2310 \, \cos \left (3 \, b x + 3 \, a\right ) + 6930 \, \cos \left (b x + a\right )}{22176 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="maxima")

[Out]

-1/22176*(63*cos(11*b*x + 11*a) + 77*cos(9*b*x + 9*a) - 495*cos(7*b*x + 7*a) - 693*cos(5*b*x + 5*a) + 2310*cos
(3*b*x + 3*a) + 6930*cos(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {32 \, {\left (63 \, \cos \left (b x + a\right )^{11} - 154 \, \cos \left (b x + a\right )^{9} + 99 \, \cos \left (b x + a\right )^{7}\right )}}{693 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

-32/693*(63*cos(b*x + a)^11 - 154*cos(b*x + a)^9 + 99*cos(b*x + a)^7)/b

Mupad [B] (verification not implemented)

Time = 19.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos (a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {32\,\left (63\,{\cos \left (a+b\,x\right )}^{11}-154\,{\cos \left (a+b\,x\right )}^9+99\,{\cos \left (a+b\,x\right )}^7\right )}{693\,b} \]

[In]

int(cos(a + b*x)*sin(2*a + 2*b*x)^5,x)

[Out]

-(32*(99*cos(a + b*x)^7 - 154*cos(a + b*x)^9 + 63*cos(a + b*x)^11))/(693*b)