\(\int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx\) [141]

Optimal result

Integrand size = 20, antiderivative size = 44 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4 \cos ^8(a+b x)}{b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {8 \cos ^{12}(a+b x)}{3 b} \]

[Out]

-4*cos(b*x+a)^8/b+32/5*cos(b*x+a)^10/b-8/3*cos(b*x+a)^12/b

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4372, 2645, 272, 45} \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {8 \cos ^{12}(a+b x)}{3 b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {4 \cos ^8(a+b x)}{b} \]

[In]

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

[Out]

(-4*Cos[a + b*x]^8)/b + (32*Cos[a + b*x]^10)/(5*b) - (8*Cos[a + b*x]^12)/(3*b)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 ^7(a+b x) \sin ^5(a+b x) \, dx \\ & = -\frac {32 \text {Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b} \\ & = -\frac {16 \text {Subst}\left (\int (1-x)^2 x^3 \, dx,x,\cos ^2(a+b x)\right )}{b} \\ & = -\frac {16 \text {Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\cos ^2(a+b x)\right )}{b} \\ & = -\frac {4 \cos ^8(a+b x)}{b}+\frac {32 \cos ^{10}(a+b x)}{5 b}-\frac {8 \cos ^{12}(a+b x)}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.55 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {600 \cos (2 (a+b x))+75 \cos (4 (a+b x))-100 \cos (6 (a+b x))-30 \cos (8 (a+b x))+12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{3840 b} \]

[In]

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

[Out]

-1/3840*(600*Cos[2*(a + b*x)] + 75*Cos[4*(a + b*x)] - 100*Cos[6*(a + b*x)] - 30*Cos[8*(a + b*x)] + 12*Cos[10*(
a + b*x)] + 5*Cos[12*(a + b*x)])/b

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68

method	result	size
parallelrisch	\(\frac {-600 \cos \left (2 x b +2 a \right )-1662-5 \cos \left (12 x b +12 a \right )-12 \cos \left (10 x b +10 a \right )+30 \cos \left (8 x b +8 a \right )+100 \cos \left (6 x b +6 a \right )-75 \cos \left (4 x b +4 a \right )}{3840 b}\)	\(74\)
default	\(-\frac {5 \cos \left (2 x b +2 a \right )}{32 b}-\frac {5 \cos \left (4 x b +4 a \right )}{256 b}+\frac {5 \cos \left (6 x b +6 a \right )}{192 b}+\frac {\cos \left (8 x b +8 a \right )}{128 b}-\frac {\cos \left (10 x b +10 a \right )}{320 b}-\frac {\cos \left (12 x b +12 a \right )}{768 b}\)	\(86\)
risch	\(-\frac {5 \cos \left (2 x b +2 a \right )}{32 b}-\frac {5 \cos \left (4 x b +4 a \right )}{256 b}+\frac {5 \cos \left (6 x b +6 a \right )}{192 b}+\frac {\cos \left (8 x b +8 a \right )}{128 b}-\frac {\cos \left (10 x b +10 a \right )}{320 b}-\frac {\cos \left (12 x b +12 a \right )}{768 b}\)	\(86\)

[In]

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

[Out]

1/3840*(-600*cos(2*b*x+2*a)-1662-5*cos(12*b*x+12*a)-12*cos(10*b*x+10*a)+30*cos(8*b*x+8*a)+100*cos(6*b*x+6*a)-7
5*cos(4*b*x+4*a))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \]

[In]

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

[Out]

-4/15*(10*cos(b*x + a)^12 - 24*cos(b*x + a)^10 + 15*cos(b*x + a)^8)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (37) = 74\).

Time = 11.30 (sec) , antiderivative size = 597, normalized size of antiderivative = 13.57 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=\begin {cases} - \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )}}{32} - \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} - \frac {5 x \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{32} - \frac {5 x \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{16} - \frac {5 x \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} - \frac {5 x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{16} + \frac {5 x \sin ^{5}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{32} + \frac {5 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac {5 x \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{32} - \frac {125 \sin ^{2}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{384 b} - \frac {2 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{3 b} - \frac {217 \sin ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{640 b} + \frac {95 \sin {\left (a + b x \right )} \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{192 b} + \frac {13 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{12 b} + \frac {109 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{192 b} - \frac {67 \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{384 b} + \frac {139 \cos ^{2}{\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{1920 b} & \text {for}\: b \neq 0 \\x \sin ^{5}{\left (2 a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-5*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**5/32 - 5*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**3*cos(2*a + 2*b
*x)**2/16 - 5*x*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**4/32 - 5*x*sin(a + b*x)*sin(2*a + 2*b*x)**4
*cos(a + b*x)*cos(2*a + 2*b*x)/16 - 5*x*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)**3/8 -
5*x*sin(a + b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**5/16 + 5*x*sin(2*a + 2*b*x)**5*cos(a + b*x)**2/32 + 5*x*sin(2*
a + 2*b*x)**3*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/16 + 5*x*sin(2*a + 2*b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**
4/32 - 125*sin(a + b*x)**2*sin(2*a + 2*b*x)**4*cos(2*a + 2*b*x)/(384*b) - 2*sin(a + b*x)**2*sin(2*a + 2*b*x)**
2*cos(2*a + 2*b*x)**3/(3*b) - 217*sin(a + b*x)**2*cos(2*a + 2*b*x)**5/(640*b) + 95*sin(a + b*x)*sin(2*a + 2*b*
x)**5*cos(a + b*x)/(192*b) + 13*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(a + b*x)*cos(2*a + 2*b*x)**2/(12*b) + 109
*sin(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**4/(192*b) - 67*sin(2*a + 2*b*x)**4*cos(a + b*x)*
*2*cos(2*a + 2*b*x)/(384*b) + 139*cos(a + b*x)**2*cos(2*a + 2*b*x)**5/(1920*b), Ne(b, 0)), (x*sin(2*a)**5*cos(
a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.64 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {5 \, \cos \left (12 \, b x + 12 \, a\right ) + 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) - 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) + 600 \, \cos \left (2 \, b x + 2 \, a\right )}{3840 \, b} \]

[In]

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

[Out]

-1/3840*(5*cos(12*b*x + 12*a) + 12*cos(10*b*x + 10*a) - 30*cos(8*b*x + 8*a) - 100*cos(6*b*x + 6*a) + 75*cos(4*
b*x + 4*a) + 600*cos(2*b*x + 2*a))/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4 \, {\left (10 \, \cos \left (b x + a\right )^{12} - 24 \, \cos \left (b x + a\right )^{10} + 15 \, \cos \left (b x + a\right )^{8}\right )}}{15 \, b} \]

[In]

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

[Out]

-4/15*(10*cos(b*x + a)^12 - 24*cos(b*x + a)^10 + 15*cos(b*x + a)^8)/b

Mupad [B] (verification not implemented)

Time = 19.37 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \cos ^2(a+b x) \sin ^5(2 a+2 b x) \, dx=-\frac {4\,{\cos \left (a+b\,x\right )}^8\,\left (10\,{\cos \left (a+b\,x\right )}^4-24\,{\cos \left (a+b\,x\right )}^2+15\right )}{15\,b} \]

[In]

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

[Out]

-(4*cos(a + b*x)^8*(10*cos(a + b*x)^4 - 24*cos(a + b*x)^2 + 15))/(15*b)