\(\int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx\) [154]

Optimal result

Integrand size = 20, antiderivative size = 46 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4 \sin ^3(a+b x)}{3 b}-\frac {8 \sin ^5(a+b x)}{5 b}+\frac {4 \sin ^7(a+b x)}{7 b} \]

[Out]

4/3*sin(b*x+a)^3/b-8/5*sin(b*x+a)^5/b+4/7*sin(b*x+a)^7/b

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4372, 2644, 276} \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4 \sin ^7(a+b x)}{7 b}-\frac {8 \sin ^5(a+b x)}{5 b}+\frac {4 \sin ^3(a+b x)}{3 b} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {(157+108 \cos (2 (a+b x))+15 \cos (4 (a+b x))) \sin ^3(a+b x)}{210 b} \]

[In]

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

[Out]

((157 + 108*Cos[2*(a + b*x)] + 15*Cos[4*(a + b*x)])*Sin[a + b*x]^3)/(210*b)

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {525 \sin \left (x b +a \right )-35 \sin \left (3 x b +3 a \right )-63 \sin \left (5 x b +5 a \right )-15 \sin \left (7 x b +7 a \right )}{1680 b}\)	\(48\)
default	\(\frac {5 \sin \left (x b +a \right )}{16 b}-\frac {\sin \left (3 x b +3 a \right )}{48 b}-\frac {3 \sin \left (5 x b +5 a \right )}{80 b}-\frac {\sin \left (7 x b +7 a \right )}{112 b}\)	\(55\)
risch	\(\frac {5 \sin \left (x b +a \right )}{16 b}-\frac {\sin \left (3 x b +3 a \right )}{48 b}-\frac {3 \sin \left (5 x b +5 a \right )}{80 b}-\frac {\sin \left (7 x b +7 a \right )}{112 b}\)	\(55\)

[In]

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

[Out]

1/1680*(525*sin(b*x+a)-35*sin(3*b*x+3*a)-63*sin(5*b*x+5*a)-15*sin(7*b*x+7*a))/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {4 \, {\left (15 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{105 \, b} \]

[In]

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

[Out]

-4/105*(15*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. 202 vs. \(2 (39) = 78\).

Time = 2.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 4.39 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\begin {cases} \frac {38 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{105 b} + \frac {32 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} + \frac {8 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} + \frac {11 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{35 b} + \frac {24 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac {12 \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (2 a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((38*sin(a + b*x)**3*sin(2*a + 2*b*x)**2/(105*b) + 32*sin(a + b*x)**3*cos(2*a + 2*b*x)**2/(105*b) + 8
*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)/(35*b) + 11*sin(a + b*x)*sin(2*a + 2*b*x)**2*c
os(a + b*x)**2/(35*b) + 24*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/(35*b) - 12*sin(2*a + 2*b*x)*cos(a
+ b*x)**3*cos(2*a + 2*b*x)/(35*b), Ne(b, 0)), (x*sin(2*a)**2*cos(a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{1680 \, b} \]

[In]

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

[Out]

-1/1680*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=-\frac {15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{1680 \, b} \]

[In]

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

[Out]

-1/1680*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \cos ^3(a+b x) \sin ^2(2 a+2 b x) \, dx=\frac {4\,\left (15\,{\sin \left (a+b\,x\right )}^7-42\,{\sin \left (a+b\,x\right )}^5+35\,{\sin \left (a+b\,x\right )}^3\right )}{105\,b} \]

[In]

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

[Out]

(4*(35*sin(a + b*x)^3 - 42*sin(a + b*x)^5 + 15*sin(a + b*x)^7))/(105*b)