\(\int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx\) [142]

Optimal result

Integrand size = 20, antiderivative size = 76 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {3 x}{16}-\frac {3 \cos (2 a+2 b x) \sin (2 a+2 b x)}{32 b}-\frac {\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}+\frac {\sin ^5(2 a+2 b x)}{20 b} \]

[Out]

3/16*x-3/32*cos(2*b*x+2*a)*sin(2*b*x+2*a)/b-1/16*cos(2*b*x+2*a)*sin(2*b*x+2*a)^3/b+1/20*sin(2*b*x+2*a)^5/b

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4370, 2715, 8, 2644, 30} \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {\sin ^5(2 a+2 b x)}{20 b}-\frac {\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{16 b}-\frac {3 \sin (2 a+2 b x) \cos (2 a+2 b x)}{32 b}+\frac {3 x}{16} \]

[In]

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

[Out]

(3*x)/16 - (3*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(32*b) - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^3)/(16*b) + Sin[2
*a + 2*b*x]^5/(20*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 4370

Int[cos[(a_.) + (b_.)*(x_)]^2*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[1/2, Int[(g*Sin[c + d*x]
)^p, x], x] + Dist[1/2, Int[Cos[c + d*x]*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c -
a*d, 0] && EqQ[d/b, 2] && IGtQ[p/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \sin ^4(2 a+2 b x) \, dx+\frac {1}{2} \int \cos (2 a+2 b x) \sin ^4(2 a+2 b x) \, dx \\ & = -\frac {\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}+\frac {3}{8} \int \sin ^2(2 a+2 b x) \, dx+\frac {\text {Subst}\left (\int x^4 \, dx,x,\sin (2 a+2 b x)\right )}{4 b} \\ & = -\frac {3 \cos (2 a+2 b x) \sin (2 a+2 b x)}{32 b}-\frac {\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}+\frac {\sin ^5(2 a+2 b x)}{20 b}+\frac {3 \int 1 \, dx}{16} \\ & = \frac {3 x}{16}-\frac {3 \cos (2 a+2 b x) \sin (2 a+2 b x)}{32 b}-\frac {\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}+\frac {\sin ^5(2 a+2 b x)}{20 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {120 b x+20 \sin (2 (a+b x))-40 \sin (4 (a+b x))-10 \sin (6 (a+b x))+5 \sin (8 (a+b x))+2 \sin (10 (a+b x))}{640 b} \]

[In]

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

[Out]

(120*b*x + 20*Sin[2*(a + b*x)] - 40*Sin[4*(a + b*x)] - 10*Sin[6*(a + b*x)] + 5*Sin[8*(a + b*x)] + 2*Sin[10*(a
+ b*x)])/(640*b)

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87

method	result	size
parallelrisch	\(\frac {120 x b +20 \sin \left (2 x b +2 a \right )-40 \sin \left (4 x b +4 a \right )-10 \sin \left (6 x b +6 a \right )+5 \sin \left (8 x b +8 a \right )+2 \sin \left (10 x b +10 a \right )}{640 b}\)	\(66\)
default	\(\frac {3 x}{16}+\frac {\sin \left (2 x b +2 a \right )}{32 b}-\frac {\sin \left (4 x b +4 a \right )}{16 b}-\frac {\sin \left (6 x b +6 a \right )}{64 b}+\frac {\sin \left (8 x b +8 a \right )}{128 b}+\frac {\sin \left (10 x b +10 a \right )}{320 b}\)	\(75\)
risch	\(\frac {3 x}{16}+\frac {\sin \left (2 x b +2 a \right )}{32 b}-\frac {\sin \left (4 x b +4 a \right )}{16 b}-\frac {\sin \left (6 x b +6 a \right )}{64 b}+\frac {\sin \left (8 x b +8 a \right )}{128 b}+\frac {\sin \left (10 x b +10 a \right )}{320 b}\)	\(75\)

[In]

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

[Out]

1/640*(120*x*b+20*sin(2*b*x+2*a)-40*sin(4*b*x+4*a)-10*sin(6*b*x+6*a)+5*sin(8*b*x+8*a)+2*sin(10*b*x+10*a))/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {15 \, b x + {\left (128 \, \cos \left (b x + a\right )^{9} - 176 \, \cos \left (b x + a\right )^{7} + 8 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{80 \, b} \]

[In]

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

[Out]

1/80*(15*b*x + (128*cos(b*x + a)^9 - 176*cos(b*x + a)^7 + 8*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 15*cos(b*x +
a))*sin(b*x + a))/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (70) = 140\).

Time = 4.91 (sec) , antiderivative size = 434, normalized size of antiderivative = 5.71 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\begin {cases} \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8} + \frac {3 x \sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{16} + \frac {3 x \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} + \frac {3 x \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8} + \frac {3 x \cos ^{2}{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{16} + \frac {7 \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{160 b} + \frac {19 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{480 b} + \frac {\sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{10 b} + \frac {2 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{5 b} + \frac {4 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{15 b} - \frac {57 \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{160 b} - \frac {109 \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{480 b} & \text {for}\: b \neq 0 \\x \sin ^{4}{\left (2 a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**4/16 + 3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*
x)**2/8 + 3*x*sin(a + b*x)**2*cos(2*a + 2*b*x)**4/16 + 3*x*sin(2*a + 2*b*x)**4*cos(a + b*x)**2/16 + 3*x*sin(2*
a + 2*b*x)**2*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/8 + 3*x*cos(a + b*x)**2*cos(2*a + 2*b*x)**4/16 + 7*sin(a + b
*x)**2*sin(2*a + 2*b*x)**3*cos(2*a + 2*b*x)/(160*b) + 19*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**3/
(480*b) + sin(a + b*x)*sin(2*a + 2*b*x)**4*cos(a + b*x)/(10*b) + 2*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(a + b*
x)*cos(2*a + 2*b*x)**2/(5*b) + 4*sin(a + b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**4/(15*b) - 57*sin(2*a + 2*b*x)**3
*cos(a + b*x)**2*cos(2*a + 2*b*x)/(160*b) - 109*sin(2*a + 2*b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(480*b),
Ne(b, 0)), (x*sin(2*a)**4*cos(a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {120 \, b x + 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) - 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) + 20 \, \sin \left (2 \, b x + 2 \, a\right )}{640 \, b} \]

[In]

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

[Out]

1/640*(120*b*x + 2*sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) - 40*sin(4*b*x + 4*a) + 20*si
n(2*b*x + 2*a))/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {120 \, b x + 120 \, a + 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) - 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) + 20 \, \sin \left (2 \, b x + 2 \, a\right )}{640 \, b} \]

[In]

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

[Out]

1/640*(120*b*x + 120*a + 2*sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) - 40*sin(4*b*x + 4*a)
+ 20*sin(2*b*x + 2*a))/b

Mupad [B] (verification not implemented)

Time = 21.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.43 \[ \int \cos ^2(a+b x) \sin ^4(2 a+2 b x) \, dx=\frac {3\,x}{16}+\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^9}{16}+\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^7}{8}+\frac {8\,{\mathrm {tan}\left (a+b\,x\right )}^5}{5}-\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^3}{8}-\frac {3\,\mathrm {tan}\left (a+b\,x\right )}{16}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^{10}+5\,{\mathrm {tan}\left (a+b\,x\right )}^8+10\,{\mathrm {tan}\left (a+b\,x\right )}^6+10\,{\mathrm {tan}\left (a+b\,x\right )}^4+5\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \]

[In]

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

[Out]

(3*x)/16 + ((8*tan(a + b*x)^5)/5 - (7*tan(a + b*x)^3)/8 - (3*tan(a + b*x))/16 + (7*tan(a + b*x)^7)/8 + (3*tan(
a + b*x)^9)/16)/(b*(5*tan(a + b*x)^2 + 10*tan(a + b*x)^4 + 10*tan(a + b*x)^6 + 5*tan(a + b*x)^8 + tan(a + b*x)
^10 + 1))