3.1106 \(\int \cos ^4(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=278 \[ -\frac{a \left (20 a^2-69 b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}-\frac{\left (-140 a^2 b^2+40 a^4+21 b^4\right ) \sin ^3(c+d x) \cos (c+d x)}{1344 b^2 d}-\frac{\left (8 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{128 d}+\frac{1}{128} x \left (8 a^2+3 b^2\right )+\frac{5 a \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}+\frac{2 a b \cos ^3(c+d x)}{35 d}-\frac{6 a b \cos (c+d x)}{35 d}-\frac{\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{8 b d} \]

[Out]

((8*a^2 + 3*b^2)*x)/128 - (6*a*b*Cos[c + d*x])/(35*d) + (2*a*b*Cos[c + d*x]^3)/(35*d) - ((8*a^2 + 3*b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(128*d) - ((40*a^4 - 140*a^2*b^2 + 21*b^4)*Cos[c + d*x]*Sin[c + d*x]^3)/(1344*b^2*d) - (
a*(20*a^2 - 69*b^2)*Cos[c + d*x]*Sin[c + d*x]^4)/(840*b*d) - ((20*a^2 - 63*b^2)*Cos[c + d*x]*Sin[c + d*x]^3*(a
+ b*Sin[c + d*x])^2)/(336*b^2*d) + (5*a*Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(56*b^2*d) - (Cos
[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(8*b*d)

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Rubi [A]  time = 0.626873, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2895, 3049, 3033, 3023, 2748, 2635, 8, 2633} \[ -\frac{a \left (20 a^2-69 b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}-\frac{\left (-140 a^2 b^2+40 a^4+21 b^4\right ) \sin ^3(c+d x) \cos (c+d x)}{1344 b^2 d}-\frac{\left (8 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{128 d}+\frac{1}{128} x \left (8 a^2+3 b^2\right )+\frac{5 a \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}+\frac{2 a b \cos ^3(c+d x)}{35 d}-\frac{6 a b \cos (c+d x)}{35 d}-\frac{\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{8 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

((8*a^2 + 3*b^2)*x)/128 - (6*a*b*Cos[c + d*x])/(35*d) + (2*a*b*Cos[c + d*x]^3)/(35*d) - ((8*a^2 + 3*b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(128*d) - ((40*a^4 - 140*a^2*b^2 + 21*b^4)*Cos[c + d*x]*Sin[c + d*x]^3)/(1344*b^2*d) - (
a*(20*a^2 - 69*b^2)*Cos[c + d*x]*Sin[c + d*x]^4)/(840*b*d) - ((20*a^2 - 63*b^2)*Cos[c + d*x]*Sin[c + d*x]^3*(a
+ b*Sin[c + d*x])^2)/(336*b^2*d) + (5*a*Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(56*b^2*d) - (Cos
[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(8*b*d)

Rule 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

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] && Integer
Q[2*n]

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}-\frac{\int \sin ^2(c+d x) (a+b \sin (c+d x))^2 \left (15 a^2-56 b^2+2 a b \sin (c+d x)-\left (20 a^2-63 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{56 b^2}\\ &=-\frac{\left (20 a^2-63 b^2\right ) \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}+\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}-\frac{\int \sin ^2(c+d x) (a+b \sin (c+d x)) \left (3 a \left (10 a^2-49 b^2\right )+b \left (2 a^2-21 b^2\right ) \sin (c+d x)-2 a \left (20 a^2-69 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{336 b^2}\\ &=-\frac{a \left (20 a^2-69 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}+\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}-\frac{\int \sin ^2(c+d x) \left (15 a^2 \left (10 a^2-49 b^2\right )-288 a b^3 \sin (c+d x)-5 \left (40 a^4-140 a^2 b^2+21 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{1680 b^2}\\ &=-\frac{\left (40 a^4-140 a^2 b^2+21 b^4\right ) \cos (c+d x) \sin ^3(c+d x)}{1344 b^2 d}-\frac{a \left (20 a^2-69 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}+\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}-\frac{\int \sin ^2(c+d x) \left (-105 b^2 \left (8 a^2+3 b^2\right )-1152 a b^3 \sin (c+d x)\right ) \, dx}{6720 b^2}\\ &=-\frac{\left (40 a^4-140 a^2 b^2+21 b^4\right ) \cos (c+d x) \sin ^3(c+d x)}{1344 b^2 d}-\frac{a \left (20 a^2-69 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}+\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}+\frac{1}{35} (6 a b) \int \sin ^3(c+d x) \, dx-\frac{1}{64} \left (-8 a^2-3 b^2\right ) \int \sin ^2(c+d x) \, dx\\ &=-\frac{\left (8 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{128 d}-\frac{\left (40 a^4-140 a^2 b^2+21 b^4\right ) \cos (c+d x) \sin ^3(c+d x)}{1344 b^2 d}-\frac{a \left (20 a^2-69 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}+\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}-\frac{1}{128} \left (-8 a^2-3 b^2\right ) \int 1 \, dx-\frac{(6 a b) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{35 d}\\ &=\frac{1}{128} \left (8 a^2+3 b^2\right ) x-\frac{6 a b \cos (c+d x)}{35 d}+\frac{2 a b \cos ^3(c+d x)}{35 d}-\frac{\left (8 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{128 d}-\frac{\left (40 a^4-140 a^2 b^2+21 b^4\right ) \cos (c+d x) \sin ^3(c+d x)}{1344 b^2 d}-\frac{a \left (20 a^2-69 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{840 b d}-\frac{\left (20 a^2-63 b^2\right ) \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{336 b^2 d}+\frac{5 a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{56 b^2 d}-\frac{\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{8 b d}\\ \end{align*}

Mathematica [A]  time = 0.581002, size = 141, normalized size = 0.51 \[ \frac{840 a^2 \sin (2 (c+d x))-840 a^2 \sin (4 (c+d x))-280 a^2 \sin (6 (c+d x))+3360 a^2 d x-5040 a b \cos (c+d x)-1680 a b \cos (3 (c+d x))+336 a b \cos (5 (c+d x))+240 a b \cos (7 (c+d x))-420 b^2 \sin (4 (c+d x))+\frac{105}{2} b^2 \sin (8 (c+d x))+1680 b^2 c+1260 b^2 d x}{53760 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

(1680*b^2*c + 3360*a^2*d*x + 1260*b^2*d*x - 5040*a*b*Cos[c + d*x] - 1680*a*b*Cos[3*(c + d*x)] + 336*a*b*Cos[5*
(c + d*x)] + 240*a*b*Cos[7*(c + d*x)] + 840*a^2*Sin[2*(c + d*x)] - 840*a^2*Sin[4*(c + d*x)] - 420*b^2*Sin[4*(c
+ d*x)] - 280*a^2*Sin[6*(c + d*x)] + (105*b^2*Sin[8*(c + d*x)])/2)/(53760*d)

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Maple [A]  time = 0.044, size = 163, normalized size = 0.6 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \right ) +2\,ab \left ( -1/7\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8}}-{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{16}}+{\frac{\sin \left ( dx+c \right ) }{64} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{128}}+{\frac{3\,c}{128}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)+2*a*b*(-
1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+b^2*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*cos(d*x+c
)^5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/128*d*x+3/128*c))

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Maxima [A]  time = 1.00985, size = 136, normalized size = 0.49 \begin{align*} \frac{560 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 6144 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a b + 105 \,{\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{107520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/107520*(560*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2 + 6144*(5*cos(d*x + c)^7 - 7*cos
(d*x + c)^5)*a*b + 105*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*b^2)/d

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Fricas [A]  time = 1.85615, size = 316, normalized size = 1.14 \begin{align*} \frac{3840 \, a b \cos \left (d x + c\right )^{7} - 5376 \, a b \cos \left (d x + c\right )^{5} + 105 \,{\left (8 \, a^{2} + 3 \, b^{2}\right )} d x + 35 \,{\left (48 \, b^{2} \cos \left (d x + c\right )^{7} - 8 \,{\left (8 \, a^{2} + 9 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (8 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13440*(3840*a*b*cos(d*x + c)^7 - 5376*a*b*cos(d*x + c)^5 + 105*(8*a^2 + 3*b^2)*d*x + 35*(48*b^2*cos(d*x + c)
^7 - 8*(8*a^2 + 9*b^2)*cos(d*x + c)^5 + 2*(8*a^2 + 3*b^2)*cos(d*x + c)^3 + 3*(8*a^2 + 3*b^2)*cos(d*x + c))*sin
(d*x + c))/d

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Sympy [A]  time = 13.5831, size = 420, normalized size = 1.51 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{2 a b \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{4 a b \cos ^{7}{\left (c + d x \right )}}{35 d} + \frac{3 b^{2} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac{3 b^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac{9 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac{3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac{3 b^{2} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac{3 b^{2} \sin ^{7}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{128 d} + \frac{11 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} - \frac{11 b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} - \frac{3 b^{2} \sin{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x*sin(c + d*x)**6/16 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**2*x*sin(c + d*x)**2*
cos(c + d*x)**4/16 + a**2*x*cos(c + d*x)**6/16 + a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + a**2*sin(c + d*x)*
*3*cos(c + d*x)**3/(6*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 2*a*b*sin(c + d*x)**2*cos(c + d*x)**5/(5
*d) - 4*a*b*cos(c + d*x)**7/(35*d) + 3*b**2*x*sin(c + d*x)**8/128 + 3*b**2*x*sin(c + d*x)**6*cos(c + d*x)**2/3
2 + 9*b**2*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 3*b**2*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 3*b**2*x*cos(c
+ d*x)**8/128 + 3*b**2*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 11*b**2*sin(c + d*x)**5*cos(c + d*x)**3/(128*d)
- 11*b**2*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) - 3*b**2*sin(c + d*x)*cos(c + d*x)**7/(128*d), Ne(d, 0)), (
x*(a + b*sin(c))**2*sin(c)**2*cos(c)**4, True))

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Giac [A]  time = 1.22418, size = 203, normalized size = 0.73 \begin{align*} \frac{1}{128} \,{\left (8 \, a^{2} + 3 \, b^{2}\right )} x + \frac{a b \cos \left (7 \, d x + 7 \, c\right )}{224 \, d} + \frac{a b \cos \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac{a b \cos \left (3 \, d x + 3 \, c\right )}{32 \, d} - \frac{3 \, a b \cos \left (d x + c\right )}{32 \, d} + \frac{b^{2} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} - \frac{{\left (2 \, a^{2} + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(8*a^2 + 3*b^2)*x + 1/224*a*b*cos(7*d*x + 7*c)/d + 1/160*a*b*cos(5*d*x + 5*c)/d - 1/32*a*b*cos(3*d*x + 3
*c)/d - 3/32*a*b*cos(d*x + c)/d + 1/1024*b^2*sin(8*d*x + 8*c)/d - 1/192*a^2*sin(6*d*x + 6*c)/d + 1/64*a^2*sin(
2*d*x + 2*c)/d - 1/128*(2*a^2 + b^2)*sin(4*d*x + 4*c)/d