3.1059 \(\int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=190 \[ \frac {\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {\left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{35 d}+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{21 d}+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac {a b \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac {a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a b x}{8} \]

[Out]

1/8*a*b*x-1/35*(7*a^2+4*b^2)*cos(d*x+c)/d+1/105*(7*a^2+4*b^2)*cos(d*x+c)^3/d-1/8*a*b*cos(d*x+c)*sin(d*x+c)/d-1
/12*a*b*cos(d*x+c)*sin(d*x+c)^3/d+1/35*(2*a^2-b^2)*cos(d*x+c)*sin(d*x+c)^4/d+1/21*a*b*cos(d*x+c)*sin(d*x+c)^5/
d+1/7*cos(d*x+c)*sin(d*x+c)^4*(a+b*sin(d*x+c))^2/d

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Rubi [A]  time = 0.38, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3050, 3033, 3023, 2748, 2633, 2635, 8} \[ \frac {\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {\left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{35 d}+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{21 d}+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac {a b \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac {a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a b x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(a*b*x)/8 - ((7*a^2 + 4*b^2)*Cos[c + d*x])/(35*d) + ((7*a^2 + 4*b^2)*Cos[c + d*x]^3)/(105*d) - (a*b*Cos[c + d*
x]*Sin[c + d*x])/(8*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^3)/(12*d) + ((2*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^4
)/(35*d) + (a*b*Cos[c + d*x]*Sin[c + d*x]^5)/(21*d) + (Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2)/(7*
d)

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]

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 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 2889

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

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 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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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])))

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \sin ^3(c+d x) (a+b \sin (c+d x))^2 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{7} \int \sin ^3(c+d x) (a+b \sin (c+d x)) \left (3 a+b \sin (c+d x)-2 a \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{42} \int \sin ^3(c+d x) \left (18 a^2+14 a b \sin (c+d x)-6 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{210} \int \sin ^3(c+d x) \left (6 \left (7 a^2+4 b^2\right )+70 a b \sin (c+d x)\right ) \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{3} (a b) \int \sin ^4(c+d x) \, dx+\frac {1}{35} \left (7 a^2+4 b^2\right ) \int \sin ^3(c+d x) \, dx\\ &=-\frac {a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{4} (a b) \int \sin ^2(c+d x) \, dx-\frac {\left (7 a^2+4 b^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{35 d}\\ &=-\frac {\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{8} (a b) \int 1 \, dx\\ &=\frac {a b x}{8}-\frac {\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 132, normalized size = 0.69 \[ \frac {-105 \left (8 a^2+5 b^2\right ) \cos (c+d x)-35 \left (4 a^2+b^2\right ) \cos (3 (c+d x))+84 a^2 \cos (5 (c+d x))-210 a b \sin (2 (c+d x))-210 a b \sin (4 (c+d x))+70 a b \sin (6 (c+d x))+840 a b c+840 a b d x+63 b^2 \cos (5 (c+d x))-15 b^2 \cos (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(840*a*b*c + 840*a*b*d*x - 105*(8*a^2 + 5*b^2)*Cos[c + d*x] - 35*(4*a^2 + b^2)*Cos[3*(c + d*x)] + 84*a^2*Cos[5
*(c + d*x)] + 63*b^2*Cos[5*(c + d*x)] - 15*b^2*Cos[7*(c + d*x)] - 210*a*b*Sin[2*(c + d*x)] - 210*a*b*Sin[4*(c
+ d*x)] + 70*a*b*Sin[6*(c + d*x)])/(6720*d)

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fricas [A]  time = 0.76, size = 104, normalized size = 0.55 \[ -\frac {120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 105 \, a b d x + 280 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 35 \, {\left (8 \, a b \cos \left (d x + c\right )^{5} - 14 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(120*b^2*cos(d*x + c)^7 - 168*(a^2 + 2*b^2)*cos(d*x + c)^5 - 105*a*b*d*x + 280*(a^2 + b^2)*cos(d*x + c)
^3 - 35*(8*a*b*cos(d*x + c)^5 - 14*a*b*cos(d*x + c)^3 + 3*a*b*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.24, size = 141, normalized size = 0.74 \[ \frac {1}{8} \, a b x - \frac {b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} + \frac {{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (8 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*b*x - 1/448*b^2*cos(7*d*x + 7*c)/d + 1/96*a*b*sin(6*d*x + 6*c)/d - 1/32*a*b*sin(4*d*x + 4*c)/d - 1/32*a*
b*sin(2*d*x + 2*c)/d + 1/320*(4*a^2 + 3*b^2)*cos(5*d*x + 5*c)/d - 1/192*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d - 1/6
4*(8*a^2 + 5*b^2)*cos(d*x + c)/d

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maple [A]  time = 0.20, size = 150, normalized size = 0.79 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+2 a b \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{7}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (d x +c \right )\right )}{105}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+2*a*b*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+
c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*c)+b^2*(-1/7*sin(d*x+c)^4*cos(d*x+c)^3-4/35*sin(d*x+c
)^2*cos(d*x+c)^3-8/105*cos(d*x+c)^3))

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maxima [A]  time = 0.40, size = 104, normalized size = 0.55 \[ \frac {224 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 32 \, {\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} b^{2}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3360*(224*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2 - 35*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x
+ 4*c))*a*b - 32*(15*cos(d*x + c)^7 - 42*cos(d*x + c)^5 + 35*cos(d*x + c)^3)*b^2)/d

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mupad [B]  time = 12.97, size = 233, normalized size = 1.23 \[ \frac {a\,b\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {8\,a^2}{3}-\frac {16\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {8\,a^2}{5}+\frac {16\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {20\,a^2}{3}+\frac {32\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {28\,a^2}{15}+\frac {16\,b^2}{15}\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {4\,a^2}{15}+\frac {16\,b^2}{105}+\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {97\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {97\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}-\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b*x)/8 - (tan(c/2 + (d*x)/2)^6*((8*a^2)/3 - (16*b^2)/3) + tan(c/2 + (d*x)/2)^4*((8*a^2)/5 + (16*b^2)/5) + t
an(c/2 + (d*x)/2)^8*((20*a^2)/3 + (32*b^2)/3) + tan(c/2 + (d*x)/2)^2*((28*a^2)/15 + (16*b^2)/15) + 4*a^2*tan(c
/2 + (d*x)/2)^10 + (4*a^2)/15 + (16*b^2)/105 + (5*a*b*tan(c/2 + (d*x)/2)^3)/3 - (97*a*b*tan(c/2 + (d*x)/2)^5)/
12 + (97*a*b*tan(c/2 + (d*x)/2)^9)/12 - (5*a*b*tan(c/2 + (d*x)/2)^11)/3 - (a*b*tan(c/2 + (d*x)/2)^13)/4 + (a*b
*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 5.91, size = 275, normalized size = 1.45 \[ \begin {cases} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a b x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {8 b^{2} \cos ^{7}{\left (c + d x \right )}}{105 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin ^{3}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 2*a**2*cos(c + d*x)**5/(15*d) + a*b*x*sin(c + d*x)**6
/8 + 3*a*b*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 3*a*b*x*sin(c + d*x)**2*cos(c + d*x)**4/8 + a*b*x*cos(c + d*x
)**6/8 + a*b*sin(c + d*x)**5*cos(c + d*x)/(8*d) - a*b*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - a*b*sin(c + d*x)
*cos(c + d*x)**5/(8*d) - b**2*sin(c + d*x)**4*cos(c + d*x)**3/(3*d) - 4*b**2*sin(c + d*x)**2*cos(c + d*x)**5/(
15*d) - 8*b**2*cos(c + d*x)**7/(105*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**3*cos(c)**2, True))

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