\(\int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1195]

Optimal result

Integrand size = 29, antiderivative size = 623 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}+\frac {3 a \left (3 b^2 (1+n)+a^2 (6+n)\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) (2+n) (4+n) (6+n) \sqrt {\cos ^2(c+d x)}}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}+\frac {3 b \left (b^2 (2+n)+3 a^2 (7+n)\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) (3+n) (5+n) (7+n) \sqrt {\cos ^2(c+d x)}}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)} \]

[Out]

-3*a*(2*a^4*(n^2+5*n+6)+3*b^4*(n^2+12*n+35)-2*a^2*b^2*(n^2+16*n+58))*cos(d*x+c)*sin(d*x+c)^(1+n)/b^2/d/(5+n)/(
6+n)/(7+n)/(n^2+6*n+8)-3*(2*a^4*(n^2+5*n+6)+b^4*(n^2+10*n+24)-2*a^2*b^2*(n^2+16*n+57))*cos(d*x+c)*sin(d*x+c)^(
2+n)/b/d/(3+n)/(4+n)/(5+n)/(6+n)/(7+n)-3*a*(a^2*(n^2+5*n+6)-b^2*(n^2+15*n+53))*cos(d*x+c)*sin(d*x+c)^(1+n)*(a+
b*sin(d*x+c))^2/b^2/d/(4+n)/(5+n)/(6+n)/(7+n)-(a^2*(2+n)*(3+n)-b^2*(6+n)*(8+n))*cos(d*x+c)*sin(d*x+c)^(1+n)*(a
+b*sin(d*x+c))^3/b^2/d/(5+n)/(6+n)/(7+n)+a*(3+n)*cos(d*x+c)*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^4/b^2/d/(6+n)/(7
+n)-cos(d*x+c)*sin(d*x+c)^(2+n)*(a+b*sin(d*x+c))^4/b/d/(7+n)+3*a*(3*b^2*(1+n)+a^2*(6+n))*cos(d*x+c)*hypergeom(
[1/2, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(1+n)/d/(6+n)/(n^3+7*n^2+14*n+8)/(cos(d*x+c)^2)^(1/2)+3*
b*(b^2*(2+n)+3*a^2*(7+n))*cos(d*x+c)*hypergeom([1/2, 1+1/2*n],[1/2*n+2],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/d/(5+n)
/(7+n)/(n^2+5*n+6)/(cos(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 623, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2974, 3128, 3112, 3102, 2827, 2722} \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 a \left (a^2 (n+6)+3 b^2 (n+1)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) (n+2) (n+4) (n+6) \sqrt {\cos ^2(c+d x)}}+\frac {3 b \left (3 a^2 (n+7)+b^2 (n+2)\right ) \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d (n+2) (n+3) (n+5) (n+7) \sqrt {\cos ^2(c+d x)}}-\frac {3 a \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+15 n+53\right )\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (n+4) (n+5) (n+6) (n+7)}-\frac {\left (a^2 (n+2) (n+3)-b^2 (n+6) (n+8)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6) (n+7)}-\frac {3 a \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+16 n+58\right )+3 b^4 \left (n^2+12 n+35\right )\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{b^2 d (n+2) (n+4) (n+5) (n+6) (n+7)}-\frac {3 \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+16 n+57\right )+b^4 \left (n^2+10 n+24\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{b d (n+3) (n+4) (n+5) (n+6) (n+7)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (n+6) (n+7)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^4}{b d (n+7)} \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^3,x]

[Out]

(-3*a*(2*a^4*(6 + 5*n + n^2) + 3*b^4*(35 + 12*n + n^2) - 2*a^2*b^2*(58 + 16*n + n^2))*Cos[c + d*x]*Sin[c + d*x
]^(1 + n))/(b^2*d*(2 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)) + (3*a*(3*b^2*(1 + n) + a^2*(6 + n))*Cos[c + d*x]*H
ypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)*(2 + n)*(4 + n)*(
6 + n)*Sqrt[Cos[c + d*x]^2]) - (3*(2*a^4*(6 + 5*n + n^2) + b^4*(24 + 10*n + n^2) - 2*a^2*b^2*(57 + 16*n + n^2)
)*Cos[c + d*x]*Sin[c + d*x]^(2 + n))/(b*d*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)) + (3*b*(b^2*(2 + n) + 3*a^2
*(7 + n))*Cos[c + d*x]*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + n))/(d*(
2 + n)*(3 + n)*(5 + n)*(7 + n)*Sqrt[Cos[c + d*x]^2]) - (3*a*(a^2*(6 + 5*n + n^2) - b^2*(53 + 15*n + n^2))*Cos[
c + d*x]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^2)/(b^2*d*(4 + n)*(5 + n)*(6 + n)*(7 + n)) - ((a^2*(2 + n)*
(3 + n) - b^2*(6 + n)*(8 + n))*Cos[c + d*x]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^3)/(b^2*d*(5 + n)*(6 + n
)*(7 + n)) + (a*(3 + n)*Cos[c + d*x]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^4)/(b^2*d*(6 + n)*(7 + n)) - (C
os[c + d*x]*Sin[c + d*x]^(2 + n)*(a + b*Sin[c + d*x])^4)/(b*d*(7 + n))

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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 2974

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 3102

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 3112

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 3128

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])))

Rubi steps \begin{align*} \text {integral}& = \frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^3 \left (a^2 (1+n) (3+n)-b^2 (6+n) (7+n)+3 a b \sin (c+d x)-\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (6+n) (7+n)} \\ & = -\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^2 \left (3 a \left (a^2 \left (3+4 n+n^2\right )-b^2 \left (54+15 n+n^2\right )\right )+3 b \left (2 a^2-b^2 (6+n)\right ) \sin (c+d x)-3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (5+n) (6+n) (7+n)} \\ & = -\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x)) \left (3 a^2 \left (2 a^2 \left (3+4 n+n^2\right )-b^2 \left (163+46 n+3 n^2\right )\right )+3 a b \left (2 a^2-b^2 \left (81+26 n+2 n^2\right )\right ) \sin (c+d x)-3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (4+n) (5+n) (6+n) (7+n)} \\ & = -\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) \left (3 a^3 (3+n) \left (2 a^2 \left (3+4 n+n^2\right )-b^2 \left (163+46 n+3 n^2\right )\right )-3 b^3 \left (24+10 n+n^2\right ) \left (b^2 (2+n)+3 a^2 (7+n)\right ) \sin (c+d x)-3 a (3+n) \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (3+n) (4+n) (5+n) (6+n) (7+n)} \\ & = -\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) \left (-3 a b^2 (3+n) \left (35+12 n+n^2\right ) \left (3 b^2 (1+n)+a^2 (6+n)\right )-3 b^3 (2+n) \left (24+10 n+n^2\right ) \left (b^2 (2+n)+3 a^2 (7+n)\right ) \sin (c+d x)\right ) \, dx}{b^2 (2+n) (3+n) (4+n) (5+n) (6+n) (7+n)} \\ & = -\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}+\frac {\left (3 a \left (3 b^2 (1+n)+a^2 (6+n)\right )\right ) \int \sin ^n(c+d x) \, dx}{(2+n) (4+n) (6+n)}+\frac {\left (3 b \left (b^2 (2+n)+3 a^2 (7+n)\right )\right ) \int \sin ^{1+n}(c+d x) \, dx}{(3+n) (5+n) (7+n)} \\ & = -\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}+\frac {3 a \left (3 b^2 (1+n)+a^2 (6+n)\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) (2+n) (4+n) (6+n) \sqrt {\cos ^2(c+d x)}}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}+\frac {3 b \left (b^2 (2+n)+3 a^2 (7+n)\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) (3+n) (5+n) (7+n) \sqrt {\cos ^2(c+d x)}}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.31 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{1+n}(c+d x) \left (\frac {a^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )}{1+n}+b \sin (c+d x) \left (\frac {3 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right )}{2+n}+b \sin (c+d x) \left (\frac {3 a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right )}{3+n}+\frac {b \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)}{4+n}\right )\right )\right )}{d} \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^3,x]

[Out]

(Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(1 + n)*((a^3*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Si
n[c + d*x]^2])/(1 + n) + b*Sin[c + d*x]*((3*a^2*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2])
/(2 + n) + b*Sin[c + d*x]*((3*a*Hypergeometric2F1[-3/2, (3 + n)/2, (5 + n)/2, Sin[c + d*x]^2])/(3 + n) + (b*Hy
pergeometric2F1[-3/2, (4 + n)/2, (6 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x])/(4 + n)))))/d

Maple [F]

\[\int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{3}d x\]

[In]

int(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x)

[Out]

int(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x)

Fricas [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-(3*a*b^2*cos(d*x + c)^6 - (a^3 + 3*a*b^2)*cos(d*x + c)^4 + (b^3*cos(d*x + c)^6 - (3*a^2*b + b^3)*cos
(d*x + c)^4)*sin(d*x + c))*sin(d*x + c)^n, x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**n*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^3*sin(d*x + c)^n*cos(d*x + c)^4, x)

Giac [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^3*sin(d*x + c)^n*cos(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3 \,d x \]

[In]

int(cos(c + d*x)^4*sin(c + d*x)^n*(a + b*sin(c + d*x))^3,x)

[Out]

int(cos(c + d*x)^4*sin(c + d*x)^n*(a + b*sin(c + d*x))^3, x)