\(\int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1196]

Optimal result

Integrand size = 29, antiderivative size = 487 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (3 b^4 (5+n)+2 a^4 \left (6+5 n+n^2\right )-2 a^2 b^2 \left (40+13 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n)}+\frac {3 \left (b^2 (1+n)+a^2 (6+n)\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) (2+n) (4+n) (6+n) \sqrt {\cos ^2(c+d x)}}-\frac {2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n)}+\frac {6 a b \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) (3+n) (5+n) \sqrt {\cos ^2(c+d x)}}-\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)} \]

[Out]

-(3*b^4*(5+n)+2*a^4*(n^2+5*n+6)-2*a^2*b^2*(n^2+13*n+40))*cos(d*x+c)*sin(d*x+c)^(1+n)/b^2/d/(5+n)/(6+n)/(n^2+6*
n+8)-2*a*(a^2*(n^2+5*n+6)-b^2*(n^2+13*n+39))*cos(d*x+c)*sin(d*x+c)^(2+n)/b/d/(3+n)/(4+n)/(5+n)/(6+n)-(a^2*(2+n
)*(3+n)-b^2*(5+n)*(7+n))*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)+a*(3+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)-cos(d*x+c)*sin(d*x+c)^(2+n)*(a+b*sin(d*x+c))^3/b/d/
(6+n)+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/(4+n)/(6+n)/(n^2+3*n+2)/(cos(d*x+c)^2)^(1/2)+6*a*b*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)/(n^2+5*n+6)/(cos(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2974, 3128, 3112, 3102, 2827, 2722} \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \left (a^2 (n+6)+b^2 (n+1)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) (n+2) (n+4) (n+6) \sqrt {\cos ^2(c+d x)}}-\frac {2 a \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{b d (n+3) (n+4) (n+5) (n+6)}-\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (n+4) (n+5) (n+6)}-\frac {\left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{b^2 d (n+2) (n+4) (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}+\frac {6 a b \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d (n+2) (n+3) (n+5) \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)} \]

[In]

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

[Out]

-(((3*b^4*(5 + n) + 2*a^4*(6 + 5*n + n^2) - 2*a^2*b^2*(40 + 13*n + n^2))*Cos[c + d*x]*Sin[c + d*x]^(1 + n))/(b
^2*d*(2 + n)*(4 + n)*(5 + n)*(6 + n))) + (3*(b^2*(1 + n) + a^2*(6 + n))*Cos[c + d*x]*Hypergeometric2F1[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]) - (2*a*(a^2*(6 + 5*n + n^2) - b^2*(39 + 13*n + n^2))*Cos[c + d*x]*Sin[c + d*x]^(2 + n))/(b*d*(3 + n)*(4 +
n)*(5 + n)*(6 + n)) + (6*a*b*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)*Sqrt[Cos[c + d*x]^2]) - ((a^2*(2 + n)*(3 + n) - b^2*(5 + n)*(7 + n)
)*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)) + (a*(3 + 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)) - (Cos[c + d*x]*Sin[c + d*x]^(2 +
n)*(a + b*Sin[c + d*x])^3)/(b*d*(6 + 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))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^2 \left (a^2 (1+n) (3+n)-b^2 (5+n) (6+n)+2 a b \sin (c+d x)-\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (5+n) (6+n)} \\ & = -\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\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)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x)) \left (a \left (2 a^2 \left (3+4 n+n^2\right )-b^2 \left (85+27 n+2 n^2\right )\right )+b \left (2 a^2-3 b^2 (5+n)\right ) \sin (c+d x)-2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (4+n) (5+n) (6+n)} \\ & = -\frac {2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 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)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\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)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)}-\frac {\int \sin ^n(c+d x) \left (a^2 (3+n) \left (2 a^2 \left (3+4 n+n^2\right )-b^2 \left (85+27 n+2 n^2\right )\right )-6 a b^3 (4+n) (6+n) \sin (c+d x)-(3+n) \left (3 b^4 (5+n)+2 a^4 \left (6+5 n+n^2\right )-2 a^2 b^2 \left (40+13 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (3+n) (4+n) (5+n) (6+n)} \\ & = -\frac {\left (3 b^4 (5+n)+2 a^4 \left (6+5 n+n^2\right )-2 a^2 b^2 \left (40+13 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)}-\frac {2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 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)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\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)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)}-\frac {\int \sin ^n(c+d x) \left (-3 b^2 \left (15+8 n+n^2\right ) \left (b^2 (1+n)+a^2 (6+n)\right )-6 a b^3 (2+n) (4+n) (6+n) \sin (c+d x)\right ) \, dx}{b^2 (2+n) (3+n) (4+n) (5+n) (6+n)} \\ & = -\frac {\left (3 b^4 (5+n)+2 a^4 \left (6+5 n+n^2\right )-2 a^2 b^2 \left (40+13 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)}-\frac {2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 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)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\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)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)}+\frac {(6 a b) \int \sin ^{1+n}(c+d x) \, dx}{15+8 n+n^2}+\frac {\left (3 \left (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^4 (5+n)+2 a^4 \left (6+5 n+n^2\right )-2 a^2 b^2 \left (40+13 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)}+\frac {3 \left (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 {2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 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)}+\frac {6 a b \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) \left (15+8 n+n^2\right ) \sqrt {\cos ^2(c+d x)}}-\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\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)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.34 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{1+n}(c+d x) \left (a^2 \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+b (1+n) \sin (c+d x) \left (2 a (3+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right )+b (2+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)\right )\right )}{d (1+n) (2+n) (3+n)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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