3.1523 \(\int \cos ^2(e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx\)

Optimal. Leaf size=552 \[ -\frac{\sqrt{2} (c+d) \cos (e+f x) \left (a^2 c d^2 \left (n^2+7 n+12\right )-2 a b d (n+4) \left (2 c^2-d^2 (n+2)\right )+b^2 c \left (6 c^2-d^2 \left (-n^2-n+3\right )\right )\right ) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{d^4 f (n+2) (n+3) (n+4) \sqrt{\sin (e+f x)+1}}-\frac{\sqrt{2} \left (c^2-d^2\right ) \cos (e+f x) \left (-a^2 d^2 \left (n^2+7 n+12\right )+4 a b c d (n+4)+b^2 \left (-\left (6 c^2+d^2 \left (n^2+4 n+3\right )\right )\right )\right ) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{d^4 f (n+2) (n+3) (n+4) \sqrt{\sin (e+f x)+1}}+\frac{\cos (e+f x) \left (2 a^2 d^2 (n+3)-4 a b c d (n+4)+b^2 \left (6 c^2-d^2 (n+3)\right )\right ) (c+d \sin (e+f x))^{n+1}}{d^3 f (n+2) (n+3) (n+4)}-\frac{b (3 b c-2 a d) \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (n+3) (n+4)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{n+1}}{d f (n+4)} \]

[Out]

((2*a^2*d^2*(3 + n) - 4*a*b*c*d*(4 + n) + b^2*(6*c^2 - d^2*(3 + n)))*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n)
)/(d^3*f*(2 + n)*(3 + n)*(4 + n)) - (b*(3*b*c - 2*a*d)*Cos[e + f*x]*Sin[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))
/(d^2*f*(3 + n)*(4 + n)) + (Cos[e + f*x]*(a + b*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(4 + n)) -
(Sqrt[2]*(c + d)*(a^2*c*d^2*(12 + 7*n + n^2) - 2*a*b*d*(4 + n)*(2*c^2 - d^2*(2 + n)) + b^2*c*(6*c^2 - d^2*(3 -
n - n^2)))*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]
*(c + d*Sin[e + f*x])^n)/(d^4*f*(2 + n)*(3 + n)*(4 + n)*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^
n) - (Sqrt[2]*(c^2 - d^2)*(4*a*b*c*d*(4 + n) - a^2*d^2*(12 + 7*n + n^2) - b^2*(6*c^2 + d^2*(3 + 4*n + n^2)))*A
ppellF1[1/2, 1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f
*x])^n)/(d^4*f*(2 + n)*(3 + n)*(4 + n)*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

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Rubi [A]  time = 1.51228, antiderivative size = 552, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {2922, 3050, 3033, 3023, 2756, 2665, 139, 138} \[ -\frac{\sqrt{2} (c+d) \cos (e+f x) \left (a^2 c d^2 \left (n^2+7 n+12\right )-2 a b d (n+4) \left (2 c^2-d^2 (n+2)\right )+b^2 \left (6 c^3-c d^2 \left (-n^2-n+3\right )\right )\right ) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{d^4 f (n+2) (n+3) (n+4) \sqrt{\sin (e+f x)+1}}-\frac{\sqrt{2} \left (c^2-d^2\right ) \cos (e+f x) \left (-a^2 d^2 \left (n^2+7 n+12\right )+4 a b c d (n+4)+b^2 \left (-\left (6 c^2+d^2 \left (n^2+4 n+3\right )\right )\right )\right ) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{d^4 f (n+2) (n+3) (n+4) \sqrt{\sin (e+f x)+1}}+\frac{\cos (e+f x) \left (2 a^2 d^2 (n+3)-4 a b c d (n+4)+b^2 \left (6 c^2-d^2 (n+3)\right )\right ) (c+d \sin (e+f x))^{n+1}}{d^3 f (n+2) (n+3) (n+4)}-\frac{b (3 b c-2 a d) \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (n+3) (n+4)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{n+1}}{d f (n+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*d^2*(3 + n) - 4*a*b*c*d*(4 + n) + b^2*(6*c^2 - d^2*(3 + n)))*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n)
)/(d^3*f*(2 + n)*(3 + n)*(4 + n)) - (b*(3*b*c - 2*a*d)*Cos[e + f*x]*Sin[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))
/(d^2*f*(3 + n)*(4 + n)) + (Cos[e + f*x]*(a + b*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(4 + n)) -
(Sqrt[2]*(c + d)*(a^2*c*d^2*(12 + 7*n + n^2) - 2*a*b*d*(4 + n)*(2*c^2 - d^2*(2 + n)) + b^2*(6*c^3 - c*d^2*(3 -
n - n^2)))*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]
*(c + d*Sin[e + f*x])^n)/(d^4*f*(2 + n)*(3 + n)*(4 + n)*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^
n) - (Sqrt[2]*(c^2 - d^2)*(4*a*b*c*d*(4 + n) - a^2*d^2*(12 + 7*n + n^2) - b^2*(6*c^2 + d^2*(3 + 4*n + n^2)))*A
ppellF1[1/2, 1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f
*x])^n)/(d^4*f*(2 + n)*(3 + n)*(4 + n)*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

Rule 2922

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

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])))

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 2756

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

Rule 2665

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \cos ^2(e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx &=\int (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^n \left (1-\sin ^2(e+f x)\right ) \, dx\\ &=\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}+\frac{\int (a+b \sin (e+f x)) (c+d \sin (e+f x))^n \left (-2 b c+3 a d+(a c+b d) \sin (e+f x)+(3 b c-2 a d) \sin ^2(e+f x)\right ) \, dx}{d (4+n)}\\ &=-\frac{b (3 b c-2 a d) \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+n) (4+n)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}+\frac{\int (c+d \sin (e+f x))^n \left (3 b^2 c^2+3 a^2 d^2 (3+n)-2 a b c d (4+n)+d \left (b^2 c n+a^2 c (3+n)+2 a b d (4+n)\right ) \sin (e+f x)-\left (2 a^2 d^2 (3+n)-4 a b c d (4+n)+b^2 \left (6 c^2-d^2 (3+n)\right )\right ) \sin ^2(e+f x)\right ) \, dx}{d^2 (3+n) (4+n)}\\ &=\frac{\left (2 a^2 d^2 (3+n)-4 a b c d (4+n)+b^2 \left (6 c^2-d^2 (3+n)\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^3 f (2+n) (3+n) (4+n)}-\frac{b (3 b c-2 a d) \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+n) (4+n)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}+\frac{\int (c+d \sin (e+f x))^n \left (d \left (2 a b c d n (4+n)+a^2 d^2 \left (12+7 n+n^2\right )-b^2 \left (3 c^2 n-d^2 \left (3+4 n+n^2\right )\right )\right )+\left (a^2 c d^2 \left (12+7 n+n^2\right )-2 a b d (4+n) \left (2 c^2-d^2 (2+n)\right )+b^2 \left (6 c^3-c d^2 \left (3-n-n^2\right )\right )\right ) \sin (e+f x)\right ) \, dx}{d^3 (2+n) (3+n) (4+n)}\\ &=\frac{\left (2 a^2 d^2 (3+n)-4 a b c d (4+n)+b^2 \left (6 c^2-d^2 (3+n)\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^3 f (2+n) (3+n) (4+n)}-\frac{b (3 b c-2 a d) \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+n) (4+n)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}+\frac{\left (a^2 c d^2 \left (12+7 n+n^2\right )-2 a b d (4+n) \left (2 c^2-d^2 (2+n)\right )+b^2 \left (6 c^3-c d^2 \left (3-n-n^2\right )\right )\right ) \int (c+d \sin (e+f x))^{1+n} \, dx}{d^4 (2+n) (3+n) (4+n)}+\frac{\left (\left (c^2-d^2\right ) \left (4 a b c d (4+n)-a^2 d^2 \left (12+7 n+n^2\right )-b^2 \left (6 c^2+d^2 \left (3+4 n+n^2\right )\right )\right )\right ) \int (c+d \sin (e+f x))^n \, dx}{d^4 (2+n) (3+n) (4+n)}\\ &=\frac{\left (2 a^2 d^2 (3+n)-4 a b c d (4+n)+b^2 \left (6 c^2-d^2 (3+n)\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^3 f (2+n) (3+n) (4+n)}-\frac{b (3 b c-2 a d) \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+n) (4+n)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}+\frac{\left (\left (a^2 c d^2 \left (12+7 n+n^2\right )-2 a b d (4+n) \left (2 c^2-d^2 (2+n)\right )+b^2 \left (6 c^3-c d^2 \left (3-n-n^2\right )\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{1+n}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{d^4 f (2+n) (3+n) (4+n) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (\left (c^2-d^2\right ) \left (4 a b c d (4+n)-a^2 d^2 \left (12+7 n+n^2\right )-b^2 \left (6 c^2+d^2 \left (3+4 n+n^2\right )\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^n}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{d^4 f (2+n) (3+n) (4+n) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=\frac{\left (2 a^2 d^2 (3+n)-4 a b c d (4+n)+b^2 \left (6 c^2-d^2 (3+n)\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^3 f (2+n) (3+n) (4+n)}-\frac{b (3 b c-2 a d) \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+n) (4+n)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}-\frac{\left ((-c-d) \left (a^2 c d^2 \left (12+7 n+n^2\right )-2 a b d (4+n) \left (2 c^2-d^2 (2+n)\right )+b^2 \left (6 c^3-c d^2 \left (3-n-n^2\right )\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^{1+n}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{d^4 f (2+n) (3+n) (4+n) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (\left (c^2-d^2\right ) \left (4 a b c d (4+n)-a^2 d^2 \left (12+7 n+n^2\right )-b^2 \left (6 c^2+d^2 \left (3+4 n+n^2\right )\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^n}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{d^4 f (2+n) (3+n) (4+n) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=\frac{\left (2 a^2 d^2 (3+n)-4 a b c d (4+n)+b^2 \left (6 c^2-d^2 (3+n)\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^3 f (2+n) (3+n) (4+n)}-\frac{b (3 b c-2 a d) \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+n) (4+n)}+\frac{\cos (e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^{1+n}}{d f (4+n)}-\frac{\sqrt{2} (c+d) \left (a^2 c d^2 \left (12+7 n+n^2\right )-2 a b d (4+n) \left (2 c^2-d^2 (2+n)\right )+b^2 \left (6 c^3-c d^2 \left (3-n-n^2\right )\right )\right ) F_1\left (\frac{1}{2};\frac{1}{2},-1-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^4 f (2+n) (3+n) (4+n) \sqrt{1+\sin (e+f x)}}-\frac{\sqrt{2} \left (c^2-d^2\right ) \left (4 a b c d (4+n)-a^2 d^2 \left (12+7 n+n^2\right )-b^2 \left (6 c^2+d^2 \left (3+4 n+n^2\right )\right )\right ) F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^4 f (2+n) (3+n) (4+n) \sqrt{1+\sin (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 8.32011, size = 0, normalized size = 0. \[ \int \cos ^2(e+f x) (a+b \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.464, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b\sin \left ( fx+e \right ) \right ) ^{2} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right ) + a\right )}^{2}{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, a b \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) -{\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right )^{2}\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right ) + a\right )}^{2}{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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