\(\int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx\) [140]

Optimal result

Integrand size = 21, antiderivative size = 294 \[ \int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx=\frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {2^{\frac {1}{2}+n} \left (9+12 n+17 n^2+6 n^3+n^4\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)} \]

[Out]

(-n^2-n+9)*cos(d*x+c)*(a+a*sin(d*x+c))^n/d/(n^4+10*n^3+35*n^2+50*n+24)-n*cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+
c))^n/d/(3+n)/(4+n)-cos(d*x+c)*sin(d*x+c)^3*(a+a*sin(d*x+c))^n/d/(4+n)-2^(1/2+n)*(n^4+6*n^3+17*n^2+12*n+9)*cos
(d*x+c)*hypergeom([1/2, 1/2-n],[3/2],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(-1/2-n)*(a+a*sin(d*x+c))^n/d/(4+n)/(n
^3+6*n^2+11*n+6)-(n^2+3*n+9)*cos(d*x+c)*(a+a*sin(d*x+c))^(1+n)/a/d/(4+n)/(n^2+5*n+6)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2862, 3062, 3047, 3102, 2830, 2731, 2730} \[ \int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {2^{n+\frac {1}{2}} \left (n^4+6 n^3+17 n^2+12 n+9\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2) (n+3) (n+4)}+\frac {\left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1) (n+2) (n+3) (n+4)}-\frac {\left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d (n+2) (n+3) (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}-\frac {n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3) (n+4)} \]

[In]

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

[Out]

((9 - n - n^2)*Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(1 + n)*(2 + n)*(3 + n)*(4 + n)) - (n*Cos[c + d*x]*Sin[
c + d*x]^2*(a + a*Sin[c + d*x])^n)/(d*(3 + n)*(4 + n)) - (Cos[c + d*x]*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^n)/
(d*(4 + n)) - (2^(1/2 + n)*(9 + 12*n + 17*n^2 + 6*n^3 + n^4)*Cos[c + d*x]*Hypergeometric2F1[1/2, 1/2 - n, 3/2,
(1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(-1/2 - n)*(a + a*Sin[c + d*x])^n)/(d*(1 + n)*(2 + n)*(3 + n)*(4 + n
)) - ((9 + 3*n + n^2)*Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(a*d*(2 + n)*(3 + n)*(4 + n))

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2830

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

Rule 2862

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

Rule 3047

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

Rule 3062

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}+\frac {\int \sin ^2(c+d x) (a+a \sin (c+d x))^n (3 a+a n \sin (c+d x)) \, dx}{a (4+n)} \\ & = -\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}+\frac {\int \sin (c+d x) (a+a \sin (c+d x))^n \left (2 a^2 n+a^2 \left (9+3 n+n^2\right ) \sin (c+d x)\right ) \, dx}{a^2 (3+n) (4+n)} \\ & = -\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}+\frac {\int (a+a \sin (c+d x))^n \left (2 a^2 n \sin (c+d x)+a^2 \left (9+3 n+n^2\right ) \sin ^2(c+d x)\right ) \, dx}{a^2 (3+n) (4+n)} \\ & = -\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}+\frac {\int (a+a \sin (c+d x))^n \left (a^3 (1+n) \left (9+3 n+n^2\right )-a^3 \left (9-n-n^2\right ) \sin (c+d x)\right ) \, dx}{a^3 (2+n) (3+n) (4+n)} \\ & = \frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}+\frac {\left (9+12 n+17 n^2+6 n^3+n^4\right ) \int (a+a \sin (c+d x))^n \, dx}{(1+n) (2+n) (3+n) (4+n)} \\ & = \frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}+\frac {\left (\left (9+12 n+17 n^2+6 n^3+n^4\right ) (1+\sin (c+d x))^{-n} (a+a \sin (c+d x))^n\right ) \int (1+\sin (c+d x))^n \, dx}{(1+n) (2+n) (3+n) (4+n)} \\ & = \frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {2^{\frac {1}{2}+n} \left (9+12 n+17 n^2+6 n^3+n^4\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2746\) vs. \(2(294)=588\).

Time = 25.23 (sec) , antiderivative size = 2746, normalized size of antiderivative = 9.34 \[ \int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx=\text {Result too large to show} \]

[In]

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

[Out]

(2*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^n*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[
Sec[c + d*x]^2]*Tan[c + d*x])*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*((15 + 16*n + 4*n^2)*Hypergeomet
ric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*(1 + 2*n)*((5 + 2*n)*Hypergeomet
ric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c +
d*x] + 2*Tan[c + d*x]^2) - 2*(3 + 2*n)*Hypergeometric2F1[5/2 + n, 5 + n, 7/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan
[c + d*x])^2]*(1 + 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 8*Tan[c + d*x]^2 + 8*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x
]^3 + 8*Tan[c + d*x]^4))))/(d*(1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sqrt[Sec[c + d*x]^2]*((4*n*(Sqrt[Sec[c + d*x]^2] +
Tan[c + d*x])*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*
x])^2*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-1 + n)*((15 + 16*n + 4*n^2)*Hypergeometric2F1[1/2 + n, 1
+ n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*(1 + 2*n)*((5 + 2*n)*Hypergeometric2F1[3/2 + n, 3
+ n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c +
d*x]^2) - 2*(3 + 2*n)*Hypergeometric2F1[5/2 + n, 5 + n, 7/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1
+ 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 8*Tan[c + d*x]^2 + 8*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^3 + 8*Tan[c + d
*x]^4))))/((1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sqrt[Sec[c + d*x]^2]) - (2*Tan[c + d*x]*(a + (a*Tan[c + d*x])/Sqrt[Se
c[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x
])^2)^n*((15 + 16*n + 4*n^2)*Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])
^2] - 8*(1 + 2*n)*((5 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])
^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2) - 2*(3 + 2*n)*Hypergeometric2F1[5/2 + n, 5 +
n, 7/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 8*Tan[c + d*x
]^2 + 8*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^3 + 8*Tan[c + d*x]^4))))/((1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sqrt[Sec[c +
d*x]^2]) + (2*n*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^(-1 + n)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*T
an[c + d*x])*(a*Sqrt[Sec[c + d*x]^2] - (a*Tan[c + d*x]^2)/Sqrt[Sec[c + d*x]^2])*(1 + (Sqrt[Sec[c + d*x]^2] + T
an[c + d*x])^2)^n*((15 + 16*n + 4*n^2)*Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan
[c + d*x])^2] - 8*(1 + 2*n)*((5 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan
[c + d*x])^2]*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2) - 2*(3 + 2*n)*Hypergeometric2F1[5/2
+ n, 5 + n, 7/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 8*T
an[c + d*x]^2 + 8*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^3 + 8*Tan[c + d*x]^4))))/((1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sq
rt[Sec[c + d*x]^2]) + (2*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*((Sec[c + d*x]^2)^(3/2) + 2*Sec[c + d*x
]^2*Tan[c + d*x] + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^2)*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*((15 +
16*n + 4*n^2)*Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*(1 + 2
*n)*((5 + 2*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 2*Sqr
t[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2) - 2*(3 + 2*n)*Hypergeometric2F1[5/2 + n, 5 + n, 7/2 + n, -(
Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(1 + 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 8*Tan[c + d*x]^2 + 8*Sqrt[S
ec[c + d*x]^2]*Tan[c + d*x]^3 + 8*Tan[c + d*x]^4))))/((1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sqrt[Sec[c + d*x]^2]) + (2
*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + (Sqrt
[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*((2*(1/2 + n)*(15 + 16*n + 4*n^2)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]
*Tan[c + d*x])*(-Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + (1 + (
Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-1 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x]) - 8*(1 + 2*n)*((5 + 2
*n)*Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(2*(Sec[c + d*x]^2)^(
3/2) + 4*Sec[c + d*x]^2*Tan[c + d*x] + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^2) - 2*(3 + 2*n)*Hypergeometric2F1[
5/2 + n, 5 + n, 7/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2]*(4*(Sec[c + d*x]^2)^(3/2) + 16*Sec[c + d*x]
^2*Tan[c + d*x] + 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^2 + 24*(Sec[c + d*x]^2)^(3/2)*Tan[c + d*x]^2 + 32*Sec[c
+ d*x]^2*Tan[c + d*x]^3 + 8*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^4) - (4*(5/2 + n)*(3 + 2*n)*(Sec[c + d*x]^2 + Sq
rt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + 4*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 8*Tan[c + d*x]^2 + 8*Sqrt[Sec[c +
d*x]^2]*Tan[c + d*x]^3 + 8*Tan[c + d*x]^4)*(-Hypergeometric2F1[5/2 + n, 5 + n, 7/2 + n, -(Sqrt[Sec[c + d*x]^2]
+ Tan[c + d*x])^2] + (1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-5 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c +
d*x]) + (2*(3/2 + n)*(5 + 2*n)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + 2*Sqrt[Sec[c + d*x]^
2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)*(-Hypergeometric2F1[3/2 + n, 3 + n, 5/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[
c + d*x])^2] + (1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-3 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x]))
))/((1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sqrt[Sec[c + d*x]^2])))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*(a*sin(d*x + c) + a)^n, x)

Sympy [F]

\[ \int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{n} \sin ^{4}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**n*sin(c + d*x)**4, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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