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

Optimal result

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

(-n^2-n+9)*cos(d*x+c)*(a+a*sin(d*x+c))^n/d/(1+n)/(2+n)/(3+n)/(4+n)-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/(1+n)/(2+n)/(3+n)/(4+n)-(n^2+3*n+9)*cos(d*x+c)*(a+a* 
sin(d*x+c))^(1+n)/a/d/(2+n)/(3+n)/(4+n)
 

Mathematica [B] (warning: unable to verify)

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

Time = 30.80 (sec) , antiderivative size = 2746, normalized size of antiderivative = 9.34 \[ \int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx=\text {Result too large to show} \] Input:

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

Output:

(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 + (Sq 
rt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n*((15 + 16*n + 4*n^2)*Hypergeometri 
c2F1[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[S 
ec[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[Se 
c[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[S...
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3262, 3042, 3462, 3042, 3447, 3042, 3502, 3042, 3230, 3042, 3131, 3042, 3130}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin ^4(c+d x) (a \sin (c+d x)+a)^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^4 (a \sin (c+d x)+a)^ndx\)

\(\Big \downarrow \) 3262

\(\displaystyle \frac {\int \sin ^2(c+d x) (\sin (c+d x) a+a)^n (n \sin (c+d x) a+3 a)dx}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin (c+d x)^2 (\sin (c+d x) a+a)^n (n \sin (c+d x) a+3 a)dx}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {\int \sin (c+d x) (\sin (c+d x) a+a)^n \left (2 n a^2+\left (n^2+3 n+9\right ) \sin (c+d x) a^2\right )dx}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sin (c+d x) (\sin (c+d x) a+a)^n \left (2 n a^2+\left (n^2+3 n+9\right ) \sin (c+d x) a^2\right )dx}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {\int (\sin (c+d x) a+a)^n \left (\left (n^2+3 n+9\right ) \sin ^2(c+d x) a^2+2 n \sin (c+d x) a^2\right )dx}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int (\sin (c+d x) a+a)^n \left (\left (n^2+3 n+9\right ) \sin (c+d x)^2 a^2+2 n \sin (c+d x) a^2\right )dx}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\frac {\int (\sin (c+d x) a+a)^n \left (a^3 (n+1) \left (n^2+3 n+9\right )-a^3 \left (-n^2-n+9\right ) \sin (c+d x)\right )dx}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int (\sin (c+d x) a+a)^n \left (a^3 (n+1) \left (n^2+3 n+9\right )-a^3 \left (-n^2-n+9\right ) \sin (c+d x)\right )dx}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {\frac {\frac {a^3 \left (n^4+6 n^3+17 n^2+12 n+9\right ) \int (\sin (c+d x) a+a)^ndx}{n+1}+\frac {a^3 \left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {a^3 \left (n^4+6 n^3+17 n^2+12 n+9\right ) \int (\sin (c+d x) a+a)^ndx}{n+1}+\frac {a^3 \left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {\frac {\frac {a^3 \left (n^4+6 n^3+17 n^2+12 n+9\right ) (\sin (c+d x)+1)^{-n} (a \sin (c+d x)+a)^n \int (\sin (c+d x)+1)^ndx}{n+1}+\frac {a^3 \left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {a^3 \left (n^4+6 n^3+17 n^2+12 n+9\right ) (\sin (c+d x)+1)^{-n} (a \sin (c+d x)+a)^n \int (\sin (c+d x)+1)^ndx}{n+1}+\frac {a^3 \left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {\frac {\frac {a^3 \left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)}-\frac {a^3 2^{n+\frac {1}{2}} \left (n^4+6 n^3+17 n^2+12 n+9\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1)}}{a (n+2)}-\frac {a \left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{d (n+2)}}{a (n+3)}-\frac {a n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}}{a (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}\)

Input:

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

Output:

-((Cos[c + d*x]*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^n)/(d*(4 + n))) + (-(( 
a*n*Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^n)/(d*(3 + n))) + (-( 
(a*(9 + 3*n + n^2)*Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(d*(2 + n))) 
 + ((a^3*(9 - n - n^2)*Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(1 + n)) - 
(2^(1/2 + n)*a^3*(9 + 12*n + 17*n^2 + 6*n^3 + n^4)*Cos[c + d*x]*Hypergeome 
tric2F1[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)))/(a*(2 + n)))/(a*(3 + n)))/(a*(4 
+ n))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3130

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]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[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 3230

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] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[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 3262

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*(a + b*Sin[e + f*x]) 
^m*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[1/(b*(m + n))   In 
t[(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] && EqQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]
 

rule 3447

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 3462

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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