\(\int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1231]

Optimal result

Integrand size = 29, antiderivative size = 170 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {2 a b \sin ^{2+n}(c+d x)}{d (2+n)}-\frac {\left (2 a^2-b^2\right ) \sin ^{3+n}(c+d x)}{d (3+n)}-\frac {4 a b \sin ^{4+n}(c+d x)}{d (4+n)}+\frac {\left (a^2-2 b^2\right ) \sin ^{5+n}(c+d x)}{d (5+n)}+\frac {2 a b \sin ^{6+n}(c+d x)}{d (6+n)}+\frac {b^2 \sin ^{7+n}(c+d x)}{d (7+n)} \] Output:

a^2*sin(d*x+c)^(1+n)/d/(1+n)+2*a*b*sin(d*x+c)^(2+n)/d/(2+n)-(2*a^2-b^2)*si 
n(d*x+c)^(3+n)/d/(3+n)-4*a*b*sin(d*x+c)^(4+n)/d/(4+n)+(a^2-2*b^2)*sin(d*x+ 
c)^(5+n)/d/(5+n)+2*a*b*sin(d*x+c)^(6+n)/d/(6+n)+b^2*sin(d*x+c)^(7+n)/d/(7+ 
n)
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.82 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {a^2}{1+n}+\frac {2 a b \sin (c+d x)}{2+n}-\frac {\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{3+n}-\frac {4 a b \sin ^3(c+d x)}{4+n}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{5+n}+\frac {2 a b \sin ^5(c+d x)}{6+n}+\frac {b^2 \sin ^6(c+d x)}{7+n}\right )}{d} \] Input:

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^2,x]
 

Output:

(Sin[c + d*x]^(1 + n)*(a^2/(1 + n) + (2*a*b*Sin[c + d*x])/(2 + n) - ((2*a^ 
2 - b^2)*Sin[c + d*x]^2)/(3 + n) - (4*a*b*Sin[c + d*x]^3)/(4 + n) + ((a^2 
- 2*b^2)*Sin[c + d*x]^4)/(5 + n) + (2*a*b*Sin[c + d*x]^5)/(6 + n) + (b^2*S 
in[c + d*x]^6)/(7 + n)))/d
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3316, 522, 2009}

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.

\(\displaystyle \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^5 \sin (c+d x)^n (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {\int \left (a^2 b^4 \sin ^n(c+d x)+2 a b^5 \sin ^{n+1}(c+d x)-b^4 \left (2 a^2-b^2\right ) \sin ^{n+2}(c+d x)-4 a b^5 \sin ^{n+3}(c+d x)+b^4 \left (a^2-2 b^2\right ) \sin ^{n+4}(c+d x)+2 a b^5 \sin ^{n+5}(c+d x)+b^6 \sin ^{n+6}(c+d x)\right )d(b \sin (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^2 b^5 \sin ^{n+1}(c+d x)}{n+1}-\frac {b^5 \left (2 a^2-b^2\right ) \sin ^{n+3}(c+d x)}{n+3}+\frac {b^5 \left (a^2-2 b^2\right ) \sin ^{n+5}(c+d x)}{n+5}+\frac {2 a b^6 \sin ^{n+2}(c+d x)}{n+2}-\frac {4 a b^6 \sin ^{n+4}(c+d x)}{n+4}+\frac {2 a b^6 \sin ^{n+6}(c+d x)}{n+6}+\frac {b^7 \sin ^{n+7}(c+d x)}{n+7}}{b^5 d}\)

Input:

Int[Cos[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^2,x]
 

Output:

((a^2*b^5*Sin[c + d*x]^(1 + n))/(1 + n) + (2*a*b^6*Sin[c + d*x]^(2 + n))/( 
2 + n) - (b^5*(2*a^2 - b^2)*Sin[c + d*x]^(3 + n))/(3 + n) - (4*a*b^6*Sin[c 
 + d*x]^(4 + n))/(4 + n) + (b^5*(a^2 - 2*b^2)*Sin[c + d*x]^(5 + n))/(5 + n 
) + (2*a*b^6*Sin[c + d*x]^(6 + n))/(6 + n) + (b^7*Sin[c + d*x]^(7 + n))/(7 
 + n))/(b^5*d)
 

Defintions of rubi rules used

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3316

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

Maple [A] (verified)

Time = 11.19 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {a^{2} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {b^{2} \sin \left (d x +c \right )^{7} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (7+n \right )}+\frac {\left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )^{5} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (5+n \right )}-\frac {\left (2 a^{2}-b^{2}\right ) \sin \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {2 a b \sin \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}-\frac {4 a b \sin \left (d x +c \right )^{4} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (4+n \right )}+\frac {2 a b \sin \left (d x +c \right )^{6} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (6+n \right )}\)	\(225\)
default	\(\frac {a^{2} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {b^{2} \sin \left (d x +c \right )^{7} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (7+n \right )}+\frac {\left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )^{5} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (5+n \right )}-\frac {\left (2 a^{2}-b^{2}\right ) \sin \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {2 a b \sin \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}-\frac {4 a b \sin \left (d x +c \right )^{4} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (4+n \right )}+\frac {2 a b \sin \left (d x +c \right )^{6} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (6+n \right )}\)	\(225\)
parallelrisch	\(\frac {\left (\frac {3 \left (\left (a^{2}+\frac {b^{2}}{4}\right ) n^{2}+\frac {2 \left (23 a^{2}+5 b^{2}\right ) n}{3}+\frac {175 a^{2}}{3}-\frac {35 b^{2}}{12}\right ) \left (4+n \right ) \left (2+n \right ) \left (1+n \right ) \left (6+n \right ) \sin \left (3 d x +3 c \right )}{2}+\frac {\left (\left (a^{2}-\frac {b^{2}}{4}\right ) n +7 a^{2}-\frac {21 b^{2}}{4}\right ) \left (3+n \right ) \left (4+n \right ) \left (2+n \right ) \left (1+n \right ) \left (6+n \right ) \sin \left (5 d x +5 c \right )}{2}+\frac {a b \left (7+n \right ) \left (5+n \right ) \left (3+n \right ) \left (1+n \right ) \left (n^{2}+6 n -120\right ) \cos \left (2 d x +2 c \right )}{2}-a b \left (n +12\right ) \left (7+n \right ) \left (5+n \right ) \left (3+n \right ) \left (2+n \right ) \left (1+n \right ) \cos \left (4 d x +4 c \right )-\frac {a b \left (7+n \right ) \left (5+n \right ) \left (4+n \right ) \left (3+n \right ) \left (2+n \right ) \left (1+n \right ) \cos \left (6 d x +6 c \right )}{2}-\frac {b^{2} \left (6+n \right ) \left (5+n \right ) \left (4+n \right ) \left (3+n \right ) \left (2+n \right ) \left (1+n \right ) \sin \left (7 d x +7 c \right )}{8}+\left (\left (a^{2}+\frac {3 b^{2}}{8}\right ) n^{3}+\left (19 a^{2}+\frac {59 b^{2}}{8}\right ) n^{2}+\left (159 a^{2}+\frac {581 b^{2}}{8}\right ) n +525 a^{2}+\frac {525 b^{2}}{8}\right ) \left (4+n \right ) \left (2+n \right ) \left (6+n \right ) \sin \left (d x +c \right )+a b \left (7+n \right ) \left (5+n \right ) \left (3+n \right ) \left (1+n \right ) \left (n^{2}+14 n +88\right )\right ) \sin \left (d x +c \right )^{n}}{8 \left (7+n \right ) \left (4+n \right ) \left (2+n \right ) \left (6+n \right ) d \left (3+n \right ) \left (1+n \right ) \left (5+n \right )}\)	\(377\)

Input:

int(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

a^2/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))+b^2/d/(7+n)*sin(d*x+c)^7*exp( 
n*ln(sin(d*x+c)))+(a^2-2*b^2)/d/(5+n)*sin(d*x+c)^5*exp(n*ln(sin(d*x+c)))-( 
2*a^2-b^2)/d/(3+n)*sin(d*x+c)^3*exp(n*ln(sin(d*x+c)))+2*a*b/d/(2+n)*sin(d* 
x+c)^2*exp(n*ln(sin(d*x+c)))-4*a*b/d/(4+n)*sin(d*x+c)^4*exp(n*ln(sin(d*x+c 
)))+2*a*b/d/(6+n)*sin(d*x+c)^6*exp(n*ln(sin(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 572 vs. \(2 (170) = 340\).

Time = 0.26 (sec) , antiderivative size = 572, normalized size of antiderivative = 3.36 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {{\left (2 \, {\left (a b n^{6} + 22 \, a b n^{5} + 190 \, a b n^{4} + 820 \, a b n^{3} + 1849 \, a b n^{2} + 2038 \, a b n + 840 \, a b\right )} \cos \left (d x + c\right )^{6} - 16 \, a b n^{4} - 256 \, a b n^{3} - 2 \, {\left (a b n^{6} + 18 \, a b n^{5} + 118 \, a b n^{4} + 348 \, a b n^{3} + 457 \, a b n^{2} + 210 \, a b n\right )} \cos \left (d x + c\right )^{4} - 1376 \, a b n^{2} - 2816 \, a b n - 8 \, {\left (a b n^{5} + 16 \, a b n^{4} + 86 \, a b n^{3} + 176 \, a b n^{2} + 105 \, a b n\right )} \cos \left (d x + c\right )^{2} - 1680 \, a b + {\left ({\left (b^{2} n^{6} + 21 \, b^{2} n^{5} + 175 \, b^{2} n^{4} + 735 \, b^{2} n^{3} + 1624 \, b^{2} n^{2} + 1764 \, b^{2} n + 720 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 8 \, {\left (a^{2} + b^{2}\right )} n^{4} - {\left ({\left (a^{2} + b^{2}\right )} n^{6} + {\left (23 \, a^{2} + 17 \, b^{2}\right )} n^{5} + 3 \, {\left (69 \, a^{2} + 37 \, b^{2}\right )} n^{4} + 5 \, {\left (185 \, a^{2} + 71 \, b^{2}\right )} n^{3} + 8 \, {\left (268 \, a^{2} + 73 \, b^{2}\right )} n^{2} + 1008 \, a^{2} + 144 \, b^{2} + 36 \, {\left (67 \, a^{2} + 13 \, b^{2}\right )} n\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (19 \, a^{2} + 13 \, b^{2}\right )} n^{3} - 64 \, {\left (16 \, a^{2} + 7 \, b^{2}\right )} n^{2} - 4 \, {\left ({\left (a^{2} + b^{2}\right )} n^{5} + 2 \, {\left (10 \, a^{2} + 7 \, b^{2}\right )} n^{4} + 3 \, {\left (49 \, a^{2} + 23 \, b^{2}\right )} n^{3} + 4 \, {\left (121 \, a^{2} + 37 \, b^{2}\right )} n^{2} + 336 \, a^{2} + 48 \, b^{2} + 4 \, {\left (173 \, a^{2} + 35 \, b^{2}\right )} n\right )} \cos \left (d x + c\right )^{2} - 2688 \, a^{2} - 384 \, b^{2} - 32 \, {\left (89 \, a^{2} + 23 \, b^{2}\right )} n\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{7} + 28 \, d n^{6} + 322 \, d n^{5} + 1960 \, d n^{4} + 6769 \, d n^{3} + 13132 \, d n^{2} + 13068 \, d n + 5040 \, d} \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

-(2*(a*b*n^6 + 22*a*b*n^5 + 190*a*b*n^4 + 820*a*b*n^3 + 1849*a*b*n^2 + 203 
8*a*b*n + 840*a*b)*cos(d*x + c)^6 - 16*a*b*n^4 - 256*a*b*n^3 - 2*(a*b*n^6 
+ 18*a*b*n^5 + 118*a*b*n^4 + 348*a*b*n^3 + 457*a*b*n^2 + 210*a*b*n)*cos(d* 
x + c)^4 - 1376*a*b*n^2 - 2816*a*b*n - 8*(a*b*n^5 + 16*a*b*n^4 + 86*a*b*n^ 
3 + 176*a*b*n^2 + 105*a*b*n)*cos(d*x + c)^2 - 1680*a*b + ((b^2*n^6 + 21*b^ 
2*n^5 + 175*b^2*n^4 + 735*b^2*n^3 + 1624*b^2*n^2 + 1764*b^2*n + 720*b^2)*c 
os(d*x + c)^6 - 8*(a^2 + b^2)*n^4 - ((a^2 + b^2)*n^6 + (23*a^2 + 17*b^2)*n 
^5 + 3*(69*a^2 + 37*b^2)*n^4 + 5*(185*a^2 + 71*b^2)*n^3 + 8*(268*a^2 + 73* 
b^2)*n^2 + 1008*a^2 + 144*b^2 + 36*(67*a^2 + 13*b^2)*n)*cos(d*x + c)^4 - 8 
*(19*a^2 + 13*b^2)*n^3 - 64*(16*a^2 + 7*b^2)*n^2 - 4*((a^2 + b^2)*n^5 + 2* 
(10*a^2 + 7*b^2)*n^4 + 3*(49*a^2 + 23*b^2)*n^3 + 4*(121*a^2 + 37*b^2)*n^2 
+ 336*a^2 + 48*b^2 + 4*(173*a^2 + 35*b^2)*n)*cos(d*x + c)^2 - 2688*a^2 - 3 
84*b^2 - 32*(89*a^2 + 23*b^2)*n)*sin(d*x + c))*sin(d*x + c)^n/(d*n^7 + 28* 
d*n^6 + 322*d*n^5 + 1960*d*n^4 + 6769*d*n^3 + 13132*d*n^2 + 13068*d*n + 50 
40*d)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18260 vs. \(2 (144) = 288\).

Time = 16.61 (sec) , antiderivative size = 18260, normalized size of antiderivative = 107.41 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)**5*sin(d*x+c)**n*(a+b*sin(d*x+c))**2,x)
 

Output:

Piecewise((x*(a + b*sin(c))**2*sin(c)**n*cos(c)**5, Eq(d, 0)), (-a**2/(6*d 
*sin(c + d*x)**2) + a**2*cos(c + d*x)**2/(6*d*sin(c + d*x)**4) - a**2*cos( 
c + d*x)**4/(6*d*sin(c + d*x)**6) - 16*a*b/(15*d*sin(c + d*x)) + 8*a*b*cos 
(c + d*x)**2/(15*d*sin(c + d*x)**3) - 2*a*b*cos(c + d*x)**4/(5*d*sin(c + d 
*x)**5) + b**2*log(sin(c + d*x))/d + b**2*cos(c + d*x)**2/(2*d*sin(c + d*x 
)**2) - b**2*cos(c + d*x)**4/(4*d*sin(c + d*x)**4), Eq(n, -7)), (-8*a**2/( 
15*d*sin(c + d*x)) + 4*a**2*cos(c + d*x)**2/(15*d*sin(c + d*x)**3) - a**2* 
cos(c + d*x)**4/(5*d*sin(c + d*x)**5) + 2*a*b*log(sin(c + d*x))/d + a*b*co 
s(c + d*x)**2/(d*sin(c + d*x)**2) - a*b*cos(c + d*x)**4/(2*d*sin(c + d*x)* 
*4) + 8*b**2*sin(c + d*x)/(3*d) + 4*b**2*cos(c + d*x)**2/(3*d*sin(c + d*x) 
) - b**2*cos(c + d*x)**4/(3*d*sin(c + d*x)**3), Eq(n, -6)), (-192*a**2*log 
(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**8/(192*d*tan(c/2 + d*x/2)**8 + 
 384*d*tan(c/2 + d*x/2)**6 + 192*d*tan(c/2 + d*x/2)**4) - 384*a**2*log(tan 
(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**6/(192*d*tan(c/2 + d*x/2)**8 + 384 
*d*tan(c/2 + d*x/2)**6 + 192*d*tan(c/2 + d*x/2)**4) - 192*a**2*log(tan(c/2 
 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(192*d*tan(c/2 + d*x/2)**8 + 384*d*t 
an(c/2 + d*x/2)**6 + 192*d*tan(c/2 + d*x/2)**4) + 192*a**2*log(tan(c/2 + d 
*x/2))*tan(c/2 + d*x/2)**8/(192*d*tan(c/2 + d*x/2)**8 + 384*d*tan(c/2 + d* 
x/2)**6 + 192*d*tan(c/2 + d*x/2)**4) + 384*a**2*log(tan(c/2 + d*x/2))*tan( 
c/2 + d*x/2)**6/(192*d*tan(c/2 + d*x/2)**8 + 384*d*tan(c/2 + d*x/2)**6 ...
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\frac {b^{2} \sin \left (d x + c\right )^{n + 7}}{n + 7} + \frac {2 \, a b \sin \left (d x + c\right )^{n + 6}}{n + 6} + \frac {a^{2} \sin \left (d x + c\right )^{n + 5}}{n + 5} - \frac {2 \, b^{2} \sin \left (d x + c\right )^{n + 5}}{n + 5} - \frac {4 \, a b \sin \left (d x + c\right )^{n + 4}}{n + 4} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {b^{2} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {2 \, a b \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a^{2} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

(b^2*sin(d*x + c)^(n + 7)/(n + 7) + 2*a*b*sin(d*x + c)^(n + 6)/(n + 6) + a 
^2*sin(d*x + c)^(n + 5)/(n + 5) - 2*b^2*sin(d*x + c)^(n + 5)/(n + 5) - 4*a 
*b*sin(d*x + c)^(n + 4)/(n + 4) - 2*a^2*sin(d*x + c)^(n + 3)/(n + 3) + b^2 
*sin(d*x + c)^(n + 3)/(n + 3) + 2*a*b*sin(d*x + c)^(n + 2)/(n + 2) + a^2*s 
in(d*x + c)^(n + 1)/(n + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.52 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\frac {a^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5}}{n + 5} + \frac {b^{2} e^{\left (n \log \left (\sin \left (d x + c\right )\right ) + 2 \, \log \left (\sin \left (d x + c\right )\right )\right )} \sin \left (d x + c\right )^{5}}{n + 7} + \frac {2 \, a b e^{\left (n \log \left (\sin \left (d x + c\right )\right ) + \log \left (\sin \left (d x + c\right )\right )\right )} \sin \left (d x + c\right )^{5}}{n + 6} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} - \frac {2 \, b^{2} e^{\left (n \log \left (\sin \left (d x + c\right )\right ) + 2 \, \log \left (\sin \left (d x + c\right )\right )\right )} \sin \left (d x + c\right )^{3}}{n + 5} - \frac {4 \, a b e^{\left (n \log \left (\sin \left (d x + c\right )\right ) + \log \left (\sin \left (d x + c\right )\right )\right )} \sin \left (d x + c\right )^{3}}{n + 4} + \frac {b^{2} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {2 \, a b \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a^{2} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

(a^2*sin(d*x + c)^n*sin(d*x + c)^5/(n + 5) + b^2*e^(n*log(sin(d*x + c)) + 
2*log(sin(d*x + c)))*sin(d*x + c)^5/(n + 7) + 2*a*b*e^(n*log(sin(d*x + c)) 
 + log(sin(d*x + c)))*sin(d*x + c)^5/(n + 6) - 2*a^2*sin(d*x + c)^n*sin(d* 
x + c)^3/(n + 3) - 2*b^2*e^(n*log(sin(d*x + c)) + 2*log(sin(d*x + c)))*sin 
(d*x + c)^3/(n + 5) - 4*a*b*e^(n*log(sin(d*x + c)) + log(sin(d*x + c)))*si 
n(d*x + c)^3/(n + 4) + b^2*sin(d*x + c)^(n + 3)/(n + 3) + 2*a*b*sin(d*x + 
c)^(n + 2)/(n + 2) + a^2*sin(d*x + c)^(n + 1)/(n + 1))/d
 

Mupad [B] (verification not implemented)

Time = 29.70 (sec) , antiderivative size = 887, normalized size of antiderivative = 5.22 \[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^5*sin(c + d*x)^n*(a + b*sin(c + d*x))^2,x)
 

Output:

(sin(c + d*x)*sin(c + d*x)^n*(44*n + 12*n^2 + n^3 + 48)*(1272*a^2*n + 581* 
b^2*n + 4200*a^2 + 525*b^2 + 152*a^2*n^2 + 8*a^2*n^3 + 59*b^2*n^2 + 3*b^2* 
n^3)*1i)/(64*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + 
 n^6*28i + n^7*1i + 5040i)) + (sin(c + d*x)^n*sin(5*c + 5*d*x)*(4*a^2*n - 
b^2*n + 28*a^2 - 21*b^2)*(324*n + 260*n^2 + 95*n^3 + 16*n^4 + n^5 + 144)*1 
i)/(64*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*2 
8i + n^7*1i + 5040i)) - (b^2*sin(c + d*x)^n*sin(7*c + 7*d*x)*(1764*n + 162 
4*n^2 + 735*n^3 + 175*n^4 + 21*n^5 + n^6 + 720)*1i)/(64*d*(n*13068i + n^2* 
13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i)) + ( 
sin(c + d*x)^n*sin(3*c + 3*d*x)*(92*n + 56*n^2 + 13*n^3 + n^4 + 48)*(184*a 
^2*n + 40*b^2*n + 700*a^2 - 35*b^2 + 12*a^2*n^2 + 3*b^2*n^2)*1i)/(64*d*(n* 
13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + n^7*1i 
+ 5040i)) + (a*b*sin(c + d*x)^n*(n*16958i + n^2*10137i + n^3*2788i + n^4*3 
98i + n^5*30i + n^6*1i + 9240i))/(8*d*(n*13068i + n^2*13132i + n^3*6769i + 
 n^4*1960i + n^5*322i + n^6*28i + n^7*1i + 5040i)) - (a*b*sin(c + d*x)^n*c 
os(6*c + 6*d*x)*(n*2038i + n^2*1849i + n^3*820i + n^4*190i + n^5*22i + n^6 
*1i + 840i))/(16*d*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*32 
2i + n^6*28i + n^7*1i + 5040i)) - (a*b*sin(c + d*x)^n*cos(4*c + 4*d*x)*(n* 
5694i + n^2*4633i + n^3*1764i + n^4*334i + n^5*30i + n^6*1i + 2520i))/(8*d 
*(n*13068i + n^2*13132i + n^3*6769i + n^4*1960i + n^5*322i + n^6*28i + ...
 

Reduce [F]

\[ \int \cos ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )d x \right ) a b +\left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{5}d x \right ) a^{2} \] Input:

int(cos(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x)
 

Output:

int(sin(c + d*x)**n*cos(c + d*x)**5*sin(c + d*x)**2,x)*b**2 + 2*int(sin(c 
+ d*x)**n*cos(c + d*x)**5*sin(c + d*x),x)*a*b + int(sin(c + d*x)**n*cos(c 
+ d*x)**5,x)*a**2