\(\int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1236]

Optimal result

Integrand size = 29, antiderivative size = 250 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (12 a^2+5 b^2\right ) x}{1024}-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {4 a b \cos ^9(c+d x)}{9 d}-\frac {2 a b \cos ^{11}(c+d x)}{11 d}+\frac {\left (12 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{1024 d}+\frac {\left (12 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{1536 d}+\frac {\left (12 a^2+5 b^2\right ) \cos ^5(c+d x) \sin (c+d x)}{1920 d}-\frac {\left (44 a^2+45 b^2\right ) \cos ^7(c+d x) \sin (c+d x)}{320 d}+\frac {\left (12 a^2+25 b^2\right ) \cos ^9(c+d x) \sin (c+d x)}{120 d}-\frac {b^2 \cos ^{11}(c+d x) \sin (c+d x)}{12 d} \] Output:

1/1024*(12*a^2+5*b^2)*x-2/7*a*b*cos(d*x+c)^7/d+4/9*a*b*cos(d*x+c)^9/d-2/11 
*a*b*cos(d*x+c)^11/d+1/1024*(12*a^2+5*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/1536* 
(12*a^2+5*b^2)*cos(d*x+c)^3*sin(d*x+c)/d+1/1920*(12*a^2+5*b^2)*cos(d*x+c)^ 
5*sin(d*x+c)/d-1/320*(44*a^2+45*b^2)*cos(d*x+c)^7*sin(d*x+c)/d+1/120*(12*a 
^2+25*b^2)*cos(d*x+c)^9*sin(d*x+c)/d-1/12*b^2*cos(d*x+c)^11*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.81 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {166320 b^2 c+332640 a^2 d x+138600 b^2 d x-554400 a b \cos (c+d x)-184800 a b \cos (3 (c+d x))+55440 a b \cos (5 (c+d x))+39600 a b \cos (7 (c+d x))-6160 a b \cos (9 (c+d x))-5040 a b \cos (11 (c+d x))+55440 a^2 \sin (2 (c+d x))-110880 a^2 \sin (4 (c+d x))-51975 b^2 \sin (4 (c+d x))-27720 a^2 \sin (6 (c+d x))+13860 a^2 \sin (8 (c+d x))+10395 b^2 \sin (8 (c+d x))+5544 a^2 \sin (10 (c+d x))-1155 b^2 \sin (12 (c+d x))}{28385280 d} \] Input:

Integrate[Cos[c + d*x]^6*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]
 

Output:

(166320*b^2*c + 332640*a^2*d*x + 138600*b^2*d*x - 554400*a*b*Cos[c + d*x] 
- 184800*a*b*Cos[3*(c + d*x)] + 55440*a*b*Cos[5*(c + d*x)] + 39600*a*b*Cos 
[7*(c + d*x)] - 6160*a*b*Cos[9*(c + d*x)] - 5040*a*b*Cos[11*(c + d*x)] + 5 
5440*a^2*Sin[2*(c + d*x)] - 110880*a^2*Sin[4*(c + d*x)] - 51975*b^2*Sin[4* 
(c + d*x)] - 27720*a^2*Sin[6*(c + d*x)] + 13860*a^2*Sin[8*(c + d*x)] + 103 
95*b^2*Sin[8*(c + d*x)] + 5544*a^2*Sin[10*(c + d*x)] - 1155*b^2*Sin[12*(c 
+ d*x)])/(28385280*d)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3390, 3042, 3045, 244, 2009, 3679, 360, 25, 1471, 27, 298, 215, 215, 215, 216}

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.

\(\displaystyle \int \sin ^4(c+d x) \cos ^6(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^4 \cos (c+d x)^6 (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3390

\(\displaystyle \int \cos ^6(c+d x) \sin ^4(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )dx+2 a b \int \cos ^6(c+d x) \sin ^5(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^6 \sin (c+d x)^4 \left (a^2+b^2 \sin (c+d x)^2\right )dx+2 a b \int \cos (c+d x)^6 \sin (c+d x)^5dx\)

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 244

\(\displaystyle \int \cos (c+d x)^6 \sin (c+d x)^4 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \int \left (\cos ^{10}(c+d x)-2 \cos ^8(c+d x)+\cos ^6(c+d x)\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \cos (c+d x)^6 \sin (c+d x)^4 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3679

\(\displaystyle \frac {\int \frac {\tan ^4(c+d x) \left (a^2+\left (a^2+b^2\right ) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right )^7}d\tan (c+d x)}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 360

\(\displaystyle \frac {-\frac {1}{12} \int -\frac {12 \left (a^2+b^2\right ) \tan ^4(c+d x)-12 b^2 \tan ^2(c+d x)+b^2}{\left (\tan ^2(c+d x)+1\right )^6}d\tan (c+d x)-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{12} \int \frac {12 \left (a^2+b^2\right ) \tan ^4(c+d x)-12 b^2 \tan ^2(c+d x)+b^2}{\left (\tan ^2(c+d x)+1\right )^6}d\tan (c+d x)-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {1}{10} \int \frac {3 \left (4 a^2+5 b^2-40 \left (a^2+b^2\right ) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right )^5}d\tan (c+d x)\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {3}{10} \int \frac {4 a^2+5 b^2-40 \left (a^2+b^2\right ) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right )^5}d\tan (c+d x)\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {3}{10} \left (\frac {\left (44 a^2+45 b^2\right ) \tan (c+d x)}{8 \left (\tan ^2(c+d x)+1\right )^4}-\frac {1}{8} \left (12 a^2+5 b^2\right ) \int \frac {1}{\left (\tan ^2(c+d x)+1\right )^4}d\tan (c+d x)\right )\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {3}{10} \left (\frac {\left (44 a^2+45 b^2\right ) \tan (c+d x)}{8 \left (\tan ^2(c+d x)+1\right )^4}-\frac {1}{8} \left (12 a^2+5 b^2\right ) \left (\frac {5}{6} \int \frac {1}{\left (\tan ^2(c+d x)+1\right )^3}d\tan (c+d x)+\frac {\tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}\right )\right )\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {3}{10} \left (\frac {\left (44 a^2+45 b^2\right ) \tan (c+d x)}{8 \left (\tan ^2(c+d x)+1\right )^4}-\frac {1}{8} \left (12 a^2+5 b^2\right ) \left (\frac {5}{6} \left (\frac {3}{4} \int \frac {1}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)+\frac {\tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}\right )+\frac {\tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}\right )\right )\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {3}{10} \left (\frac {\left (44 a^2+45 b^2\right ) \tan (c+d x)}{8 \left (\tan ^2(c+d x)+1\right )^4}-\frac {1}{8} \left (12 a^2+5 b^2\right ) \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)+\frac {\tan (c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )+\frac {\tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}\right )+\frac {\tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}\right )\right )\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{12} \left (\frac {\left (12 a^2+25 b^2\right ) \tan (c+d x)}{10 \left (\tan ^2(c+d x)+1\right )^5}-\frac {3}{10} \left (\frac {\left (44 a^2+45 b^2\right ) \tan (c+d x)}{8 \left (\tan ^2(c+d x)+1\right )^4}-\frac {1}{8} \left (12 a^2+5 b^2\right ) \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \arctan (\tan (c+d x))+\frac {\tan (c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )+\frac {\tan (c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}\right )+\frac {\tan (c+d x)}{6 \left (\tan ^2(c+d x)+1\right )^3}\right )\right )\right )-\frac {b^2 \tan (c+d x)}{12 \left (\tan ^2(c+d x)+1\right )^6}}{d}-\frac {2 a b \left (\frac {1}{11} \cos ^{11}(c+d x)-\frac {2}{9} \cos ^9(c+d x)+\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

Input:

Int[Cos[c + d*x]^6*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]
 

Output:

(-2*a*b*(Cos[c + d*x]^7/7 - (2*Cos[c + d*x]^9)/9 + Cos[c + d*x]^11/11))/d 
+ (-1/12*(b^2*Tan[c + d*x])/(1 + Tan[c + d*x]^2)^6 + (((12*a^2 + 25*b^2)*T 
an[c + d*x])/(10*(1 + Tan[c + d*x]^2)^5) - (3*(((44*a^2 + 45*b^2)*Tan[c + 
d*x])/(8*(1 + Tan[c + d*x]^2)^4) - ((12*a^2 + 5*b^2)*(Tan[c + d*x]/(6*(1 + 
 Tan[c + d*x]^2)^3) + (5*(Tan[c + d*x]/(4*(1 + Tan[c + d*x]^2)^2) + (3*(Ar 
cTan[Tan[c + d*x]]/2 + Tan[c + d*x]/(2*(1 + Tan[c + d*x]^2))))/4))/6))/8)) 
/10)/12)/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

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 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3390

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[2*a*(b/d) 
Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e + f* 
x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, 
e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0]
 

rule 3679

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_)*((a_) + (b_.) 
*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[ 
e + f*x], x]}, Simp[ff^(n + 1)/f   Subst[Int[x^n*((a + (a + b)*ff^2*x^2)^p/ 
(1 + ff^2*x^2)^((m + n)/2 + p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[ 
{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[n/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.95

\[\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{10}-\frac {3 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{80}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{7}}{11}-\frac {4 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{99}-\frac {8 \cos \left (d x +c \right )^{7}}{693}\right )+b^{2} \left (-\frac {\sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{7}}{12}-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{24}-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{64}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{384}+\frac {5 d x}{1024}+\frac {5 c}{1024}\right )}{d}\]

Input:

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

Output:

1/d*(a^2*(-1/10*sin(d*x+c)^3*cos(d*x+c)^7-3/80*cos(d*x+c)^7*sin(d*x+c)+1/1 
60*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+3/256*d*x+3/ 
256*c)+2*a*b*(-1/11*sin(d*x+c)^4*cos(d*x+c)^7-4/99*cos(d*x+c)^7*sin(d*x+c) 
^2-8/693*cos(d*x+c)^7)+b^2*(-1/12*sin(d*x+c)^5*cos(d*x+c)^7-1/24*sin(d*x+c 
)^3*cos(d*x+c)^7-1/64*cos(d*x+c)^7*sin(d*x+c)+1/384*(cos(d*x+c)^5+5/4*cos( 
d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/1024*d*x+5/1024*c))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.73 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {645120 \, a b \cos \left (d x + c\right )^{11} - 1576960 \, a b \cos \left (d x + c\right )^{9} + 1013760 \, a b \cos \left (d x + c\right )^{7} - 3465 \, {\left (12 \, a^{2} + 5 \, b^{2}\right )} d x + 231 \, {\left (1280 \, b^{2} \cos \left (d x + c\right )^{11} - 128 \, {\left (12 \, a^{2} + 25 \, b^{2}\right )} \cos \left (d x + c\right )^{9} + 48 \, {\left (44 \, a^{2} + 45 \, b^{2}\right )} \cos \left (d x + c\right )^{7} - 8 \, {\left (12 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (12 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (12 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3548160 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

-1/3548160*(645120*a*b*cos(d*x + c)^11 - 1576960*a*b*cos(d*x + c)^9 + 1013 
760*a*b*cos(d*x + c)^7 - 3465*(12*a^2 + 5*b^2)*d*x + 231*(1280*b^2*cos(d*x 
 + c)^11 - 128*(12*a^2 + 25*b^2)*cos(d*x + c)^9 + 48*(44*a^2 + 45*b^2)*cos 
(d*x + c)^7 - 8*(12*a^2 + 5*b^2)*cos(d*x + c)^5 - 10*(12*a^2 + 5*b^2)*cos( 
d*x + c)^3 - 15*(12*a^2 + 5*b^2)*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (231) = 462\).

Time = 2.54 (sec) , antiderivative size = 656, normalized size of antiderivative = 2.62 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)**6*sin(d*x+c)**4*(a+b*sin(d*x+c))**2,x)
 

Output:

Piecewise((3*a**2*x*sin(c + d*x)**10/256 + 15*a**2*x*sin(c + d*x)**8*cos(c 
 + d*x)**2/256 + 15*a**2*x*sin(c + d*x)**6*cos(c + d*x)**4/128 + 15*a**2*x 
*sin(c + d*x)**4*cos(c + d*x)**6/128 + 15*a**2*x*sin(c + d*x)**2*cos(c + d 
*x)**8/256 + 3*a**2*x*cos(c + d*x)**10/256 + 3*a**2*sin(c + d*x)**9*cos(c 
+ d*x)/(256*d) + 7*a**2*sin(c + d*x)**7*cos(c + d*x)**3/(128*d) + a**2*sin 
(c + d*x)**5*cos(c + d*x)**5/(10*d) - 7*a**2*sin(c + d*x)**3*cos(c + d*x)* 
*7/(128*d) - 3*a**2*sin(c + d*x)*cos(c + d*x)**9/(256*d) - 2*a*b*sin(c + d 
*x)**4*cos(c + d*x)**7/(7*d) - 8*a*b*sin(c + d*x)**2*cos(c + d*x)**9/(63*d 
) - 16*a*b*cos(c + d*x)**11/(693*d) + 5*b**2*x*sin(c + d*x)**12/1024 + 15* 
b**2*x*sin(c + d*x)**10*cos(c + d*x)**2/512 + 75*b**2*x*sin(c + d*x)**8*co 
s(c + d*x)**4/1024 + 25*b**2*x*sin(c + d*x)**6*cos(c + d*x)**6/256 + 75*b* 
*2*x*sin(c + d*x)**4*cos(c + d*x)**8/1024 + 15*b**2*x*sin(c + d*x)**2*cos( 
c + d*x)**10/512 + 5*b**2*x*cos(c + d*x)**12/1024 + 5*b**2*sin(c + d*x)**1 
1*cos(c + d*x)/(1024*d) + 85*b**2*sin(c + d*x)**9*cos(c + d*x)**3/(3072*d) 
 + 33*b**2*sin(c + d*x)**7*cos(c + d*x)**5/(512*d) - 33*b**2*sin(c + d*x)* 
*5*cos(c + d*x)**7/(512*d) - 85*b**2*sin(c + d*x)**3*cos(c + d*x)**9/(3072 
*d) - 5*b**2*sin(c + d*x)*cos(c + d*x)**11/(1024*d), Ne(d, 0)), (x*(a + b* 
sin(c))**2*sin(c)**4*cos(c)**6, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.55 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {2772 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 81920 \, {\left (63 \, \cos \left (d x + c\right )^{11} - 154 \, \cos \left (d x + c\right )^{9} + 99 \, \cos \left (d x + c\right )^{7}\right )} a b + 1155 \, {\left (4 \, \sin \left (4 \, d x + 4 \, c\right )^{3} + 120 \, d x + 120 \, c + 9 \, \sin \left (8 \, d x + 8 \, c\right ) - 48 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{28385280 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

1/28385280*(2772*(32*sin(2*d*x + 2*c)^5 + 120*d*x + 120*c + 5*sin(8*d*x + 
8*c) - 40*sin(4*d*x + 4*c))*a^2 - 81920*(63*cos(d*x + c)^11 - 154*cos(d*x 
+ c)^9 + 99*cos(d*x + c)^7)*a*b + 1155*(4*sin(4*d*x + 4*c)^3 + 120*d*x + 1 
20*c + 9*sin(8*d*x + 8*c) - 48*sin(4*d*x + 4*c))*b^2)/d
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.90 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{1024} \, {\left (12 \, a^{2} + 5 \, b^{2}\right )} x - \frac {a b \cos \left (11 \, d x + 11 \, c\right )}{5632 \, d} - \frac {a b \cos \left (9 \, d x + 9 \, c\right )}{4608 \, d} + \frac {5 \, a b \cos \left (7 \, d x + 7 \, c\right )}{3584 \, d} + \frac {a b \cos \left (5 \, d x + 5 \, c\right )}{512 \, d} - \frac {5 \, a b \cos \left (3 \, d x + 3 \, c\right )}{768 \, d} - \frac {5 \, a b \cos \left (d x + c\right )}{256 \, d} - \frac {b^{2} \sin \left (12 \, d x + 12 \, c\right )}{24576 \, d} + \frac {a^{2} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{1024 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac {{\left (4 \, a^{2} + 3 \, b^{2}\right )} \sin \left (8 \, d x + 8 \, c\right )}{8192 \, d} - \frac {{\left (32 \, a^{2} + 15 \, b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{8192 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

1/1024*(12*a^2 + 5*b^2)*x - 1/5632*a*b*cos(11*d*x + 11*c)/d - 1/4608*a*b*c 
os(9*d*x + 9*c)/d + 5/3584*a*b*cos(7*d*x + 7*c)/d + 1/512*a*b*cos(5*d*x + 
5*c)/d - 5/768*a*b*cos(3*d*x + 3*c)/d - 5/256*a*b*cos(d*x + c)/d - 1/24576 
*b^2*sin(12*d*x + 12*c)/d + 1/5120*a^2*sin(10*d*x + 10*c)/d - 1/1024*a^2*s 
in(6*d*x + 6*c)/d + 1/512*a^2*sin(2*d*x + 2*c)/d + 1/8192*(4*a^2 + 3*b^2)* 
sin(8*d*x + 8*c)/d - 1/8192*(32*a^2 + 15*b^2)*sin(4*d*x + 4*c)/d
 

Mupad [B] (verification not implemented)

Time = 25.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.83 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {6930\,a^2\,\sin \left (2\,c+2\,d\,x\right )-13860\,a^2\,\sin \left (4\,c+4\,d\,x\right )-3465\,a^2\,\sin \left (6\,c+6\,d\,x\right )+\frac {3465\,a^2\,\sin \left (8\,c+8\,d\,x\right )}{2}+693\,a^2\,\sin \left (10\,c+10\,d\,x\right )-\frac {51975\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {10395\,b^2\,\sin \left (8\,c+8\,d\,x\right )}{8}-\frac {1155\,b^2\,\sin \left (12\,c+12\,d\,x\right )}{8}-69300\,a\,b\,\cos \left (c+d\,x\right )-23100\,a\,b\,\cos \left (3\,c+3\,d\,x\right )+6930\,a\,b\,\cos \left (5\,c+5\,d\,x\right )+4950\,a\,b\,\cos \left (7\,c+7\,d\,x\right )-770\,a\,b\,\cos \left (9\,c+9\,d\,x\right )-630\,a\,b\,\cos \left (11\,c+11\,d\,x\right )+41580\,a^2\,d\,x+17325\,b^2\,d\,x}{3548160\,d} \] Input:

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

Output:

(6930*a^2*sin(2*c + 2*d*x) - 13860*a^2*sin(4*c + 4*d*x) - 3465*a^2*sin(6*c 
 + 6*d*x) + (3465*a^2*sin(8*c + 8*d*x))/2 + 693*a^2*sin(10*c + 10*d*x) - ( 
51975*b^2*sin(4*c + 4*d*x))/8 + (10395*b^2*sin(8*c + 8*d*x))/8 - (1155*b^2 
*sin(12*c + 12*d*x))/8 - 69300*a*b*cos(c + d*x) - 23100*a*b*cos(3*c + 3*d* 
x) + 6930*a*b*cos(5*c + 5*d*x) + 4950*a*b*cos(7*c + 7*d*x) - 770*a*b*cos(9 
*c + 9*d*x) - 630*a*b*cos(11*c + 11*d*x) + 41580*a^2*d*x + 17325*b^2*d*x)/ 
(3548160*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.32 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {295680 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{11} b^{2}+645120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{10} a b +354816 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9} a^{2}-739200 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9} b^{2}-1648640 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} a b -931392 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} a^{2}+498960 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} b^{2}+1157120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a b +687456 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a^{2}-9240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{2}-30720 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a b -27720 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2}-11550 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}-40960 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a b -41580 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}-17325 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-81920 \cos \left (d x +c \right ) a b +41580 a^{2} d x +81920 a b +17325 b^{2} d x}{3548160 d} \] Input:

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

Output:

(295680*cos(c + d*x)*sin(c + d*x)**11*b**2 + 645120*cos(c + d*x)*sin(c + d 
*x)**10*a*b + 354816*cos(c + d*x)*sin(c + d*x)**9*a**2 - 739200*cos(c + d* 
x)*sin(c + d*x)**9*b**2 - 1648640*cos(c + d*x)*sin(c + d*x)**8*a*b - 93139 
2*cos(c + d*x)*sin(c + d*x)**7*a**2 + 498960*cos(c + d*x)*sin(c + d*x)**7* 
b**2 + 1157120*cos(c + d*x)*sin(c + d*x)**6*a*b + 687456*cos(c + d*x)*sin( 
c + d*x)**5*a**2 - 9240*cos(c + d*x)*sin(c + d*x)**5*b**2 - 30720*cos(c + 
d*x)*sin(c + d*x)**4*a*b - 27720*cos(c + d*x)*sin(c + d*x)**3*a**2 - 11550 
*cos(c + d*x)*sin(c + d*x)**3*b**2 - 40960*cos(c + d*x)*sin(c + d*x)**2*a* 
b - 41580*cos(c + d*x)*sin(c + d*x)*a**2 - 17325*cos(c + d*x)*sin(c + d*x) 
*b**2 - 81920*cos(c + d*x)*a*b + 41580*a**2*d*x + 81920*a*b + 17325*b**2*d 
*x)/(3548160*d)