\(\int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [224]

Optimal result

Integrand size = 32, antiderivative size = 140 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {1}{8} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {3 a^3 A \cos ^5(c+d x)}{5 d}-\frac {a^3 A \cos ^7(c+d x)}{7 d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^3 A \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac {a^3 A \cos (c+d x) \sin ^5(c+d x)}{3 d} \] Output:

1/8*a^3*A*x-2/3*a^3*A*cos(d*x+c)^3/d+3/5*a^3*A*cos(d*x+c)^5/d-1/7*a^3*A*co 
s(d*x+c)^7/d-1/8*a^3*A*cos(d*x+c)*sin(d*x+c)/d-1/12*a^3*A*cos(d*x+c)*sin(d 
*x+c)^3/d+1/3*a^3*A*cos(d*x+c)*sin(d*x+c)^5/d
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A (840 c+840 d x-1365 \cos (c+d x)-175 \cos (3 (c+d x))+147 \cos (5 (c+d x))-15 \cos (7 (c+d x))-210 \sin (2 (c+d x))-210 \sin (4 (c+d x))+70 \sin (6 (c+d x)))}{6720 d} \] Input:

Integrate[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]
 

Output:

(a^3*A*(840*c + 840*d*x - 1365*Cos[c + d*x] - 175*Cos[3*(c + d*x)] + 147*C 
os[5*(c + d*x)] - 15*Cos[7*(c + d*x)] - 210*Sin[2*(c + d*x)] - 210*Sin[4*( 
c + d*x)] + 70*Sin[6*(c + d*x)]))/(6720*d)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3042, 3445, 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.

\(\displaystyle \int \sin ^3(c+d x) (a \sin (c+d x)+a)^3 (A-A \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^3 (a \sin (c+d x)+a)^3 (A-A \sin (c+d x))dx\)

\(\Big \downarrow \) 3445

\(\displaystyle \int \left (-a^3 A \sin ^7(c+d x)-2 a^3 A \sin ^6(c+d x)+2 a^3 A \sin ^4(c+d x)+a^3 A \sin ^3(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^3 A \cos ^7(c+d x)}{7 d}+\frac {3 a^3 A \cos ^5(c+d x)}{5 d}-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac {a^3 A \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac {a^3 A \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a^3 A x\)

Input:

Int[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]
 

Output:

(a^3*A*x)/8 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (3*a^3*A*Cos[c + d*x]^5)/(5 
*d) - (a^3*A*Cos[c + d*x]^7)/(7*d) - (a^3*A*Cos[c + d*x]*Sin[c + d*x])/(8* 
d) - (a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(12*d) + (a^3*A*Cos[c + d*x]*Sin[ 
c + d*x]^5)/(3*d)
 

Defintions of rubi rules used

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 3445

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[si 
n[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; FreeQ[{ 
a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ 
[m] && IntegerQ[n]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.13

\[\frac {\frac {a^{3} A \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{7}-2 a^{3} A \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a^{3} A \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a^{3} A \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\]

Input:

int(sin(d*x+c)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)
                                                                                    
                                                                                    
 

Output:

1/d*(1/7*a^3*A*(16/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d 
*x+c)-2*a^3*A*(-1/6*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d* 
x+c)+5/16*d*x+5/16*c)+2*a^3*A*(-1/4*(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+ 
c)+3/8*d*x+3/8*c)-1/3*a^3*A*(2+sin(d*x+c)^2)*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.75 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {120 \, A a^{3} \cos \left (d x + c\right )^{7} - 504 \, A a^{3} \cos \left (d x + c\right )^{5} + 560 \, A a^{3} \cos \left (d x + c\right )^{3} - 105 \, A a^{3} d x - 35 \, {\left (8 \, A a^{3} \cos \left (d x + c\right )^{5} - 14 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

-1/840*(120*A*a^3*cos(d*x + c)^7 - 504*A*a^3*cos(d*x + c)^5 + 560*A*a^3*co 
s(d*x + c)^3 - 105*A*a^3*d*x - 35*(8*A*a^3*cos(d*x + c)^5 - 14*A*a^3*cos(d 
*x + c)^3 + 3*A*a^3*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. 440 vs. \(2 (131) = 262\).

Time = 0.54 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.14 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\begin {cases} - \frac {5 A a^{3} x \sin ^{6}{\left (c + d x \right )}}{8} - \frac {15 A a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{4} - \frac {15 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} - \frac {5 A a^{3} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {A a^{3} \sin ^{6}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {11 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {8 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {16 A a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {2 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (- A \sin {\left (c \right )} + A\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \sin ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(sin(d*x+c)**3*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)
 

Output:

Piecewise((-5*A*a**3*x*sin(c + d*x)**6/8 - 15*A*a**3*x*sin(c + d*x)**4*cos 
(c + d*x)**2/8 + 3*A*a**3*x*sin(c + d*x)**4/4 - 15*A*a**3*x*sin(c + d*x)** 
2*cos(c + d*x)**4/8 + 3*A*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/2 - 5*A*a 
**3*x*cos(c + d*x)**6/8 + 3*A*a**3*x*cos(c + d*x)**4/4 + A*a**3*sin(c + d* 
x)**6*cos(c + d*x)/d + 11*A*a**3*sin(c + d*x)**5*cos(c + d*x)/(8*d) + 2*A* 
a**3*sin(c + d*x)**4*cos(c + d*x)**3/d + 5*A*a**3*sin(c + d*x)**3*cos(c + 
d*x)**3/(3*d) - 5*A*a**3*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 8*A*a**3*sin 
(c + d*x)**2*cos(c + d*x)**5/(5*d) - A*a**3*sin(c + d*x)**2*cos(c + d*x)/d 
 + 5*A*a**3*sin(c + d*x)*cos(c + d*x)**5/(8*d) - 3*A*a**3*sin(c + d*x)*cos 
(c + d*x)**3/(4*d) + 16*A*a**3*cos(c + d*x)**7/(35*d) - 2*A*a**3*cos(c + d 
*x)**3/(3*d), Ne(d, 0)), (x*(-A*sin(c) + A)*(a*sin(c) + a)**3*sin(c)**3, T 
rue))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.12 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 21 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3} - 35 \, \cos \left (d x + c\right )\right )} A a^{3} - 1120 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3}}{3360 \, d} \] Input:

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/3360*(96*(5*cos(d*x + c)^7 - 21*cos(d*x + c)^5 + 35*cos(d*x + c)^3 - 35 
*cos(d*x + c))*A*a^3 - 1120*(cos(d*x + c)^3 - 3*cos(d*x + c))*A*a^3 + 35*( 
4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 
 2*c))*A*a^3 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c)) 
*A*a^3)/d
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {1}{8} \, A a^{3} x - \frac {A a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {7 \, A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {5 \, A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {13 \, A a^{3} \cos \left (d x + c\right )}{64 \, d} + \frac {A a^{3} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {A a^{3} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \] Input:

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/8*A*a^3*x - 1/448*A*a^3*cos(7*d*x + 7*c)/d + 7/320*A*a^3*cos(5*d*x + 5*c 
)/d - 5/192*A*a^3*cos(3*d*x + 3*c)/d - 13/64*A*a^3*cos(d*x + c)/d + 1/96*A 
*a^3*sin(6*d*x + 6*c)/d - 1/32*A*a^3*sin(4*d*x + 4*c)/d - 1/32*A*a^3*sin(2 
*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 37.81 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.14 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A\,a^3\,\left (105\,c-210\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2464\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1400\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4032\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6790\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2240\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-14560\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-6790\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-3360\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+1400\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+210\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+105\,d\,x+735\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (c+d\,x\right )+2205\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (c+d\,x\right )+3675\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (c+d\,x\right )+3675\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (c+d\,x\right )+2205\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (c+d\,x\right )+735\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (c+d\,x\right )+105\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (c+d\,x\right )-352\right )}{840\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \] Input:

int(sin(c + d*x)^3*(A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)
 

Output:

(A*a^3*(105*c - 210*tan(c/2 + (d*x)/2) - 2464*tan(c/2 + (d*x)/2)^2 - 1400* 
tan(c/2 + (d*x)/2)^3 - 4032*tan(c/2 + (d*x)/2)^4 + 6790*tan(c/2 + (d*x)/2) 
^5 + 2240*tan(c/2 + (d*x)/2)^6 - 14560*tan(c/2 + (d*x)/2)^8 - 6790*tan(c/2 
 + (d*x)/2)^9 - 3360*tan(c/2 + (d*x)/2)^10 + 1400*tan(c/2 + (d*x)/2)^11 + 
210*tan(c/2 + (d*x)/2)^13 + 105*d*x + 735*tan(c/2 + (d*x)/2)^2*(c + d*x) + 
 2205*tan(c/2 + (d*x)/2)^4*(c + d*x) + 3675*tan(c/2 + (d*x)/2)^6*(c + d*x) 
 + 3675*tan(c/2 + (d*x)/2)^8*(c + d*x) + 2205*tan(c/2 + (d*x)/2)^10*(c + d 
*x) + 735*tan(c/2 + (d*x)/2)^12*(c + d*x) + 105*tan(c/2 + (d*x)/2)^14*(c + 
 d*x) - 352))/(840*d*(tan(c/2 + (d*x)/2)^2 + 1)^7)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^{4} \left (120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+144 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-88 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right )-176 \cos \left (d x +c \right )+105 d x +176\right )}{840 d} \] Input:

int(sin(d*x+c)^3*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)
 

Output:

(a**4*(120*cos(c + d*x)*sin(c + d*x)**6 + 280*cos(c + d*x)*sin(c + d*x)**5 
 + 144*cos(c + d*x)*sin(c + d*x)**4 - 70*cos(c + d*x)*sin(c + d*x)**3 - 88 
*cos(c + d*x)*sin(c + d*x)**2 - 105*cos(c + d*x)*sin(c + d*x) - 176*cos(c 
+ d*x) + 105*d*x + 176))/(840*d)