Integrand size = 30, antiderivative size = 96 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {1}{4} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos ^5(c+d x)}{5 d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^3 A \cos (c+d x) \sin ^3(c+d x)}{2 d} \]
1/4*a^3*A*x-2/3*a^3*A*cos(d*x+c)^3/d+1/5*a^3*A*cos(d*x+c)^5/d-1/4*a^3*A*co s(d*x+c)*sin(d*x+c)/d+1/2*a^3*A*cos(d*x+c)*sin(d*x+c)^3/d
Time = 0.72 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \cos (c+d x) \left (-30 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(c+d x)} \left (-28-15 \sin (c+d x)+16 \sin ^2(c+d x)+30 \sin ^3(c+d x)+12 \sin ^4(c+d x)\right )\right )}{60 d \sqrt {\cos ^2(c+d x)}} \]
(a^3*A*Cos[c + d*x]*(-30*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]] + Sqrt[Cos [c + d*x]^2]*(-28 - 15*Sin[c + d*x] + 16*Sin[c + d*x]^2 + 30*Sin[c + d*x]^ 3 + 12*Sin[c + d*x]^4)))/(60*d*Sqrt[Cos[c + d*x]^2])
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
| \(\displaystyle \int \sin (c+d x) (a \sin (c+d x)+a)^3 (A-A \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) (a \sin (c+d x)+a)^3 (A-A \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3445 |
| \(\displaystyle \int \left (-a^3 A \sin ^5(c+d x)-2 a^3 A \sin ^4(c+d x)+2 a^3 A \sin ^2(c+d x)+a^3 A \sin (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 A \cos ^5(c+d x)}{5 d}-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac {a^3 A \sin (c+d x) \cos (c+d x)}{4 d}+\frac {1}{4} a^3 A x\) |
(a^3*A*x)/4 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (a^3*A*Cos[c + d*x]^5)/(5*d ) - (a^3*A*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(2*d)
3.3.26.3.1 Defintions of rubi rules used
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]
Time = 1.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(-\frac {A \,a^{3} \left (-60 d x +90 \cos \left (d x +c \right )-3 \cos \left (5 d x +5 c \right )+15 \sin \left (4 d x +4 c \right )+25 \cos \left (3 d x +3 c \right )+112\right )}{240 d}\) | \(57\) |
| risch | \(\frac {a^{3} A x}{4}-\frac {3 a^{3} A \cos \left (d x +c \right )}{8 d}+\frac {A \,a^{3} \cos \left (5 d x +5 c \right )}{80 d}-\frac {A \,a^{3} \sin \left (4 d x +4 c \right )}{16 d}-\frac {5 A \,a^{3} \cos \left (3 d x +3 c \right )}{48 d}\) | \(78\) |
| derivativedivides | \(\frac {\frac {A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}-2 A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A \,a^{3} \cos \left (d x +c \right )}{d}\) | \(117\) |
| default | \(\frac {\frac {A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}-2 A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A \,a^{3} \cos \left (d x +c \right )}{d}\) | \(117\) |
| parts | \(-\frac {a^{3} A \cos \left (d x +c \right )}{d}+\frac {2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5 d}\) | \(125\) |
| norman | \(\frac {-\frac {14 A \,a^{3}}{15 d}+\frac {a^{3} A x}{4}-\frac {4 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 A \,a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {3 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{3} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{3} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a^{3} A x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(282\) |
-1/240*A*a^3*(-60*d*x+90*cos(d*x+c)-3*cos(5*d*x+5*c)+15*sin(4*d*x+4*c)+25* cos(3*d*x+3*c)+112)/d
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {12 \, A a^{3} \cos \left (d x + c\right )^{5} - 40 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{3} d x - 15 \, {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} - A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(12*A*a^3*cos(d*x + c)^5 - 40*A*a^3*cos(d*x + c)^3 + 15*A*a^3*d*x - 1 5*(2*A*a^3*cos(d*x + c)^3 - A*a^3*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (88) = 176\).
Time = 0.25 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.78 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\begin {cases} - \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{4} - \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + A a^{3} x \sin ^{2}{\left (c + d x \right )} - \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{4} + A a^{3} x \cos ^{2}{\left (c + d x \right )} + \frac {A a^{3} \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {4 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {8 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {A a^{3} \cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (- A \sin {\left (c \right )} + A\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \sin {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((-3*A*a**3*x*sin(c + d*x)**4/4 - 3*A*a**3*x*sin(c + d*x)**2*cos( c + d*x)**2/2 + A*a**3*x*sin(c + d*x)**2 - 3*A*a**3*x*cos(c + d*x)**4/4 + A*a**3*x*cos(c + d*x)**2 + A*a**3*sin(c + d*x)**4*cos(c + d*x)/d + 5*A*a** 3*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 4*A*a**3*sin(c + d*x)**2*cos(c + d* x)**3/(3*d) + 3*A*a**3*sin(c + d*x)*cos(c + d*x)**3/(4*d) - A*a**3*sin(c + d*x)*cos(c + d*x)/d + 8*A*a**3*cos(c + d*x)**5/(15*d) - A*a**3*cos(c + d* x)/d, Ne(d, 0)), (x*(-A*sin(c) + A)*(a*sin(c) + a)**3*sin(c), True))
Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} A a^{3} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 120 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 240 \, A a^{3} \cos \left (d x + c\right )}{240 \, d} \]
1/240*(16*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*A*a^3 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*A*a^3 + 120*(2 *d*x + 2*c - sin(2*d*x + 2*c))*A*a^3 - 240*A*a^3*cos(d*x + c))/d
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {1}{4} \, A a^{3} x + \frac {A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, A a^{3} \cos \left (d x + c\right )}{8 \, d} - \frac {A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} \]
1/4*A*a^3*x + 1/80*A*a^3*cos(5*d*x + 5*c)/d - 5/48*A*a^3*cos(3*d*x + 3*c)/ d - 3/8*A*a^3*cos(d*x + c)/d - 1/16*A*a^3*sin(4*d*x + 4*c)/d
Time = 14.73 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.04 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A\,a^3\,x}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{12}-\frac {A\,a^3\,\left (75\,c+75\,d\,x-120\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{12}-\frac {A\,a^3\,\left (75\,c+75\,d\,x-160\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (150\,c+150\,d\,x-80\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{6}-\frac {A\,a^3\,\left (150\,c+150\,d\,x-480\right )}{60}\right )+\frac {A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {A\,a^3\,\left (15\,c+15\,d\,x\right )}{60}-\frac {A\,a^3\,\left (15\,c+15\,d\,x-56\right )}{60}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]
(A*a^3*x)/4 - (tan(c/2 + (d*x)/2)^8*((A*a^3*(15*c + 15*d*x))/12 - (A*a^3*( 75*c + 75*d*x - 120))/60) + tan(c/2 + (d*x)/2)^2*((A*a^3*(15*c + 15*d*x))/ 12 - (A*a^3*(75*c + 75*d*x - 160))/60) + tan(c/2 + (d*x)/2)^4*((A*a^3*(15* c + 15*d*x))/6 - (A*a^3*(150*c + 150*d*x - 80))/60) + tan(c/2 + (d*x)/2)^6 *((A*a^3*(15*c + 15*d*x))/6 - (A*a^3*(150*c + 150*d*x - 480))/60) + (A*a^3 *tan(c/2 + (d*x)/2))/2 - 3*A*a^3*tan(c/2 + (d*x)/2)^3 + 3*A*a^3*tan(c/2 + (d*x)/2)^7 - (A*a^3*tan(c/2 + (d*x)/2)^9)/2 + (A*a^3*(15*c + 15*d*x))/60 - (A*a^3*(15*c + 15*d*x - 56))/60)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)