Integrand size = 36, antiderivative size = 89 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\frac {1}{8} (4 a+3 c) x-\frac {b \cos (c+d x)}{d}+\frac {b \cos ^3(c+d x)}{3 d}-\frac {(4 a+3 c) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {c \cos (c+d x) \sin ^3(c+d x)}{4 d} \]
1/8*(4*a+3*c)*x-b*cos(d*x+c)/d+1/3*b*cos(d*x+c)^3/d-1/8*(4*a+3*c)*cos(d*x+ c)*sin(d*x+c)/d-1/4*c*cos(d*x+c)*sin(d*x+c)^3/d
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.18 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\frac {a (c+d x)}{2 d}+\frac {3 c (c+d x)}{8 d}-\frac {3 b \cos (c+d x)}{4 d}+\frac {b \cos (3 (c+d x))}{12 d}-\frac {a \sin (2 (c+d x))}{4 d}-\frac {c \sin (2 (c+d x))}{4 d}+\frac {c \sin (4 (c+d x))}{32 d} \]
(a*(c + d*x))/(2*d) + (3*c*(c + d*x))/(8*d) - (3*b*Cos[c + d*x])/(4*d) + ( b*Cos[3*(c + d*x)])/(12*d) - (a*Sin[2*(c + d*x)])/(4*d) - (c*Sin[2*(c + d* x)])/(4*d) + (c*Sin[4*(c + d*x)])/(32*d)
Time = 0.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {3042, 4725, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 3115, 24}
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.
\(\displaystyle \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \left (a \sin (c+d x)+b \sin (c+d x)^2+c \sin (c+d x)^3\right )dx\) |
\(\Big \downarrow \) 4725 |
\(\displaystyle \int \sin ^2(c+d x) \left (a+b \sin (c+d x)+c \sin ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \left (a+b \sin (c+d x)+c \sin (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{4} \int \sin ^2(c+d x) (4 a+3 c+4 b \sin (c+d x))dx-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \sin (c+d x)^2 (4 a+3 c+4 b \sin (c+d x))dx-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{4} \left ((4 a+3 c) \int \sin ^2(c+d x)dx+4 b \int \sin ^3(c+d x)dx\right )-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left ((4 a+3 c) \int \sin (c+d x)^2dx+4 b \int \sin (c+d x)^3dx\right )-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {1}{4} \left ((4 a+3 c) \int \sin (c+d x)^2dx-\frac {4 b \int \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\right )-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left ((4 a+3 c) \int \sin (c+d x)^2dx-\frac {4 b \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{4} \left ((4 a+3 c) \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 b \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} \left ((4 a+3 c) \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 b \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )-\frac {c \sin ^3(c+d x) \cos (c+d x)}{4 d}\) |
-1/4*(c*Cos[c + d*x]*Sin[c + d*x]^3)/d + ((-4*b*(Cos[c + d*x] - Cos[c + d* x]^3/3))/d + (4*a + 3*c)*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4
3.10.36.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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]
Int[(u_)*((A_.)*sin[(a_.) + (b_.)*(x_)]^(n_.) + (B_.)*sin[(a_.) + (b_.)*(x_ )]^(n1_) + (C_.)*sin[(a_.) + (b_.)*(x_)]^(n2_)), x_Symbol] :> Int[ActivateT rig[u]*Sin[a + b*x]^n*(A + B*Sin[a + b*x] + C*Sin[a + b*x]^2), x] /; FreeQ[ {a, b, A, B, C, n}, x] && EqQ[n1, n + 1] && EqQ[n2, n + 2]
Time = 1.89 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79
method | result | size |
parallelrisch | \(\frac {\left (-24 a -24 c \right ) \sin \left (2 d x +2 c \right )+48 d x a +36 c d x -72 \cos \left (d x +c \right ) b +8 b \cos \left (3 d x +3 c \right )+3 c \sin \left (4 d x +4 c \right )-64 b}{96 d}\) | \(70\) |
risch | \(\frac {a x}{2}+\frac {3 c x}{8}-\frac {3 b \cos \left (d x +c \right )}{4 d}+\frac {c \sin \left (4 d x +4 c \right )}{32 d}+\frac {b \cos \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (2 d x +2 c \right ) a}{4 d}-\frac {\sin \left (2 d x +2 c \right ) c}{4 d}\) | \(82\) |
derivativedivides | \(\frac {c \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(84\) |
default | \(\frac {c \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(84\) |
parts | \(\frac {a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}+\frac {c \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(89\) |
norman | \(\frac {\left (\frac {a}{2}+\frac {3 c}{8}\right ) x +\left (2 a +\frac {3 c}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (2 a +\frac {3 c}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 a +\frac {9 c}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {a}{2}+\frac {3 c}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {4 b}{3 d}-\frac {\left (4 a +3 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (4 a +3 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {\left (4 a +11 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {\left (4 a +11 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {4 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {16 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(237\) |
1/96*((-24*a-24*c)*sin(2*d*x+2*c)+48*d*x*a+36*c*d*x-72*cos(d*x+c)*b+8*b*co s(3*d*x+3*c)+3*c*sin(4*d*x+4*c)-64*b)/d
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.81 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\frac {8 \, b \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a + 3 \, c\right )} d x - 24 \, b \cos \left (d x + c\right ) + 3 \, {\left (2 \, c \cos \left (d x + c\right )^{3} - {\left (4 \, a + 5 \, c\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(8*b*cos(d*x + c)^3 + 3*(4*a + 3*c)*d*x - 24*b*cos(d*x + c) + 3*(2*c* cos(d*x + c)^3 - (4*a + 5*c)*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (76) = 152\).
Time = 0.18 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.26 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {b \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 b \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 c x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 c x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 c x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {5 c \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 c \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + b \sin ^{2}{\left (c \right )} + c \sin ^{3}{\left (c \right )}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a*x*sin(c + d*x)**2/2 + a*x*cos(c + d*x)**2/2 - a*sin(c + d*x)* cos(c + d*x)/(2*d) - b*sin(c + d*x)**2*cos(c + d*x)/d - 2*b*cos(c + d*x)** 3/(3*d) + 3*c*x*sin(c + d*x)**4/8 + 3*c*x*sin(c + d*x)**2*cos(c + d*x)**2/ 4 + 3*c*x*cos(c + d*x)**4/8 - 5*c*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*c *sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a*sin(c) + b*sin(c)**2 + c*sin(c)**3)*sin(c), True))
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a + 32 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} c}{96 \, d} \]
1/96*(24*(2*d*x + 2*c - sin(2*d*x + 2*c))*a + 32*(cos(d*x + c)^3 - 3*cos(d *x + c))*b + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*c)/ d
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, a + 3 \, c\right )} x + \frac {b \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {3 \, b \cos \left (d x + c\right )}{4 \, d} + \frac {c \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {{\left (a + c\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]
1/8*(4*a + 3*c)*x + 1/12*b*cos(3*d*x + 3*c)/d - 3/4*b*cos(d*x + c)/d + 1/3 2*c*sin(4*d*x + 4*c)/d - 1/4*(a + c)*sin(2*d*x + 2*c)/d
Time = 27.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.82 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right ) \, dx=\frac {2\,b\,\cos \left (3\,c+3\,d\,x\right )-18\,b\,\cos \left (c+d\,x\right )-6\,a\,\sin \left (2\,c+2\,d\,x\right )-6\,c\,\sin \left (2\,c+2\,d\,x\right )+\frac {3\,c\,\sin \left (4\,c+4\,d\,x\right )}{4}+12\,a\,d\,x+9\,c\,d\,x}{24\,d} \]