Integrand size = 28, antiderivative size = 77 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3 b x}{8}-\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos (c+d x) \sin ^3(c+d x)}{4 d} \]
3/8*b*x-a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d-3/8*b*cos(d*x+c)*sin(d*x+c)/d- 1/4*b*cos(d*x+c)*sin(d*x+c)^3/d
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3 b (c+d x)}{8 d}-\frac {3 a \cos (c+d x)}{4 d}+\frac {a \cos (3 (c+d x))}{12 d}-\frac {b \sin (2 (c+d x))}{4 d}+\frac {b \sin (4 (c+d x))}{32 d} \]
(3*b*(c + d*x))/(8*d) - (3*a*Cos[c + d*x])/(4*d) + (a*Cos[3*(c + d*x)])/(1 2*d) - (b*Sin[2*(c + d*x)])/(4*d) + (b*Sin[4*(c + d*x)])/(32*d)
Time = 0.47 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4893, 3042, 3227, 3042, 3113, 2009, 3115, 3042, 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.
\(\displaystyle \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \left (a \sin (c+d x)^2+b \sin (c+d x)^3\right )dx\) |
\(\Big \downarrow \) 4893 |
\(\displaystyle \int \sin ^3(c+d x) (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^3 (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle a \int \sin ^3(c+d x)dx+b \int \sin ^4(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin (c+d x)^3dx+b \int \sin (c+d x)^4dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle b \int \sin (c+d x)^4dx-\frac {a \int \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b \int \sin (c+d x)^4dx-\frac {a \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle b \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle b \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\) |
-((a*(Cos[c + d*x] - Cos[c + d*x]^3/3))/d) + b*(-1/4*(Cos[c + d*x]*Sin[c + d*x]^3)/d + (3*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)
3.10.34.3.1 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[(u_)*((a_)*(F_)[(c_.) + (d_.)*(x_)]^(p_.) + (b_.)*(F_)[(c_.) + (d_.)*(x _)]^(q_.))^(n_.), x_Symbol] :> Int[ActivateTrig[u*F[c + d*x]^(n*p)*(a + b*F [c + d*x]^(q - p))^n], x] /; FreeQ[{a, b, c, d, p, q}, x] && InertTrigQ[F] && IntegerQ[n] && PosQ[q - p]
Time = 1.99 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\) | \(60\) |
default | \(\frac {b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\) | \(60\) |
parallelrisch | \(\frac {36 d x b -72 \cos \left (d x +c \right ) a +3 b \sin \left (4 d x +4 c \right )+8 a \cos \left (3 d x +3 c \right )-24 b \sin \left (2 d x +2 c \right )-64 a}{96 d}\) | \(60\) |
parts | \(-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}+\frac {b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(62\) |
risch | \(\frac {3 x b}{8}-\frac {3 a \cos \left (d x +c \right )}{4 d}+\frac {b \sin \left (4 d x +4 c \right )}{32 d}+\frac {a \cos \left (3 d x +3 c \right )}{12 d}-\frac {b \sin \left (2 d x +2 c \right )}{4 d}\) | \(63\) |
norman | \(\frac {\frac {3 x b}{8}-\frac {4 a}{3 d}-\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {11 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {11 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {3 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {9 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {3 x b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {16 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(188\) |
1/d*(b*(-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 *(2+sin(d*x+c)^2)*cos(d*x+c))
Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {8 \, a \cos \left (d x + c\right )^{3} + 9 \, b d x - 24 \, a \cos \left (d x + c\right ) + 3 \, {\left (2 \, b \cos \left (d x + c\right )^{3} - 5 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(8*a*cos(d*x + c)^3 + 9*b*d*x - 24*a*cos(d*x + c) + 3*(2*b*cos(d*x + c)^3 - 5*b*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (70) = 140\).
Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.95 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\begin {cases} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 a \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 b x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {5 b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \sin ^{2}{\left (c \right )} + b \sin ^{3}{\left (c \right )}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((-a*sin(c + d*x)**2*cos(c + d*x)/d - 2*a*cos(c + d*x)**3/(3*d) + 3*b*x*sin(c + d*x)**4/8 + 3*b*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*b*x *cos(c + d*x)**4/8 - 5*b*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*b*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a*sin(c)**2 + b*sin(c)**3)*sin( c), True))
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {32 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{96 \, d} \]
1/96*(32*(cos(d*x + c)^3 - 3*cos(d*x + c))*a + 3*(12*d*x + 12*c + sin(4*d* x + 4*c) - 8*sin(2*d*x + 2*c))*b)/d
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3}{8} \, b x + \frac {a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {3 \, a \cos \left (d x + c\right )}{4 \, d} + \frac {b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {b \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]
3/8*b*x + 1/12*a*cos(3*d*x + 3*c)/d - 3/4*a*cos(d*x + c)/d + 1/32*b*sin(4* d*x + 4*c)/d - 1/4*b*sin(2*d*x + 2*c)/d
Time = 29.71 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.44 \[ \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right ) \, dx=\frac {3\,b\,x}{8}-\frac {-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {4\,a}{3}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]