Integrand size = 29, antiderivative size = 51 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {(A-B) (a+a \sin (c+d x))^4}{4 a d}+\frac {B (a+a \sin (c+d x))^5}{5 a^2 d} \]
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (1+\sin (c+d x))^4 (5 A-B+4 B \sin (c+d x))}{20 d} \]
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {(\sin (c+d x) a+a)^3 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (\sin (c+d x) a+a)^3 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (B (\sin (c+d x) a+a)^4+a (A-B) (\sin (c+d x) a+a)^3\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} a (A-B) (a \sin (c+d x)+a)^4+\frac {1}{5} B (a \sin (c+d x)+a)^5}{a^2 d}\) |
3.10.89.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84
method | result | size |
parallelrisch | \(-\frac {\left (\frac {\left (7 A +5 B \right ) \cos \left (2 d x +2 c \right )}{2}+\frac {\left (-A -3 B \right ) \cos \left (4 d x +4 c \right )}{8}+\left (A +\frac {5 B}{4}\right ) \sin \left (3 d x +3 c \right )-\frac {B \sin \left (5 d x +5 c \right )}{20}+7 \left (-A -\frac {B}{2}\right ) \sin \left (d x +c \right )-\frac {27 A}{8}-\frac {17 B}{8}\right ) a^{3}}{4 d}\) | \(94\) |
derivativedivides | \(\frac {\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) A \,a^{3}}{d}\) | \(98\) |
default | \(\frac {\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) A \,a^{3}}{d}\) | \(98\) |
risch | \(\frac {7 \sin \left (d x +c \right ) A \,a^{3}}{4 d}+\frac {7 a^{3} B \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (5 d x +5 c \right ) B \,a^{3}}{80 d}+\frac {a^{3} \cos \left (4 d x +4 c \right ) A}{32 d}+\frac {3 a^{3} \cos \left (4 d x +4 c \right ) B}{32 d}-\frac {\sin \left (3 d x +3 c \right ) A \,a^{3}}{4 d}-\frac {5 \sin \left (3 d x +3 c \right ) B \,a^{3}}{16 d}-\frac {7 a^{3} \cos \left (2 d x +2 c \right ) A}{8 d}-\frac {5 a^{3} \cos \left (2 d x +2 c \right ) B}{8 d}\) | \(158\) |
norman | \(\frac {\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{3}+9 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{3}+9 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (2 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (2 A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (5 A +4 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(244\) |
-1/4*(1/2*(7*A+5*B)*cos(2*d*x+2*c)+1/8*(-A-3*B)*cos(4*d*x+4*c)+(A+5/4*B)*s in(3*d*x+3*c)-1/20*B*sin(5*d*x+5*c)+7*(-A-1/2*B)*sin(d*x+c)-27/8*A-17/8*B) *a^3/d
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 40 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 4 \, {\left (B a^{3} \cos \left (d x + c\right )^{4} - {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{20 \, d} \]
1/20*(5*(A + 3*B)*a^3*cos(d*x + c)^4 - 40*(A + B)*a^3*cos(d*x + c)^2 + 4*( B*a^3*cos(d*x + c)^4 - (5*A + 7*B)*a^3*cos(d*x + c)^2 + 2*(5*A + 3*B)*a^3) *sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (41) = 82\).
Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.96 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {A a^{3} \sin {\left (c + d x \right )}}{d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((A*a**3*sin(c + d*x)**4/(4*d) + A*a**3*sin(c + d*x)**3/d + 3*A*a **3*sin(c + d*x)**2/(2*d) + A*a**3*sin(c + d*x)/d + B*a**3*sin(c + d*x)**5 /(5*d) + 3*B*a**3*sin(c + d*x)**4/(4*d) + B*a**3*sin(c + d*x)**3/d + B*a** 3*sin(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*co s(c), True))
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.65 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {4 \, B a^{3} \sin \left (d x + c\right )^{5} + 5 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (A + B\right )} a^{3} \sin \left (d x + c\right )^{3} + 10 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} + 20 \, A a^{3} \sin \left (d x + c\right )}{20 \, d} \]
1/20*(4*B*a^3*sin(d*x + c)^5 + 5*(A + 3*B)*a^3*sin(d*x + c)^4 + 20*(A + B) *a^3*sin(d*x + c)^3 + 10*(3*A + B)*a^3*sin(d*x + c)^2 + 20*A*a^3*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (47) = 94\).
Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.27 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {4 \, B a^{3} \sin \left (d x + c\right )^{5} + 5 \, A a^{3} \sin \left (d x + c\right )^{4} + 15 \, B a^{3} \sin \left (d x + c\right )^{4} + 20 \, A a^{3} \sin \left (d x + c\right )^{3} + 20 \, B a^{3} \sin \left (d x + c\right )^{3} + 30 \, A a^{3} \sin \left (d x + c\right )^{2} + 10 \, B a^{3} \sin \left (d x + c\right )^{2} + 20 \, A a^{3} \sin \left (d x + c\right )}{20 \, d} \]
1/20*(4*B*a^3*sin(d*x + c)^5 + 5*A*a^3*sin(d*x + c)^4 + 15*B*a^3*sin(d*x + c)^4 + 20*A*a^3*sin(d*x + c)^3 + 20*B*a^3*sin(d*x + c)^3 + 30*A*a^3*sin(d *x + c)^2 + 10*B*a^3*sin(d*x + c)^2 + 20*A*a^3*sin(d*x + c))/d
Time = 11.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.59 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (A+3\,B\right )}{4}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^5}{5}+A\,a^3\,\sin \left (c+d\,x\right )+a^3\,{\sin \left (c+d\,x\right )}^3\,\left (A+B\right )}{d} \]