Integrand size = 27, antiderivative size = 49 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \sin (c+d x)}{d}+\frac {a (A+B) \sin ^2(c+d x)}{2 d}+\frac {a B \sin ^3(c+d x)}{3 d} \] Output:
a*A*sin(d*x+c)/d+1/2*a*(A+B)*sin(d*x+c)^2/d+1/3*a*B*sin(d*x+c)^3/d
Time = 0.50 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (9 A+3 B-3 (A+B) \cos (2 (c+d x))+3 (4 A+B) \sin (c+d x)-B \sin (3 (c+d x)))}{12 d} \] Input:
Integrate[Cos[c + d*x]*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*(9*A + 3*B - 3*(A + B)*Cos[2*(c + d*x)] + 3*(4*A + B)*Sin[c + d*x] - B* Sin[3*(c + d*x)]))/(12*d)
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {(\sin (c+d x) a+a) (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (\sin (c+d x) a+a) (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (B \sin ^2(c+d x) a^2+A a^2+(A+B) \sin (c+d x) a^2\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} a^3 (A+B) \sin ^2(c+d x)+a^3 A \sin (c+d x)+\frac {1}{3} a^3 B \sin ^3(c+d x)}{a^2 d}\) |
Input:
Int[Cos[c + d*x]*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a^3*A*Sin[c + d*x] + (a^3*(A + B)*Sin[c + d*x]^2)/2 + (a^3*B*Sin[c + d*x] ^3)/3)/(a^2*d)
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 = 1.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\frac {a B \sin \left (d x +c \right )^{3}}{3}+\frac {\left (a A +a B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right ) a}{d}\) | \(44\) |
default | \(\frac {\frac {a B \sin \left (d x +c \right )^{3}}{3}+\frac {\left (a A +a B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right ) a}{d}\) | \(44\) |
parallelrisch | \(-\frac {\left (\left (A +B \right ) \cos \left (2 d x +2 c \right )+\frac {B \sin \left (3 d x +3 c \right )}{3}+\left (-4 A -B \right ) \sin \left (d x +c \right )-A -B \right ) a}{4 d}\) | \(53\) |
risch | \(\frac {a A \sin \left (d x +c \right )}{d}+\frac {a B \sin \left (d x +c \right )}{4 d}-\frac {\sin \left (3 d x +3 c \right ) a B}{12 d}-\frac {a \cos \left (2 d x +2 c \right ) A}{4 d}-\frac {a \cos \left (2 d x +2 c \right ) B}{4 d}\) | \(75\) |
norman | \(\frac {\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 a \left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(124\) |
orering | \(-\frac {49 \left (-d \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) \left (A +B \sin \left (d x +c \right )\right )+\cos \left (d x +c \right )^{2} a d \left (A +B \sin \left (d x +c \right )\right )+\cos \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) B d \right )}{36 d^{2}}-\frac {7 \left (d^{3} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) \left (A +B \sin \left (d x +c \right )\right )-4 d^{3} \cos \left (d x +c \right )^{2} a \left (A +B \sin \left (d x +c \right )\right )-4 d^{3} \cos \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) B +3 d^{3} \sin \left (d x +c \right )^{2} a \left (A +B \sin \left (d x +c \right )\right )-12 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} a \,d^{3}+3 d^{3} \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) B \right )}{18 d^{4}}-\frac {-d^{5} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) \left (A +B \sin \left (d x +c \right )\right )+16 d^{5} \cos \left (d x +c \right )^{2} a \left (A +B \sin \left (d x +c \right )\right )+16 d^{5} \cos \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) B -15 d^{5} \sin \left (d x +c \right )^{2} a \left (A +B \sin \left (d x +c \right )\right )+150 d^{5} \sin \left (d x +c \right ) a \cos \left (d x +c \right )^{2} B -15 d^{5} \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) B -30 d^{5} \sin \left (d x +c \right )^{3} a B}{36 d^{6}}\) | \(401\) |
Input:
int(cos(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*a*B*sin(d*x+c)^3+1/2*(A*a+B*a)*sin(d*x+c)^2+A*sin(d*x+c)*a)
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (B a \cos \left (d x + c\right )^{2} - {\left (3 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="frica s")
Output:
-1/6*(3*(A + B)*a*cos(d*x + c)^2 + 2*(B*a*cos(d*x + c)^2 - (3*A + B)*a)*si n(d*x + c))/d
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.53 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a \sin {\left (c + d x \right )}}{d} - \frac {A a \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {B a \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {B a \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
Piecewise((A*a*sin(c + d*x)/d - A*a*cos(c + d*x)**2/(2*d) + B*a*sin(c + d* x)**3/(3*d) - B*a*cos(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*s in(c) + a)*cos(c), True))
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {2 \, B a \sin \left (d x + c\right )^{3} + 3 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{2} + 6 \, A a \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxim a")
Output:
1/6*(2*B*a*sin(d*x + c)^3 + 3*(A + B)*a*sin(d*x + c)^2 + 6*A*a*sin(d*x + c ))/d
Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {2 \, B a \sin \left (d x + c\right )^{3} + 3 \, A a \sin \left (d x + c\right )^{2} + 3 \, B a \sin \left (d x + c\right )^{2} + 6 \, A a \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac" )
Output:
1/6*(2*B*a*sin(d*x + c)^3 + 3*A*a*sin(d*x + c)^2 + 3*B*a*sin(d*x + c)^2 + 6*A*a*sin(d*x + c))/d
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\frac {B\,a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^2}{2}+A\,a\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)*(A + B*sin(c + d*x))*(a + a*sin(c + d*x)),x)
Output:
(A*a*sin(c + d*x) + (a*sin(c + d*x)^2*(A + B))/2 + (B*a*sin(c + d*x)^3)/3) /d
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (-3 \cos \left (d x +c \right )^{2} a -3 \cos \left (d x +c \right )^{2} b +2 \sin \left (d x +c \right )^{3} b +6 \sin \left (d x +c \right ) a \right )}{6 d} \] Input:
int(cos(d*x+c)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(a*( - 3*cos(c + d*x)**2*a - 3*cos(c + d*x)**2*b + 2*sin(c + d*x)**3*b + 6 *sin(c + d*x)*a))/(6*d)