Integrand size = 31, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {(A-B) (a+a \sin (c+d x))^4}{2 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {B (a+a \sin (c+d x))^6}{6 a^4 d} \] Output:
1/2*(A-B)*(a+a*sin(d*x+c))^4/a^2/d-1/5*(A-3*B)*(a+a*sin(d*x+c))^5/a^3/d-1/ 6*B*(a+a*sin(d*x+c))^6/a^4/d
Time = 0.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^8 (18 A-9 B+5 B \cos (2 (c+d x))-4 (3 A-4 B) \sin (c+d x))}{60 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8*(18*A - 9*B + 5*B*Cos[2*(c + d*x)] - 4*(3*A - 4*B)*Sin[c + d*x]))/(60*d)
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.
\(\displaystyle \int \cos ^3(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^3 (a \sin (c+d x)+a)^2 (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (\sin (c+d x) a+a)^3 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x)) (\sin (c+d x) a+a)^3 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-B (\sin (c+d x) a+a)^5-a (A-3 B) (\sin (c+d x) a+a)^4+2 a^2 (A-B) (\sin (c+d x) a+a)^3\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} a^2 (A-B) (a \sin (c+d x)+a)^4-\frac {1}{5} a (A-3 B) (a \sin (c+d x)+a)^5-\frac {1}{6} B (a \sin (c+d x)+a)^6}{a^4 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
((a^2*(A - B)*(a + a*Sin[c + d*x])^4)/2 - (a*(A - 3*B)*(a + a*Sin[c + d*x] )^5)/5 - (B*(a + a*Sin[c + d*x])^6)/6)/(a^4*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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 83.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(-\frac {a^{2} \left (\frac {B \sin \left (d x +c \right )^{6}}{6}+\frac {\left (A +2 B \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4} A}{2}-\frac {2 B \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-B -2 A \right ) \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(84\) |
default | \(-\frac {a^{2} \left (\frac {B \sin \left (d x +c \right )^{6}}{6}+\frac {\left (A +2 B \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4} A}{2}-\frac {2 B \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-B -2 A \right ) \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(84\) |
parallelrisch | \(-\frac {a^{2} \left (-840 A \sin \left (d x +c \right )-60 A \sin \left (3 d x +3 c \right )+240 A \cos \left (2 d x +2 c \right )-300 A +60 A \cos \left (4 d x +4 c \right )+12 A \sin \left (5 d x +5 c \right )-240 B \sin \left (d x +c \right )+40 B \sin \left (3 d x +3 c \right )-190 B +165 B \cos \left (2 d x +2 c \right )+30 B \cos \left (4 d x +4 c \right )-5 B \cos \left (6 d x +6 c \right )+24 B \sin \left (5 d x +5 c \right )\right )}{960 d}\) | \(142\) |
risch | \(\frac {7 \sin \left (d x +c \right ) A \,a^{2}}{8 d}+\frac {\sin \left (d x +c \right ) a^{2} B}{4 d}+\frac {B \,a^{2} \cos \left (6 d x +6 c \right )}{192 d}-\frac {\sin \left (5 d x +5 c \right ) A \,a^{2}}{80 d}-\frac {\sin \left (5 d x +5 c \right ) a^{2} B}{40 d}-\frac {a^{2} \cos \left (4 d x +4 c \right ) A}{16 d}-\frac {a^{2} \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{2}}{16 d}-\frac {\sin \left (3 d x +3 c \right ) a^{2} B}{24 d}-\frac {a^{2} \cos \left (2 d x +2 c \right ) A}{4 d}-\frac {11 a^{2} \cos \left (2 d x +2 c \right ) B}{64 d}\) | \(194\) |
norman | \(\frac {\frac {\left (8 A \,a^{2}+8 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (8 A \,a^{2}+8 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 \left (2 A \,a^{2}+a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 \left (2 A \,a^{2}+a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {4 \left (6 A \,a^{2}+a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 a^{2} \left (15 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 a^{2} \left (15 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {4 a^{2} \left (17 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {4 a^{2} \left (17 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\) | \(300\) |
orering | \(\text {Expression too large to display}\) | \(2462\) |
Input:
int(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-a^2/d*(1/6*B*sin(d*x+c)^6+1/5*(A+2*B)*sin(d*x+c)^5+1/2*sin(d*x+c)^4*A-2/3 *B*sin(d*x+c)^3+1/2*(-B-2*A)*sin(d*x+c)^2-A*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.17 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {5 \, B a^{2} \cos \left (d x + c\right )^{6} - 15 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
1/30*(5*B*a^2*cos(d*x + c)^6 - 15*(A + B)*a^2*cos(d*x + c)^4 - 2*(3*(A + 2 *B)*a^2*cos(d*x + c)^4 - 2*(3*A + B)*a^2*cos(d*x + c)^2 - 4*(3*A + B)*a^2) *sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (68) = 136\).
Time = 0.35 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.92 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {B a^{2} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {4 B a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Piecewise((2*A*a**2*sin(c + d*x)**5/(15*d) + A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 2*A*a**2*sin(c + d*x)**3/(3*d) + A*a**2*sin(c + d*x)*cos (c + d*x)**2/d - A*a**2*cos(c + d*x)**4/(2*d) + B*a**2*sin(c + d*x)**6/(12 *d) + 4*B*a**2*sin(c + d*x)**5/(15*d) + B*a**2*sin(c + d*x)**4*cos(c + d*x )**2/(4*d) + 2*B*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - B*a**2*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c)**3 , True))
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {5 \, B a^{2} \sin \left (d x + c\right )^{6} + 6 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} + 15 \, A a^{2} \sin \left (d x + c\right )^{4} - 20 \, B a^{2} \sin \left (d x + c\right )^{3} - 15 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} - 30 \, A a^{2} \sin \left (d x + c\right )}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
-1/30*(5*B*a^2*sin(d*x + c)^6 + 6*(A + 2*B)*a^2*sin(d*x + c)^5 + 15*A*a^2* sin(d*x + c)^4 - 20*B*a^2*sin(d*x + c)^3 - 15*(2*A + B)*a^2*sin(d*x + c)^2 - 30*A*a^2*sin(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {5 \, B a^{2} \sin \left (d x + c\right )^{6} + 6 \, A a^{2} \sin \left (d x + c\right )^{5} + 12 \, B a^{2} \sin \left (d x + c\right )^{5} + 15 \, A a^{2} \sin \left (d x + c\right )^{4} - 20 \, B a^{2} \sin \left (d x + c\right )^{3} - 30 \, A a^{2} \sin \left (d x + c\right )^{2} - 15 \, B a^{2} \sin \left (d x + c\right )^{2} - 30 \, A a^{2} \sin \left (d x + c\right )}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
-1/30*(5*B*a^2*sin(d*x + c)^6 + 6*A*a^2*sin(d*x + c)^5 + 12*B*a^2*sin(d*x + c)^5 + 15*A*a^2*sin(d*x + c)^4 - 20*B*a^2*sin(d*x + c)^3 - 30*A*a^2*sin( d*x + c)^2 - 15*B*a^2*sin(d*x + c)^2 - 30*A*a^2*sin(d*x + c))/d
Time = 33.87 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\frac {A\,a^2\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^5\,\left (A+2\,B\right )}{5}-\frac {2\,B\,a^2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^6}{6}-A\,a^2\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^3*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2,x)
Output:
-((A*a^2*sin(c + d*x)^4)/2 - (a^2*sin(c + d*x)^2*(2*A + B))/2 + (a^2*sin(c + d*x)^5*(A + 2*B))/5 - (2*B*a^2*sin(c + d*x)^3)/3 + (B*a^2*sin(c + d*x)^ 6)/6 - A*a^2*sin(c + d*x))/d
Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.17 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a^{2} \left (-5 \sin \left (d x +c \right )^{5} b -6 \sin \left (d x +c \right )^{4} a -12 \sin \left (d x +c \right )^{4} b -15 \sin \left (d x +c \right )^{3} a +20 \sin \left (d x +c \right )^{2} b +30 \sin \left (d x +c \right ) a +15 \sin \left (d x +c \right ) b +30 a \right )}{30 d} \] Input:
int(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(sin(c + d*x)*a**2*( - 5*sin(c + d*x)**5*b - 6*sin(c + d*x)**4*a - 12*sin( c + d*x)**4*b - 15*sin(c + d*x)**3*a + 20*sin(c + d*x)**2*b + 30*sin(c + d *x)*a + 15*sin(c + d*x)*b + 30*a))/(30*d)