Integrand size = 31, antiderivative size = 105 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {4 (A-B) (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {2 (A-2 B) (a+a \sin (c+d x))^6}{3 a^4 d}+\frac {(A-5 B) (a+a \sin (c+d x))^7}{7 a^5 d}+\frac {B (a+a \sin (c+d x))^8}{8 a^6 d} \] Output:
4/5*(A-B)*(a+a*sin(d*x+c))^5/a^3/d-2/3*(A-2*B)*(a+a*sin(d*x+c))^6/a^4/d+1/ 7*(A-5*B)*(a+a*sin(d*x+c))^7/a^5/d+1/8*B*(a+a*sin(d*x+c))^8/a^6/d
Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (1+\sin (c+d x))^5 \left (232 A-47 B-5 (64 A-47 B) \sin (c+d x)+15 (8 A-19 B) \sin ^2(c+d x)+105 B \sin ^3(c+d x)\right )}{840 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^2*(1 + Sin[c + d*x])^5*(232*A - 47*B - 5*(64*A - 47*B)*Sin[c + d*x] + 1 5*(8*A - 19*B)*Sin[c + d*x]^2 + 105*B*Sin[c + d*x]^3))/(840*d)
Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)^2 (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (B (\sin (c+d x) a+a)^7+a (A-5 B) (\sin (c+d x) a+a)^6-4 a^2 (A-2 B) (\sin (c+d x) a+a)^5+4 a^3 (A-B) (\sin (c+d x) a+a)^4\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{5} a^3 (A-B) (a \sin (c+d x)+a)^5-\frac {2}{3} a^2 (A-2 B) (a \sin (c+d x)+a)^6+\frac {1}{7} a (A-5 B) (a \sin (c+d x)+a)^7+\frac {1}{8} B (a \sin (c+d x)+a)^8}{a^6 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
((4*a^3*(A - B)*(a + a*Sin[c + d*x])^5)/5 - (2*a^2*(A - 2*B)*(a + a*Sin[c + d*x])^6)/3 + (a*(A - 5*B)*(a + a*Sin[c + d*x])^7)/7 + (B*(a + a*Sin[c + d*x])^8)/8)/(a^6*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 = 0.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.20
\[\frac {a^{2} \left (\frac {B \sin \left (d x +c \right )^{8}}{8}+\frac {\left (A +2 B \right ) \sin \left (d x +c \right )^{7}}{7}+\frac {\left (2 A -B \right ) \sin \left (d x +c \right )^{6}}{6}+\frac {\left (-4 B -A \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\left (-4 A -B \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (2 B -A \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (2 A +B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )\right )}{d}\]
Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
a^2/d*(1/8*B*sin(d*x+c)^8+1/7*(A+2*B)*sin(d*x+c)^7+1/6*(2*A-B)*sin(d*x+c)^ 6+1/5*(-4*B-A)*sin(d*x+c)^5+1/4*(-4*A-B)*sin(d*x+c)^4+1/3*(2*B-A)*sin(d*x+ c)^3+1/2*(2*A+B)*sin(d*x+c)^2+A*sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {105 \, B a^{2} \cos \left (d x + c\right )^{8} - 280 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{6} - 8 \, {\left (15 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 6 \, {\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 8 \, {\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 16 \, {\left (4 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
1/840*(105*B*a^2*cos(d*x + c)^8 - 280*(A + B)*a^2*cos(d*x + c)^6 - 8*(15*( A + 2*B)*a^2*cos(d*x + c)^6 - 6*(4*A + B)*a^2*cos(d*x + c)^4 - 8*(4*A + B) *a^2*cos(d*x + c)^2 - 16*(4*A + B)*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (99) = 198\).
Time = 0.69 (sec) , antiderivative size = 335, normalized size of antiderivative = 3.19 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {8 A a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {4 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac {16 B a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {B a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac {8 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Piecewise((8*A*a**2*sin(c + d*x)**7/(105*d) + 4*A*a**2*sin(c + d*x)**5*cos (c + d*x)**2/(15*d) + 8*A*a**2*sin(c + d*x)**5/(15*d) + A*a**2*sin(c + d*x )**3*cos(c + d*x)**4/(3*d) + 4*A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d ) + A*a**2*sin(c + d*x)*cos(c + d*x)**4/d - A*a**2*cos(c + d*x)**6/(3*d) + B*a**2*sin(c + d*x)**8/(24*d) + 16*B*a**2*sin(c + d*x)**7/(105*d) + B*a** 2*sin(c + d*x)**6*cos(c + d*x)**2/(6*d) + 8*B*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + B*a**2*sin(c + d*x)**4*cos(c + d*x)**4/(4*d) + 2*B*a**2* sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - B*a**2*cos(c + d*x)**6/(6*d), Ne(d , 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.35 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {105 \, B a^{2} \sin \left (d x + c\right )^{8} + 120 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{7} + 140 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{6} - 168 \, {\left (A + 4 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} - 210 \, {\left (4 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{4} - 280 \, {\left (A - 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{3} + 420 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} + 840 \, A a^{2} \sin \left (d x + c\right )}{840 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/840*(105*B*a^2*sin(d*x + c)^8 + 120*(A + 2*B)*a^2*sin(d*x + c)^7 + 140*( 2*A - B)*a^2*sin(d*x + c)^6 - 168*(A + 4*B)*a^2*sin(d*x + c)^5 - 210*(4*A + B)*a^2*sin(d*x + c)^4 - 280*(A - 2*B)*a^2*sin(d*x + c)^3 + 420*(2*A + B) *a^2*sin(d*x + c)^2 + 840*A*a^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (97) = 194\).
Time = 0.19 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.90 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {105 \, B a^{2} \sin \left (d x + c\right )^{8} + 120 \, A a^{2} \sin \left (d x + c\right )^{7} + 240 \, B a^{2} \sin \left (d x + c\right )^{7} + 280 \, A a^{2} \sin \left (d x + c\right )^{6} - 140 \, B a^{2} \sin \left (d x + c\right )^{6} - 168 \, A a^{2} \sin \left (d x + c\right )^{5} - 672 \, B a^{2} \sin \left (d x + c\right )^{5} - 840 \, A a^{2} \sin \left (d x + c\right )^{4} - 210 \, B a^{2} \sin \left (d x + c\right )^{4} - 280 \, A a^{2} \sin \left (d x + c\right )^{3} + 560 \, B a^{2} \sin \left (d x + c\right )^{3} + 840 \, A a^{2} \sin \left (d x + c\right )^{2} + 420 \, B a^{2} \sin \left (d x + c\right )^{2} + 840 \, A a^{2} \sin \left (d x + c\right )}{840 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
1/840*(105*B*a^2*sin(d*x + c)^8 + 120*A*a^2*sin(d*x + c)^7 + 240*B*a^2*sin (d*x + c)^7 + 280*A*a^2*sin(d*x + c)^6 - 140*B*a^2*sin(d*x + c)^6 - 168*A* a^2*sin(d*x + c)^5 - 672*B*a^2*sin(d*x + c)^5 - 840*A*a^2*sin(d*x + c)^4 - 210*B*a^2*sin(d*x + c)^4 - 280*A*a^2*sin(d*x + c)^3 + 560*B*a^2*sin(d*x + c)^3 + 840*A*a^2*sin(d*x + c)^2 + 420*B*a^2*sin(d*x + c)^2 + 840*A*a^2*si n(d*x + c))/d
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}-\frac {a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A-2\,B\right )}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^4\,\left (4\,A+B\right )}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5\,\left (A+4\,B\right )}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^7\,\left (A+2\,B\right )}{7}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {a^2\,{\sin \left (c+d\,x\right )}^6\,\left (2\,A-B\right )}{6}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^5*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2,x)
Output:
((a^2*sin(c + d*x)^2*(2*A + B))/2 - (a^2*sin(c + d*x)^3*(A - 2*B))/3 - (a^ 2*sin(c + d*x)^4*(4*A + B))/4 - (a^2*sin(c + d*x)^5*(A + 4*B))/5 + (a^2*si n(c + d*x)^7*(A + 2*B))/7 + (B*a^2*sin(c + d*x)^8)/8 + (a^2*sin(c + d*x)^6 *(2*A - B))/6 + A*a^2*sin(c + d*x))/d
Time = 0.18 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.50 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a^{2} \left (105 \sin \left (d x +c \right )^{7} b +120 \sin \left (d x +c \right )^{6} a +240 \sin \left (d x +c \right )^{6} b +280 \sin \left (d x +c \right )^{5} a -140 \sin \left (d x +c \right )^{5} b -168 \sin \left (d x +c \right )^{4} a -672 \sin \left (d x +c \right )^{4} b -840 \sin \left (d x +c \right )^{3} a -210 \sin \left (d x +c \right )^{3} b -280 \sin \left (d x +c \right )^{2} a +560 \sin \left (d x +c \right )^{2} b +840 \sin \left (d x +c \right ) a +420 \sin \left (d x +c \right ) b +840 a \right )}{840 d} \] Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(sin(c + d*x)*a**2*(105*sin(c + d*x)**7*b + 120*sin(c + d*x)**6*a + 240*si n(c + d*x)**6*b + 280*sin(c + d*x)**5*a - 140*sin(c + d*x)**5*b - 168*sin( c + d*x)**4*a - 672*sin(c + d*x)**4*b - 840*sin(c + d*x)**3*a - 210*sin(c + d*x)**3*b - 280*sin(c + d*x)**2*a + 560*sin(c + d*x)**2*b + 840*sin(c + d*x)*a + 420*sin(c + d*x)*b + 840*a))/(840*d)