Integrand size = 29, antiderivative size = 51 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {(A-B) (a+a \sin (c+d x))^3}{3 a d}+\frac {B (a+a \sin (c+d x))^4}{4 a^2 d} \] Output:
1/3*(A-B)*(a+a*sin(d*x+c))^3/a/d+1/4*B*(a+a*sin(d*x+c))^4/a^2/d
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {1}{3} (A-B) (a+a \sin (c+d x))^3+\frac {B (a+a \sin (c+d x))^4}{4 a}}{a d} \] Input:
Integrate[Cos[c + d*x]*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(((A - B)*(a + a*Sin[c + d*x])^3)/3 + (B*(a + a*Sin[c + d*x])^4)/(4*a))/(a *d)
Time = 0.27 (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.
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {(\sin (c+d x) a+a)^2 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (\sin (c+d x) a+a)^2 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (B (\sin (c+d x) a+a)^3+a (A-B) (\sin (c+d x) a+a)^2\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} a (A-B) (a \sin (c+d x)+a)^3+\frac {1}{4} B (a \sin (c+d x)+a)^4}{a^2 d}\) |
Input:
Int[Cos[c + d*x]*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
((a*(A - B)*(a + a*Sin[c + d*x])^3)/3 + (B*(a + a*Sin[c + d*x])^4)/4)/(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 = 5.68 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45
method | result | size |
parallelrisch | \(-\frac {\left (\left (6 A +\frac {9 B}{2}\right ) \cos \left (2 d x +2 c \right )+\left (A +2 B \right ) \sin \left (3 d x +3 c \right )-\frac {3 B \cos \left (4 d x +4 c \right )}{8}+\left (-15 A -6 B \right ) \sin \left (d x +c \right )-6 A -\frac {33 B}{8}\right ) a^{2}}{12 d}\) | \(74\) |
derivativedivides | \(\frac {\frac {a^{2} B \sin \left (d x +c \right )^{4}}{4}+\frac {\left (A \,a^{2}+2 a^{2} B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (2 A \,a^{2}+a^{2} B \right ) \sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right ) A \,a^{2}}{d}\) | \(75\) |
default | \(\frac {\frac {a^{2} B \sin \left (d x +c \right )^{4}}{4}+\frac {\left (A \,a^{2}+2 a^{2} B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (2 A \,a^{2}+a^{2} B \right ) \sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right ) A \,a^{2}}{d}\) | \(75\) |
risch | \(\frac {5 \sin \left (d x +c \right ) A \,a^{2}}{4 d}+\frac {\sin \left (d x +c \right ) a^{2} B}{2 d}+\frac {a^{2} \cos \left (4 d x +4 c \right ) B}{32 d}-\frac {\sin \left (3 d x +3 c \right ) A \,a^{2}}{12 d}-\frac {\sin \left (3 d x +3 c \right ) a^{2} B}{6 d}-\frac {a^{2} \cos \left (2 d x +2 c \right ) A}{2 d}-\frac {3 a^{2} \cos \left (2 d x +2 c \right ) B}{8 d}\) | \(122\) |
norman | \(\frac {\frac {2 \left (2 A \,a^{2}+a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 \left (2 A \,a^{2}+a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {2 \left (4 A \,a^{2}+4 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{2} \left (13 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 a^{2} \left (13 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(193\) |
orering | \(\text {Expression too large to display}\) | \(978\) |
Input:
int(cos(d*x+c)*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
-1/12*((6*A+9/2*B)*cos(2*d*x+2*c)+(A+2*B)*sin(3*d*x+3*c)-3/8*B*cos(4*d*x+4 *c)+(-15*A-6*B)*sin(d*x+c)-6*A-33/8*B)*a^2/d
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.41 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{2} \cos \left (d x + c\right )^{4} - 12 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
1/12*(3*B*a^2*cos(d*x + c)^4 - 12*(A + B)*a^2*cos(d*x + c)^2 - 4*((A + 2*B )*a^2*cos(d*x + c)^2 - 2*(2*A + B)*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (41) = 82\).
Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.29 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Piecewise((A*a**2*sin(c + d*x)**3/(3*d) + A*a**2*sin(c + d*x)/d - A*a**2*c os(c + d*x)**2/d + B*a**2*sin(c + d*x)**4/(4*d) + 2*B*a**2*sin(c + d*x)**3 /(3*d) - B*a**2*cos(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin (c) + a)**2*cos(c), True))
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{2} \sin \left (d x + c\right )^{4} + 4 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} + 12 \, A a^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
1/12*(3*B*a^2*sin(d*x + c)^4 + 4*(A + 2*B)*a^2*sin(d*x + c)^3 + 6*(2*A + B )*a^2*sin(d*x + c)^2 + 12*A*a^2*sin(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.73 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{2} \sin \left (d x + c\right )^{4} + 4 \, A a^{2} \sin \left (d x + c\right )^{3} + 8 \, B a^{2} \sin \left (d x + c\right )^{3} + 12 \, A a^{2} \sin \left (d x + c\right )^{2} + 6 \, B a^{2} \sin \left (d x + c\right )^{2} + 12 \, A a^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
1/12*(3*B*a^2*sin(d*x + c)^4 + 4*A*a^2*sin(d*x + c)^3 + 8*B*a^2*sin(d*x + c)^3 + 12*A*a^2*sin(d*x + c)^2 + 6*B*a^2*sin(d*x + c)^2 + 12*A*a^2*sin(d*x + c))/d
Time = 34.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.29 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A+2\,B\right )}{3}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^4}{4}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^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 + (B* a^2*sin(c + d*x)^4)/4 + A*a^2*sin(c + d*x))/d
Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^{2} \left (-12 \cos \left (d x +c \right )^{2} a -6 \cos \left (d x +c \right )^{2} b +3 \sin \left (d x +c \right )^{4} b +4 \sin \left (d x +c \right )^{3} a +8 \sin \left (d x +c \right )^{3} b +12 \sin \left (d x +c \right ) a \right )}{12 d} \] Input:
int(cos(d*x+c)*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(a**2*( - 12*cos(c + d*x)**2*a - 6*cos(c + d*x)**2*b + 3*sin(c + d*x)**4*b + 4*sin(c + d*x)**3*a + 8*sin(c + d*x)**3*b + 12*sin(c + d*x)*a))/(12*d)