Integrand size = 29, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {2 (A-B) (a+a \sin (c+d x))^3}{3 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^4}{4 a^3 d}-\frac {B (a+a \sin (c+d x))^5}{5 a^4 d} \] Output:
2/3*(A-B)*(a+a*sin(d*x+c))^3/a^2/d-1/4*(A-3*B)*(a+a*sin(d*x+c))^4/a^3/d-1/ 5*B*(a+a*sin(d*x+c))^5/a^4/d
Time = 0.71 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (A \sin (c+d x)+\frac {1}{2} (A+B) \sin ^2(c+d x)-\frac {1}{3} (A-B) \sin ^3(c+d x)-\frac {1}{4} (A+B) \sin ^4(c+d x)-\frac {1}{5} B \sin ^5(c+d x)\right )}{d} \] Input:
Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*(A*Sin[c + d*x] + ((A + B)*Sin[c + d*x]^2)/2 - ((A - B)*Sin[c + d*x]^3) /3 - ((A + B)*Sin[c + d*x]^4)/4 - (B*Sin[c + d*x]^5)/5))/d
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
\(\displaystyle \int \cos ^3(c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^3 (a \sin (c+d x)+a) (A+B \sin (c+d x))dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (\sin (c+d x) a+a)^2 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-B (\sin (c+d x) a+a)^4-a (A-3 B) (\sin (c+d x) a+a)^3+2 a^2 (A-B) (\sin (c+d x) a+a)^2\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2}{3} a^2 (A-B) (a \sin (c+d x)+a)^3-\frac {1}{4} a (A-3 B) (a \sin (c+d x)+a)^4-\frac {1}{5} B (a \sin (c+d x)+a)^5}{a^4 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
((2*a^2*(A - B)*(a + a*Sin[c + d*x])^3)/3 - (a*(A - 3*B)*(a + a*Sin[c + d* x])^4)/4 - (B*(a + a*Sin[c + d*x])^5)/5)/(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 = 6.54 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(-\frac {a \left (\frac {B \sin \left (d x +c \right )^{5}}{5}+\frac {\left (A +B \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (A -B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-A -B \right ) \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(73\) |
default | \(-\frac {a \left (\frac {B \sin \left (d x +c \right )^{5}}{5}+\frac {\left (A +B \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (A -B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-A -B \right ) \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(73\) |
parallelrisch | \(\frac {a \left (\frac {3 \left (-A -B \right ) \cos \left (2 d x +2 c \right )}{2}+\frac {3 \left (-A -B \right ) \cos \left (4 d x +4 c \right )}{8}+\left (A -\frac {B}{4}\right ) \sin \left (3 d x +3 c \right )-\frac {3 B \sin \left (5 d x +5 c \right )}{20}+3 \left (3 A +\frac {B}{2}\right ) \sin \left (d x +c \right )+\frac {15 A}{8}+\frac {15 B}{8}\right )}{12 d}\) | \(92\) |
risch | \(\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {a B \sin \left (d x +c \right )}{8 d}-\frac {\sin \left (5 d x +5 c \right ) a B}{80 d}-\frac {a \cos \left (4 d x +4 c \right ) A}{32 d}-\frac {a \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) a B}{48 d}-\frac {a \cos \left (2 d x +2 c \right ) A}{8 d}-\frac {a \cos \left (2 d x +2 c \right ) B}{8 d}\) | \(140\) |
norman | \(\frac {\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (2 a A +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {8 a \left (2 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 a \left (2 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 a \left (25 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(214\) |
orering | \(\text {Expression too large to display}\) | \(1174\) |
Input:
int(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
-a/d*(1/5*B*sin(d*x+c)^5+1/4*(A+B)*sin(d*x+c)^4+1/3*(A-B)*sin(d*x+c)^3+1/2 *(-A-B)*sin(d*x+c)^2-A*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {15 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B a \cos \left (d x + c\right )^{4} - {\left (5 \, A + B\right )} a \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/60*(15*(A + B)*a*cos(d*x + c)^4 + 4*(3*B*a*cos(d*x + c)^4 - (5*A + B)*a *cos(d*x + c)^2 - 2*(5*A + B)*a)*sin(d*x + c))/d
Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.64 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
Piecewise((2*A*a*sin(c + d*x)**3/(3*d) + A*a*sin(c + d*x)*cos(c + d*x)**2/ d - A*a*cos(c + d*x)**4/(4*d) + 2*B*a*sin(c + d*x)**5/(15*d) + B*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - B*a*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x* (A + B*sin(c))*(a*sin(c) + a)*cos(c)**3, True))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B a \sin \left (d x + c\right )^{5} + 15 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{4} + 20 \, {\left (A - B\right )} a \sin \left (d x + c\right )^{3} - 30 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
-1/60*(12*B*a*sin(d*x + c)^5 + 15*(A + B)*a*sin(d*x + c)^4 + 20*(A - B)*a* sin(d*x + c)^3 - 30*(A + B)*a*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.28 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B a \sin \left (d x + c\right )^{5} + 15 \, A a \sin \left (d x + c\right )^{4} + 15 \, B a \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B a \sin \left (d x + c\right )^{3} - 30 \, A a \sin \left (d x + c\right )^{2} - 30 \, B a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
-1/60*(12*B*a*sin(d*x + c)^5 + 15*A*a*sin(d*x + c)^4 + 15*B*a*sin(d*x + c) ^4 + 20*A*a*sin(d*x + c)^3 - 20*B*a*sin(d*x + c)^3 - 30*A*a*sin(d*x + c)^2 - 30*B*a*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d
Time = 33.88 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\frac {B\,a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,\left (A-B\right )\,{\sin \left (c+d\,x\right )}^3}{3}-\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^2}{2}-A\,a\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^3*(A + B*sin(c + d*x))*(a + a*sin(c + d*x)),x)
Output:
-((a*sin(c + d*x)^4*(A + B))/4 - (a*sin(c + d*x)^2*(A + B))/2 - A*a*sin(c + d*x) + (a*sin(c + d*x)^3*(A - B))/3 + (B*a*sin(c + d*x)^5)/5)/d
Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.14 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a \left (-12 \sin \left (d x +c \right )^{4} b -15 \sin \left (d x +c \right )^{3} a -15 \sin \left (d x +c \right )^{3} b -20 \sin \left (d x +c \right )^{2} a +20 \sin \left (d x +c \right )^{2} b +30 \sin \left (d x +c \right ) a +30 \sin \left (d x +c \right ) b +60 a \right )}{60 d} \] Input:
int(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(sin(c + d*x)*a*( - 12*sin(c + d*x)**4*b - 15*sin(c + d*x)**3*a - 15*sin(c + d*x)**3*b - 20*sin(c + d*x)**2*a + 20*sin(c + d*x)**2*b + 30*sin(c + d* x)*a + 30*sin(c + d*x)*b + 60*a))/(60*d)