Integrand size = 27, antiderivative size = 65 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d} \] Output:
1/3*a*sin(d*x+c)^3/d+1/4*a*sin(d*x+c)^4/d-1/5*a*sin(d*x+c)^5/d-1/6*a*sin(d *x+c)^6/d
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-45 \cos (2 (c+d x))+5 \cos (6 (c+d x))+32 (7+3 \cos (2 (c+d x))) \sin ^3(c+d x)\right )}{960 d} \] Input:
Integrate[Cos[c + d*x]^3*Sin[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
(a*(-45*Cos[2*(c + d*x)] + 5*Cos[6*(c + d*x)] + 32*(7 + 3*Cos[2*(c + d*x)] )*Sin[c + d*x]^3))/(960*d)
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 84, 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.
\(\displaystyle \int \sin ^2(c+d x) \cos ^3(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^3 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^2(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^2 \sin ^2(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (-\sin ^5(c+d x) a^5-\sin ^4(c+d x) a^5+\sin ^3(c+d x) a^5+\sin ^2(c+d x) a^5\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{6} a^6 \sin ^6(c+d x)-\frac {1}{5} a^6 \sin ^5(c+d x)+\frac {1}{4} a^6 \sin ^4(c+d x)+\frac {1}{3} a^6 \sin ^3(c+d x)}{a^5 d}\) |
Input:
Int[Cos[c + d*x]^3*Sin[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
((a^6*Sin[c + d*x]^3)/3 + (a^6*Sin[c + d*x]^4)/4 - (a^6*Sin[c + d*x]^5)/5 - (a^6*Sin[c + d*x]^6)/6)/(a^5*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
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 = 6.44 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(-\frac {a \left (\frac {\sin \left (d x +c \right )^{6}}{6}+\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}\) | \(48\) |
default | \(-\frac {a \left (\frac {\sin \left (d x +c \right )^{6}}{6}+\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}\) | \(48\) |
risch | \(\frac {a \sin \left (d x +c \right )}{8 d}+\frac {a \cos \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \sin \left (3 d x +3 c \right )}{48 d}-\frac {3 a \cos \left (2 d x +2 c \right )}{64 d}\) | \(74\) |
parallelrisch | \(-\frac {a \left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (28+12 \cos \left (2 d x +2 c \right )+5 \sin \left (3 d x +3 c \right )+15 \sin \left (d x +c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}\) | \(81\) |
norman | \(\frac {\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\) | \(137\) |
orering | \(-\frac {643 \left (-3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right ) d +2 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) d +\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2} a d \right )}{450 d^{2}}-\frac {625 \left (-6 d^{3} \sin \left (d x +c \right )^{5} \left (a +a \sin \left (d x +c \right )\right )+75 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right ) d^{3}+27 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{4} a \,d^{3}-44 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) d^{3}-58 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2} a \,d^{3}+6 \cos \left (d x +c \right )^{6} d^{3} a \right )}{1296 d^{4}}-\frac {601 \left (-1923 d^{5} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}-1725 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{4} a \,d^{5}+180 d^{5} \sin \left (d x +c \right )^{5} \left (a +a \sin \left (d x +c \right )\right )+60 \sin \left (d x +c \right )^{6} a \,d^{5}+1022 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) d^{5}+2626 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2} a \,d^{5}-240 \cos \left (d x +c \right )^{6} a \,d^{5}\right )}{10800 d^{6}}-\frac {-24824 d^{7} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{4}-106408 d^{7} \sin \left (d x +c \right )^{2} a \cos \left (d x +c \right )^{4}+48555 d^{7} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}+82677 d^{7} \sin \left (d x +c \right )^{4} a \cos \left (d x +c \right )^{2}-4746 d^{7} \sin \left (d x +c \right )^{5} \left (a +a \sin \left (d x +c \right )\right )-3990 d^{7} \sin \left (d x +c \right )^{6} a +8736 \cos \left (d x +c \right )^{6} d^{7} a}{432 d^{8}}-\frac {614042 d^{9} \cos \left (d x +c \right )^{4} \left (a +a \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )-314880 \cos \left (d x +c \right )^{6} a \,d^{9}+4109926 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2} a \,d^{9}-1218243 d^{9} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}-3505725 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{4} a \,d^{9}+120840 d^{9} \sin \left (d x +c \right )^{5} \left (a +a \sin \left (d x +c \right )\right )+194040 \sin \left (d x +c \right )^{6} a \,d^{9}}{32400 d^{10}}\) | \(713\) |
Input:
int(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-a/d*(1/6*sin(d*x+c)^6+1/5*sin(d*x+c)^5-1/4*sin(d*x+c)^4-1/3*sin(d*x+c)^3)
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {10 \, a \cos \left (d x + c\right )^{6} - 15 \, a \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/60*(10*a*cos(d*x + c)^6 - 15*a*cos(d*x + c)^4 - 4*(3*a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - 2*a)*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {2 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)
Output:
Piecewise((a*sin(c + d*x)**6/(12*d) + 2*a*sin(c + d*x)**5/(15*d) + a*sin(c + d*x)**4*cos(c + d*x)**2/(4*d) + a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) , Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**2*cos(c)**3, True))
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 2 0*a*sin(d*x + c)^3)/d
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 2 0*a*sin(d*x + c)^3)/d
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {-\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \] Input:
int(cos(c + d*x)^3*sin(c + d*x)^2*(a + a*sin(c + d*x)),x)
Output:
((a*sin(c + d*x)^3)/3 + (a*sin(c + d*x)^4)/4 - (a*sin(c + d*x)^5)/5 - (a*s in(c + d*x)^6)/6)/d
Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68 \[ \int \cos ^3(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right )^{3} a \left (-10 \sin \left (d x +c \right )^{3}-12 \sin \left (d x +c \right )^{2}+15 \sin \left (d x +c \right )+20\right )}{60 d} \] Input:
int(cos(d*x+c)^3*sin(d*x+c)^2*(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**3*a*( - 10*sin(c + d*x)**3 - 12*sin(c + d*x)**2 + 15*sin(c + d*x) + 20))/(60*d)