Integrand size = 27, antiderivative size = 97 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^6(c+d x)}{6 d}-\frac {2 a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}+\frac {a \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d} \] Output:
1/5*a*sin(d*x+c)^5/d+1/6*a*sin(d*x+c)^6/d-2/7*a*sin(d*x+c)^7/d-1/4*a*sin(d *x+c)^8/d+1/9*a*sin(d*x+c)^9/d+1/10*a*sin(d*x+c)^10/d
Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (3150 \cos (2 (c+d x))-525 \cos (6 (c+d x))+63 \cos (10 (c+d x))-7560 \sin (c+d x)+1680 \sin (3 (c+d x))+1008 \sin (5 (c+d x))-180 \sin (7 (c+d x))-140 \sin (9 (c+d x)))}{322560 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^4*(a + a*Sin[c + d*x]),x]
Output:
-1/322560*(a*(3150*Cos[2*(c + d*x)] - 525*Cos[6*(c + d*x)] + 63*Cos[10*(c + d*x)] - 7560*Sin[c + d*x] + 1680*Sin[3*(c + d*x)] + 1008*Sin[5*(c + d*x) ] - 180*Sin[7*(c + d*x)] - 140*Sin[9*(c + d*x)]))/d
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, 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, 99, 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.
\(\displaystyle \int \sin ^4(c+d x) \cos ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^4 \cos (c+d x)^5 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^4(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^4 \sin ^4(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^9 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\sin ^9(c+d x) a^9+\sin ^8(c+d x) a^9-2 \sin ^7(c+d x) a^9-2 \sin ^6(c+d x) a^9+\sin ^5(c+d x) a^9+\sin ^4(c+d x) a^9\right )d(a \sin (c+d x))}{a^9 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{10} a^{10} \sin ^{10}(c+d x)+\frac {1}{9} a^{10} \sin ^9(c+d x)-\frac {1}{4} a^{10} \sin ^8(c+d x)-\frac {2}{7} a^{10} \sin ^7(c+d x)+\frac {1}{6} a^{10} \sin ^6(c+d x)+\frac {1}{5} a^{10} \sin ^5(c+d x)}{a^9 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^4*(a + a*Sin[c + d*x]),x]
Output:
((a^10*Sin[c + d*x]^5)/5 + (a^10*Sin[c + d*x]^6)/6 - (2*a^10*Sin[c + d*x]^ 7)/7 - (a^10*Sin[c + d*x]^8)/4 + (a^10*Sin[c + d*x]^9)/9 + (a^10*Sin[c + d *x]^10)/10)/(a^9*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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 195.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{10}}{10}+\frac {\sin \left (d x +c \right )^{9}}{9}-\frac {\sin \left (d x +c \right )^{8}}{4}-\frac {2 \sin \left (d x +c \right )^{7}}{7}+\frac {\sin \left (d x +c \right )^{6}}{6}+\frac {\sin \left (d x +c \right )^{5}}{5}\right )}{d}\) | \(67\) |
default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{10}}{10}+\frac {\sin \left (d x +c \right )^{9}}{9}-\frac {\sin \left (d x +c \right )^{8}}{4}-\frac {2 \sin \left (d x +c \right )^{7}}{7}+\frac {\sin \left (d x +c \right )^{6}}{6}+\frac {\sin \left (d x +c \right )^{5}}{5}\right )}{d}\) | \(67\) |
risch | \(\frac {3 a \sin \left (d x +c \right )}{128 d}-\frac {a \cos \left (10 d x +10 c \right )}{5120 d}+\frac {a \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a \sin \left (7 d x +7 c \right )}{1792 d}+\frac {5 a \cos \left (6 d x +6 c \right )}{3072 d}-\frac {a \sin \left (5 d x +5 c \right )}{320 d}-\frac {a \sin \left (3 d x +3 c \right )}{192 d}-\frac {5 a \cos \left (2 d x +2 c \right )}{512 d}\) | \(119\) |
parallelrisch | \(\frac {a \left (\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (63 \sin \left (5 d x +5 c \right )+880 \cos \left (2 d x +2 c \right )+996+420 \sin \left (d x +c \right )+315 \sin \left (3 d x +3 c \right )+140 \cos \left (4 d x +4 c \right )\right ) \left (\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80640 d}\) | \(125\) |
orering | \(\text {Expression too large to display}\) | \(1867\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
a/d*(1/10*sin(d*x+c)^10+1/9*sin(d*x+c)^9-1/4*sin(d*x+c)^8-2/7*sin(d*x+c)^7 +1/6*sin(d*x+c)^6+1/5*sin(d*x+c)^5)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {126 \, a \cos \left (d x + c\right )^{10} - 315 \, a \cos \left (d x + c\right )^{8} + 210 \, a \cos \left (d x + c\right )^{6} - 4 \, {\left (35 \, a \cos \left (d x + c\right )^{8} - 50 \, a \cos \left (d x + c\right )^{6} + 3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/1260*(126*a*cos(d*x + c)^10 - 315*a*cos(d*x + c)^8 + 210*a*cos(d*x + c) ^6 - 4*(35*a*cos(d*x + c)^8 - 50*a*cos(d*x + c)^6 + 3*a*cos(d*x + c)^4 + 4 *a*cos(d*x + c)^2 + 8*a)*sin(d*x + c))/d
Time = 1.47 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a \sin ^{10}{\left (c + d x \right )}}{60 d} + \frac {8 a \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {a \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{12 d} + \frac {4 a \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {a \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{6 d} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{4}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**4*(a+a*sin(d*x+c)),x)
Output:
Piecewise((a*sin(c + d*x)**10/(60*d) + 8*a*sin(c + d*x)**9/(315*d) + a*sin (c + d*x)**8*cos(c + d*x)**2/(12*d) + 4*a*sin(c + d*x)**7*cos(c + d*x)**2/ (35*d) + a*sin(c + d*x)**6*cos(c + d*x)**4/(6*d) + a*sin(c + d*x)**5*cos(c + d*x)**4/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**4*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {126 \, a \sin \left (d x + c\right )^{10} + 140 \, a \sin \left (d x + c\right )^{9} - 315 \, a \sin \left (d x + c\right )^{8} - 360 \, a \sin \left (d x + c\right )^{7} + 210 \, a \sin \left (d x + c\right )^{6} + 252 \, a \sin \left (d x + c\right )^{5}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/1260*(126*a*sin(d*x + c)^10 + 140*a*sin(d*x + c)^9 - 315*a*sin(d*x + c)^ 8 - 360*a*sin(d*x + c)^7 + 210*a*sin(d*x + c)^6 + 252*a*sin(d*x + c)^5)/d
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {126 \, a \sin \left (d x + c\right )^{10} + 140 \, a \sin \left (d x + c\right )^{9} - 315 \, a \sin \left (d x + c\right )^{8} - 360 \, a \sin \left (d x + c\right )^{7} + 210 \, a \sin \left (d x + c\right )^{6} + 252 \, a \sin \left (d x + c\right )^{5}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/1260*(126*a*sin(d*x + c)^10 + 140*a*sin(d*x + c)^9 - 315*a*sin(d*x + c)^ 8 - 360*a*sin(d*x + c)^7 + 210*a*sin(d*x + c)^6 + 252*a*sin(d*x + c)^5)/d
Time = 18.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{4}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^4*(a + a*sin(c + d*x)),x)
Output:
((a*sin(c + d*x)^5)/5 + (a*sin(c + d*x)^6)/6 - (2*a*sin(c + d*x)^7)/7 - (a *sin(c + d*x)^8)/4 + (a*sin(c + d*x)^9)/9 + (a*sin(c + d*x)^10)/10)/d
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right )^{5} a \left (126 \sin \left (d x +c \right )^{5}+140 \sin \left (d x +c \right )^{4}-315 \sin \left (d x +c \right )^{3}-360 \sin \left (d x +c \right )^{2}+210 \sin \left (d x +c \right )+252\right )}{1260 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**5*a*(126*sin(c + d*x)**5 + 140*sin(c + d*x)**4 - 315*sin(c + d*x)**3 - 360*sin(c + d*x)**2 + 210*sin(c + d*x) + 252))/(1260*d)