Integrand size = 27, antiderivative size = 97 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}+\frac {b \sin ^6(c+d x)}{6 d}-\frac {2 a \sin ^7(c+d x)}{7 d}-\frac {b \sin ^8(c+d x)}{4 d}+\frac {a \sin ^9(c+d x)}{9 d}+\frac {b \sin ^{10}(c+d x)}{10 d} \] Output:
1/5*a*sin(d*x+c)^5/d+1/6*b*sin(d*x+c)^6/d-2/7*a*sin(d*x+c)^7/d-1/4*b*sin(d *x+c)^8/d+1/9*a*sin(d*x+c)^9/d+1/10*b*sin(d*x+c)^10/d
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-3150 b \cos (2 (c+d x))+525 b \cos (6 (c+d x))-63 b \cos (10 (c+d x))+7560 a \sin (c+d x)-1680 a \sin (3 (c+d x))-1008 a \sin (5 (c+d x))+180 a \sin (7 (c+d x))+140 a \sin (9 (c+d x))}{322560 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^4*(a + b*Sin[c + d*x]),x]
Output:
(-3150*b*Cos[2*(c + d*x)] + 525*b*Cos[6*(c + d*x)] - 63*b*Cos[10*(c + d*x) ] + 7560*a*Sin[c + d*x] - 1680*a*Sin[3*(c + d*x)] - 1008*a*Sin[5*(c + d*x) ] + 180*a*Sin[7*(c + d*x)] + 140*a*Sin[9*(c + d*x)])/(322560*d)
Time = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 522, 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.
\(\displaystyle \int \sin ^4(c+d x) \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^4 \cos (c+d x)^5 (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \sin ^4(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int b^4 \sin ^4(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^9 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\sin ^9(c+d x) b^9-2 \sin ^7(c+d x) b^9+\sin ^5(c+d x) b^9+a \sin ^8(c+d x) b^8-2 a \sin ^6(c+d x) b^8+a \sin ^4(c+d x) b^8\right )d(b \sin (c+d x))}{b^9 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{9} a b^9 \sin ^9(c+d x)-\frac {2}{7} a b^9 \sin ^7(c+d x)+\frac {1}{5} a b^9 \sin ^5(c+d x)+\frac {1}{10} b^{10} \sin ^{10}(c+d x)-\frac {1}{4} b^{10} \sin ^8(c+d x)+\frac {1}{6} b^{10} \sin ^6(c+d x)}{b^9 d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^4*(a + b*Sin[c + d*x]),x]
Output:
((a*b^9*Sin[c + d*x]^5)/5 + (b^10*Sin[c + d*x]^6)/6 - (2*a*b^9*Sin[c + d*x ]^7)/7 - (b^10*Sin[c + d*x]^8)/4 + (a*b^9*Sin[c + d*x]^9)/9 + (b^10*Sin[c + d*x]^10)/10)/(b^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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 201.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(\frac {\frac {b \sin \left (d x +c \right )^{10}}{10}+\frac {a \sin \left (d x +c \right )^{9}}{9}-\frac {b \sin \left (d x +c \right )^{8}}{4}-\frac {2 a \sin \left (d x +c \right )^{7}}{7}+\frac {b \sin \left (d x +c \right )^{6}}{6}+\frac {a \sin \left (d x +c \right )^{5}}{5}}{d}\) | \(72\) |
default | \(\frac {\frac {b \sin \left (d x +c \right )^{10}}{10}+\frac {a \sin \left (d x +c \right )^{9}}{9}-\frac {b \sin \left (d x +c \right )^{8}}{4}-\frac {2 a \sin \left (d x +c \right )^{7}}{7}+\frac {b \sin \left (d x +c \right )^{6}}{6}+\frac {a \sin \left (d x +c \right )^{5}}{5}}{d}\) | \(72\) |
risch | \(\frac {3 a \sin \left (d x +c \right )}{128 d}-\frac {b \cos \left (10 d x +10 c \right )}{5120 d}+\frac {a \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a \sin \left (7 d x +7 c \right )}{1792 d}+\frac {5 b \cos \left (6 d x +6 c \right )}{3072 d}-\frac {a \sin \left (5 d x +5 c \right )}{320 d}-\frac {a \sin \left (3 d x +3 c \right )}{192 d}-\frac {5 b \cos \left (2 d x +2 c \right )}{512 d}\) | \(119\) |
parallelrisch | \(\frac {\left (\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (140 a \cos \left (4 d x +4 c \right )+880 a \cos \left (2 d x +2 c \right )+315 b \sin \left (3 d x +3 c \right )+420 b \sin \left (d x +c \right )+63 b \sin \left (5 d x +5 c \right )+996 a \right ) \left (\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80640 d}\) | \(131\) |
orering | \(\text {Expression too large to display}\) | \(1867\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/10*b*sin(d*x+c)^10+1/9*a*sin(d*x+c)^9-1/4*b*sin(d*x+c)^8-2/7*a*sin( d*x+c)^7+1/6*b*sin(d*x+c)^6+1/5*a*sin(d*x+c)^5)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {126 \, b \cos \left (d x + c\right )^{10} - 315 \, b \cos \left (d x + c\right )^{8} + 210 \, b \cos \left (d x + c\right )^{6} - 4 \, {\left (35 \, a \cos \left (d x + c\right )^{8} - 50 \, a \cos \left (d x + c\right )^{6} + 3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/1260*(126*b*cos(d*x + c)^10 - 315*b*cos(d*x + c)^8 + 210*b*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 = 1.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 a \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {b \sin ^{10}{\left (c + d x \right )}}{60 d} + \frac {b \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{12 d} + \frac {b \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin ^{4}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**4*(a+b*sin(d*x+c)),x)
Output:
Piecewise((8*a*sin(c + d*x)**9/(315*d) + 4*a*sin(c + d*x)**7*cos(c + d*x)* *2/(35*d) + a*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + b*sin(c + d*x)**10/( 60*d) + b*sin(c + d*x)**8*cos(c + d*x)**2/(12*d) + b*sin(c + d*x)**6*cos(c + d*x)**4/(6*d), Ne(d, 0)), (x*(a + b*sin(c))*sin(c)**4*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {126 \, b \sin \left (d x + c\right )^{10} + 140 \, a \sin \left (d x + c\right )^{9} - 315 \, b \sin \left (d x + c\right )^{8} - 360 \, a \sin \left (d x + c\right )^{7} + 210 \, b \sin \left (d x + c\right )^{6} + 252 \, a \sin \left (d x + c\right )^{5}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/1260*(126*b*sin(d*x + c)^10 + 140*a*sin(d*x + c)^9 - 315*b*sin(d*x + c)^ 8 - 360*a*sin(d*x + c)^7 + 210*b*sin(d*x + c)^6 + 252*a*sin(d*x + c)^5)/d
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {126 \, b \sin \left (d x + c\right )^{10} + 140 \, a \sin \left (d x + c\right )^{9} - 315 \, b \sin \left (d x + c\right )^{8} - 360 \, a \sin \left (d x + c\right )^{7} + 210 \, b \sin \left (d x + c\right )^{6} + 252 \, a \sin \left (d x + c\right )^{5}}{1260 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/1260*(126*b*sin(d*x + c)^10 + 140*a*sin(d*x + c)^9 - 315*b*sin(d*x + c)^ 8 - 360*a*sin(d*x + c)^7 + 210*b*sin(d*x + c)^6 + 252*a*sin(d*x + c)^5)/d
Time = 22.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\frac {b\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {b\,{\sin \left (c+d\,x\right )}^8}{4}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {b\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^4*(a + b*sin(c + d*x)),x)
Output:
((a*sin(c + d*x)^5)/5 - (2*a*sin(c + d*x)^7)/7 + (a*sin(c + d*x)^9)/9 + (b *sin(c + d*x)^6)/6 - (b*sin(c + d*x)^8)/4 + (b*sin(c + d*x)^10)/10)/d
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \cos ^5(c+d x) \sin ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right )^{5} \left (126 \sin \left (d x +c \right )^{5} b +140 \sin \left (d x +c \right )^{4} a -315 \sin \left (d x +c \right )^{3} b -360 \sin \left (d x +c \right )^{2} a +210 \sin \left (d x +c \right ) b +252 a \right )}{1260 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x)
Output:
(sin(c + d*x)**5*(126*sin(c + d*x)**5*b + 140*sin(c + d*x)**4*a - 315*sin( c + d*x)**3*b - 360*sin(c + d*x)**2*a + 210*sin(c + d*x)*b + 252*a))/(1260 *d)