Integrand size = 27, antiderivative size = 97 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin ^6(c+d x)}{6 d}+\frac {b \sin ^7(c+d x)}{7 d}-\frac {a \sin ^8(c+d x)}{4 d}-\frac {2 b \sin ^9(c+d x)}{9 d}+\frac {a \sin ^{10}(c+d x)}{10 d}+\frac {b \sin ^{11}(c+d x)}{11 d} \] Output:
1/6*a*sin(d*x+c)^6/d+1/7*b*sin(d*x+c)^7/d-1/4*a*sin(d*x+c)^8/d-2/9*b*sin(d *x+c)^9/d+1/10*a*sin(d*x+c)^10/d+1/11*b*sin(d*x+c)^11/d
Time = 0.43 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-34650 a \cos (2 (c+d x))+5775 a \cos (6 (c+d x))-693 a \cos (10 (c+d x))+34650 b \sin (c+d x)-11550 b \sin (3 (c+d x))-3465 b \sin (5 (c+d x))+2475 b \sin (7 (c+d x))+385 b \sin (9 (c+d x))-315 b \sin (11 (c+d x))}{3548160 d} \] Input:
Integrate[Cos[c + d*x]^5*Sin[c + d*x]^5*(a + b*Sin[c + d*x]),x]
Output:
(-34650*a*Cos[2*(c + d*x)] + 5775*a*Cos[6*(c + d*x)] - 693*a*Cos[10*(c + d *x)] + 34650*b*Sin[c + d*x] - 11550*b*Sin[3*(c + d*x)] - 3465*b*Sin[5*(c + d*x)] + 2475*b*Sin[7*(c + d*x)] + 385*b*Sin[9*(c + d*x)] - 315*b*Sin[11*( c + d*x)])/(3548160*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.
\(\displaystyle \int \sin ^5(c+d x) \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^5 \cos (c+d x)^5 (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \sin ^5(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int b^5 \sin ^5(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^{10} d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\sin ^{10}(c+d x) b^{10}-2 \sin ^8(c+d x) b^{10}+\sin ^6(c+d x) b^{10}+a \sin ^9(c+d x) b^9-2 a \sin ^7(c+d x) b^9+a \sin ^5(c+d x) b^9\right )d(b \sin (c+d x))}{b^{10} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{10} a b^{10} \sin ^{10}(c+d x)-\frac {1}{4} a b^{10} \sin ^8(c+d x)+\frac {1}{6} a b^{10} \sin ^6(c+d x)+\frac {1}{11} b^{11} \sin ^{11}(c+d x)-\frac {2}{9} b^{11} \sin ^9(c+d x)+\frac {1}{7} b^{11} \sin ^7(c+d x)}{b^{10} d}\) |
Input:
Int[Cos[c + d*x]^5*Sin[c + d*x]^5*(a + b*Sin[c + d*x]),x]
Output:
((a*b^10*Sin[c + d*x]^6)/6 + (b^11*Sin[c + d*x]^7)/7 - (a*b^10*Sin[c + d*x ]^8)/4 - (2*b^11*Sin[c + d*x]^9)/9 + (a*b^10*Sin[c + d*x]^10)/10 + (b^11*S in[c + d*x]^11)/11)/(b^10*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 = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74
\[\frac {\frac {b \sin \left (d x +c \right )^{11}}{11}+\frac {a \sin \left (d x +c \right )^{10}}{10}-\frac {2 b \sin \left (d x +c \right )^{9}}{9}-\frac {a \sin \left (d x +c \right )^{8}}{4}+\frac {b \sin \left (d x +c \right )^{7}}{7}+\frac {a \sin \left (d x +c \right )^{6}}{6}}{d}\]
Input:
int(cos(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c)),x)
Output:
1/d*(1/11*b*sin(d*x+c)^11+1/10*a*sin(d*x+c)^10-2/9*b*sin(d*x+c)^9-1/4*a*si n(d*x+c)^8+1/7*b*sin(d*x+c)^7+1/6*a*sin(d*x+c)^6)
Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {1386 \, a \cos \left (d x + c\right )^{10} - 3465 \, a \cos \left (d x + c\right )^{8} + 2310 \, a \cos \left (d x + c\right )^{6} + 20 \, {\left (63 \, b \cos \left (d x + c\right )^{10} - 161 \, b \cos \left (d x + c\right )^{8} + 113 \, b \cos \left (d x + c\right )^{6} - 3 \, b \cos \left (d x + c\right )^{4} - 4 \, b \cos \left (d x + c\right )^{2} - 8 \, b\right )} \sin \left (d x + c\right )}{13860 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/13860*(1386*a*cos(d*x + c)^10 - 3465*a*cos(d*x + c)^8 + 2310*a*cos(d*x + c)^6 + 20*(63*b*cos(d*x + c)^10 - 161*b*cos(d*x + c)^8 + 113*b*cos(d*x + c)^6 - 3*b*cos(d*x + c)^4 - 4*b*cos(d*x + c)^2 - 8*b)*sin(d*x + c))/d
Time = 1.78 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {a \sin ^{10}{\left (c + d x \right )}}{60 d} + \frac {a \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{12 d} + \frac {a \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{6 d} + \frac {8 b \sin ^{11}{\left (c + d x \right )}}{693 d} + \frac {4 b \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{63 d} + \frac {b \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin ^{5}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**5*(a+b*sin(d*x+c)),x)
Output:
Piecewise((a*sin(c + d*x)**10/(60*d) + a*sin(c + d*x)**8*cos(c + d*x)**2/( 12*d) + a*sin(c + d*x)**6*cos(c + d*x)**4/(6*d) + 8*b*sin(c + d*x)**11/(69 3*d) + 4*b*sin(c + d*x)**9*cos(c + d*x)**2/(63*d) + b*sin(c + d*x)**7*cos( c + d*x)**4/(7*d), Ne(d, 0)), (x*(a + b*sin(c))*sin(c)**5*cos(c)**5, True) )
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {1260 \, b \sin \left (d x + c\right )^{11} + 1386 \, a \sin \left (d x + c\right )^{10} - 3080 \, b \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 1980 \, b \sin \left (d x + c\right )^{7} + 2310 \, a \sin \left (d x + c\right )^{6}}{13860 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/13860*(1260*b*sin(d*x + c)^11 + 1386*a*sin(d*x + c)^10 - 3080*b*sin(d*x + c)^9 - 3465*a*sin(d*x + c)^8 + 1980*b*sin(d*x + c)^7 + 2310*a*sin(d*x + c)^6)/d
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {1260 \, b \sin \left (d x + c\right )^{11} + 1386 \, a \sin \left (d x + c\right )^{10} - 3080 \, b \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 1980 \, b \sin \left (d x + c\right )^{7} + 2310 \, a \sin \left (d x + c\right )^{6}}{13860 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/13860*(1260*b*sin(d*x + c)^11 + 1386*a*sin(d*x + c)^10 - 3080*b*sin(d*x + c)^9 - 3465*a*sin(d*x + c)^8 + 1980*b*sin(d*x + c)^7 + 2310*a*sin(d*x + c)^6)/d
Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\frac {b\,{\sin \left (c+d\,x\right )}^{11}}{11}+\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}-\frac {2\,b\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{4}+\frac {b\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}}{d} \] Input:
int(cos(c + d*x)^5*sin(c + d*x)^5*(a + b*sin(c + d*x)),x)
Output:
((a*sin(c + d*x)^6)/6 - (a*sin(c + d*x)^8)/4 + (a*sin(c + d*x)^10)/10 + (b *sin(c + d*x)^7)/7 - (2*b*sin(c + d*x)^9)/9 + (b*sin(c + d*x)^11)/11)/d
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \cos ^5(c+d x) \sin ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right )^{6} \left (1260 \sin \left (d x +c \right )^{5} b +1386 \sin \left (d x +c \right )^{4} a -3080 \sin \left (d x +c \right )^{3} b -3465 \sin \left (d x +c \right )^{2} a +1980 \sin \left (d x +c \right ) b +2310 a \right )}{13860 d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c)),x)
Output:
(sin(c + d*x)**6*(1260*sin(c + d*x)**5*b + 1386*sin(c + d*x)**4*a - 3080*s in(c + d*x)**3*b - 3465*sin(c + d*x)**2*a + 1980*sin(c + d*x)*b + 2310*a)) /(13860*d)