Integrand size = 25, antiderivative size = 87 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b x}{16}-\frac {a \cos ^5(c+d x)}{5 d}+\frac {b \cos (c+d x) \sin (c+d x)}{16 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {b \cos ^5(c+d x) \sin (c+d x)}{6 d} \] Output:
1/16*b*x-1/5*a*cos(d*x+c)^5/d+1/16*b*cos(d*x+c)*sin(d*x+c)/d+1/24*b*cos(d* x+c)^3*sin(d*x+c)/d-1/6*b*cos(d*x+c)^5*sin(d*x+c)/d
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {-60 b d x+120 a \cos (c+d x)+60 a \cos (3 (c+d x))+12 a \cos (5 (c+d x))-15 b \sin (2 (c+d x))+15 b \sin (4 (c+d x))+5 b \sin (6 (c+d x))}{960 d} \] Input:
Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x]),x]
Output:
-1/960*(-60*b*d*x + 120*a*Cos[c + d*x] + 60*a*Cos[3*(c + d*x)] + 12*a*Cos[ 5*(c + d*x)] - 15*b*Sin[2*(c + d*x)] + 15*b*Sin[4*(c + d*x)] + 5*b*Sin[6*( c + d*x)])/d
Time = 0.50 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3317, 3042, 3045, 15, 3048, 3042, 3115, 3042, 3115, 24}
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.
| \(\displaystyle \int \sin (c+d x) \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \cos (c+d x)^4 (a+b \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \cos ^4(c+d x) \sin (c+d x)dx+b \int \cos ^4(c+d x) \sin ^2(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \cos (c+d x)^4 \sin (c+d x)dx+b \int \cos (c+d x)^4 \sin (c+d x)^2dx\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle b \int \cos (c+d x)^4 \sin (c+d x)^2dx-\frac {a \int \cos ^4(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle b \int \cos (c+d x)^4 \sin (c+d x)^2dx-\frac {a \cos ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3048 |
| \(\displaystyle b \left (\frac {1}{6} \int \cos ^4(c+d x)dx-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {1}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle b \left (\frac {1}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {1}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle b \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle b \left (\frac {1}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\) |
Input:
Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x]),x]
Output:
-1/5*(a*Cos[c + d*x]^5)/d + b*(-1/6*(Cos[c + d*x]^5*Sin[c + d*x])/d + ((Co s[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/( 2*d)))/4)/6)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Time = 7.00 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {-\frac {a \cos \left (d x +c \right )^{5}}{5}+b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\) | \(68\) |
| default | \(\frac {-\frac {a \cos \left (d x +c \right )^{5}}{5}+b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\) | \(68\) |
| parallelrisch | \(\frac {60 b x d -60 a \cos \left (3 d x +3 c \right )-120 a \cos \left (d x +c \right )+15 b \sin \left (2 d x +2 c \right )-5 b \sin \left (6 d x +6 c \right )-12 a \cos \left (5 d x +5 c \right )-15 b \sin \left (4 d x +4 c \right )-192 a}{960 d}\) | \(84\) |
| risch | \(\frac {b x}{16}-\frac {a \cos \left (d x +c \right )}{8 d}-\frac {b \sin \left (6 d x +6 c \right )}{192 d}-\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {b \sin \left (4 d x +4 c \right )}{64 d}-\frac {a \cos \left (3 d x +3 c \right )}{16 d}+\frac {b \sin \left (2 d x +2 c \right )}{64 d}\) | \(93\) |
| norman | \(\frac {\frac {b x}{16}-\frac {2 a}{5 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {47 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {13 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {13 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {47 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {3 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {5 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {3 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\) | \(303\) |
| orering | \(\text {Expression too large to display}\) | \(1734\) |
Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/5*a*cos(d*x+c)^5+b*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^ 3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c))
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {48 \, a \cos \left (d x + c\right )^{5} - 15 \, b d x + 5 \, {\left (8 \, b \cos \left (d x + c\right )^{5} - 2 \, b \cos \left (d x + c\right )^{3} - 3 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/240*(48*a*cos(d*x + c)^5 - 15*b*d*x + 5*(8*b*cos(d*x + c)^5 - 2*b*cos(d *x + c)^3 - 3*b*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (76) = 152\).
Time = 0.34 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.92 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} - \frac {a \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin {\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)*(a+b*sin(d*x+c)),x)
Output:
Piecewise((-a*cos(c + d*x)**5/(5*d) + b*x*sin(c + d*x)**6/16 + 3*b*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*b*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + b*x*cos(c + d*x)**6/16 + b*sin(c + d*x)**5*cos(c + d*x)/(16*d) + b*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - b*sin(c + d*x)*cos(c + d*x)**5/(16*d), N e(d, 0)), (x*(a + b*sin(c))*sin(c)*cos(c)**4, True))
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {192 \, a \cos \left (d x + c\right )^{5} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b}{960 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/960*(192*a*cos(d*x + c)^5 - 5*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3 *sin(4*d*x + 4*c))*b)/d
Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {1}{16} \, b x - \frac {a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac {a \cos \left (d x + c\right )}{8 \, d} - \frac {b \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/16*b*x - 1/80*a*cos(5*d*x + 5*c)/d - 1/16*a*cos(3*d*x + 3*c)/d - 1/8*a*c os(d*x + c)/d - 1/192*b*sin(6*d*x + 6*c)/d - 1/64*b*sin(4*d*x + 4*c)/d + 1 /64*b*sin(2*d*x + 2*c)/d
Time = 36.87 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.08 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b\,x}{16}-\frac {-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {47\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {13\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {13\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {47\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {2\,a}{5}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \] Input:
int(cos(c + d*x)^4*sin(c + d*x)*(a + b*sin(c + d*x)),x)
Output:
(b*x)/16 - ((2*a)/5 + (b*tan(c/2 + (d*x)/2))/8 + (2*a*tan(c/2 + (d*x)/2)^2 )/5 + 4*a*tan(c/2 + (d*x)/2)^4 + 4*a*tan(c/2 + (d*x)/2)^6 + 2*a*tan(c/2 + (d*x)/2)^8 + 2*a*tan(c/2 + (d*x)/2)^10 - (47*b*tan(c/2 + (d*x)/2)^3)/24 + (13*b*tan(c/2 + (d*x)/2)^5)/4 - (13*b*tan(c/2 + (d*x)/2)^7)/4 + (47*b*tan( c/2 + (d*x)/2)^9)/24 - (b*tan(c/2 + (d*x)/2)^11)/8)/(d*(tan(c/2 + (d*x)/2) ^2 + 1)^6)
Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.22 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b -48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a +70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -48 \cos \left (d x +c \right ) a +48 a +15 b d x}{240 d} \] Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c)),x)
Output:
( - 40*cos(c + d*x)*sin(c + d*x)**5*b - 48*cos(c + d*x)*sin(c + d*x)**4*a + 70*cos(c + d*x)*sin(c + d*x)**3*b + 96*cos(c + d*x)*sin(c + d*x)**2*a - 15*cos(c + d*x)*sin(c + d*x)*b - 48*cos(c + d*x)*a + 48*a + 15*b*d*x)/(240 *d)