Integrand size = 27, antiderivative size = 194 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3}{128} b \left (8 a^2+b^2\right ) x-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{560 d}+\frac {3 b \left (8 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{128 d}+\frac {b \left (8 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{112 d}-\frac {3 a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{56 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d} \] Output:
3/128*b*(8*a^2+b^2)*x-1/560*a*(2*a^2+61*b^2)*cos(d*x+c)^5/d+3/128*b*(8*a^2 +b^2)*cos(d*x+c)*sin(d*x+c)/d+1/64*b*(8*a^2+b^2)*cos(d*x+c)^3*sin(d*x+c)/d -1/112*(2*a^2+7*b^2)*cos(d*x+c)^5*(a+b*sin(d*x+c))/d-3/56*a*cos(d*x+c)^5*( a+b*sin(d*x+c))^2/d-1/8*cos(d*x+c)^5*(a+b*sin(d*x+c))^3/d
Time = 0.84 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.97 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3360 a^2 b c+840 b^3 c+3360 a^2 b d x+420 b^3 d x-280 a \left (8 a^2+9 b^2\right ) \cos (c+d x)-280 \left (4 a^3+3 a b^2\right ) \cos (3 (c+d x))-224 a^3 \cos (5 (c+d x))+168 a b^2 \cos (5 (c+d x))+120 a b^2 \cos (7 (c+d x))+840 a^2 b \sin (2 (c+d x))-840 a^2 b \sin (4 (c+d x))-140 b^3 \sin (4 (c+d x))-280 a^2 b \sin (6 (c+d x))+\frac {35}{2} b^3 \sin (8 (c+d x))}{17920 d} \] Input:
Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x])^3,x]
Output:
(3360*a^2*b*c + 840*b^3*c + 3360*a^2*b*d*x + 420*b^3*d*x - 280*a*(8*a^2 + 9*b^2)*Cos[c + d*x] - 280*(4*a^3 + 3*a*b^2)*Cos[3*(c + d*x)] - 224*a^3*Cos [5*(c + d*x)] + 168*a*b^2*Cos[5*(c + d*x)] + 120*a*b^2*Cos[7*(c + d*x)] + 840*a^2*b*Sin[2*(c + d*x)] - 840*a^2*b*Sin[4*(c + d*x)] - 140*b^3*Sin[4*(c + d*x)] - 280*a^2*b*Sin[6*(c + d*x)] + (35*b^3*Sin[8*(c + d*x)])/2)/(1792 0*d)
Time = 1.00 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 3341, 27, 3042, 3341, 3042, 3341, 3042, 3148, 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.
| \(\displaystyle \int \sin (c+d x) \cos ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \cos (c+d x)^4 (a+b \sin (c+d x))^3dx\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {1}{8} \int 3 \cos ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2dx-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{8} \int \cos ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2dx-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{8} \int \cos (c+d x)^4 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2dx-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \int \cos ^4(c+d x) (a+b \sin (c+d x)) \left (9 a b+\left (2 a^2+7 b^2\right ) \sin (c+d x)\right )dx-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \int \cos (c+d x)^4 (a+b \sin (c+d x)) \left (9 a b+\left (2 a^2+7 b^2\right ) \sin (c+d x)\right )dx-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3341 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \int \cos ^4(c+d x) \left (7 b \left (8 a^2+b^2\right )+a \left (2 a^2+61 b^2\right ) \sin (c+d x)\right )dx-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \int \cos (c+d x)^4 \left (7 b \left (8 a^2+b^2\right )+a \left (2 a^2+61 b^2\right ) \sin (c+d x)\right )dx-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \left (7 b \left (8 a^2+b^2\right ) \int \cos ^4(c+d x)dx-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \left (7 b \left (8 a^2+b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \left (7 b \left (8 a^2+b^2\right ) \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \left (7 b \left (8 a^2+b^2\right ) \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \left (7 b \left (8 a^2+b^2\right ) \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {3}{8} \left (\frac {1}{7} \left (\frac {1}{6} \left (7 b \left (8 a^2+b^2\right ) \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a \left (2 a^2+61 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {\left (2 a^2+7 b^2\right ) \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^3}{8 d}\) |
Input:
Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Sin[c + d*x])^3,x]
Output:
-1/8*(Cos[c + d*x]^5*(a + b*Sin[c + d*x])^3)/d + (3*(-1/7*(a*Cos[c + d*x]^ 5*(a + b*Sin[c + d*x])^2)/d + (-1/6*((2*a^2 + 7*b^2)*Cos[c + d*x]^5*(a + b *Sin[c + d*x]))/d + (-1/5*(a*(2*a^2 + 61*b^2)*Cos[c + d*x]^5)/d + 7*b*(8*a ^2 + b^2)*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*S in[c + d*x])/(2*d)))/4))/6)/7))/8
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[1/(m + p + 1) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && !LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S implerQ[c + d*x, a + b*x])
Time = 0.39 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.93
\[\frac {-\frac {a^{3} \cos \left (d x +c \right )^{5}}{5}+3 a^{2} b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{5}}{8}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{16}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )}{d}\]
Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^3,x)
Output:
1/d*(-1/5*a^3*cos(d*x+c)^5+3*a^2*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)+3*a*b^2*(-1/7*sin(d* x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+b^3*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5 -1/16*sin(d*x+c)*cos(d*x+c)^5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c )+3/128*d*x+3/128*c))
Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {1920 \, a b^{2} \cos \left (d x + c\right )^{7} - 896 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 105 \, {\left (8 \, a^{2} b + b^{3}\right )} d x + 35 \, {\left (16 \, b^{3} \cos \left (d x + c\right )^{7} - 8 \, {\left (8 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (8 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4480 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
1/4480*(1920*a*b^2*cos(d*x + c)^7 - 896*(a^3 + 3*a*b^2)*cos(d*x + c)^5 + 1 05*(8*a^2*b + b^3)*d*x + 35*(16*b^3*cos(d*x + c)^7 - 8*(8*a^2*b + 3*b^3)*c os(d*x + c)^5 + 2*(8*a^2*b + b^3)*cos(d*x + c)^3 + 3*(8*a^2*b + b^3)*cos(d *x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (177) = 354\).
Time = 0.70 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.35 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=\begin {cases} - \frac {a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a^{2} b x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{2} b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{2} b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{2} b x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a^{2} b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a^{2} b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {3 a b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {6 a b^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} + \frac {3 b^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {3 b^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {9 b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 b^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 b^{3} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {11 b^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} - \frac {11 b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} - \frac {3 b^{3} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{3} \sin {\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)*(a+b*sin(d*x+c))**3,x)
Output:
Piecewise((-a**3*cos(c + d*x)**5/(5*d) + 3*a**2*b*x*sin(c + d*x)**6/16 + 9 *a**2*b*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a**2*b*x*sin(c + d*x)**2* cos(c + d*x)**4/16 + 3*a**2*b*x*cos(c + d*x)**6/16 + 3*a**2*b*sin(c + d*x) **5*cos(c + d*x)/(16*d) + a**2*b*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) - 3 *a**2*b*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 3*a*b**2*sin(c + d*x)**2*cos (c + d*x)**5/(5*d) - 6*a*b**2*cos(c + d*x)**7/(35*d) + 3*b**3*x*sin(c + d* x)**8/128 + 3*b**3*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 9*b**3*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 3*b**3*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 3*b**3*x*cos(c + d*x)**8/128 + 3*b**3*sin(c + d*x)**7*cos(c + d*x)/(128* d) + 11*b**3*sin(c + d*x)**5*cos(c + d*x)**3/(128*d) - 11*b**3*sin(c + d*x )**3*cos(c + d*x)**5/(128*d) - 3*b**3*sin(c + d*x)*cos(c + d*x)**7/(128*d) , Ne(d, 0)), (x*(a + b*sin(c))**3*sin(c)*cos(c)**4, True))
Time = 0.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.60 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {7168 \, a^{3} \cos \left (d x + c\right )^{5} - 560 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} b - 3072 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a b^{2} - 35 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{3}}{35840 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
-1/35840*(7168*a^3*cos(d*x + c)^5 - 560*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 1 2*c - 3*sin(4*d*x + 4*c))*a^2*b - 3072*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^ 5)*a*b^2 - 35*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*b^3) /d
Time = 0.19 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, a b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {b^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{2} b \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} + \frac {3 \, a^{2} b \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {3}{128} \, {\left (8 \, a^{2} b + b^{3}\right )} x - \frac {{\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (8 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} - \frac {{\left (6 \, a^{2} b + b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
3/448*a*b^2*cos(7*d*x + 7*c)/d + 1/1024*b^3*sin(8*d*x + 8*c)/d - 1/64*a^2* b*sin(6*d*x + 6*c)/d + 3/64*a^2*b*sin(2*d*x + 2*c)/d + 3/128*(8*a^2*b + b^ 3)*x - 1/320*(4*a^3 - 3*a*b^2)*cos(5*d*x + 5*c)/d - 1/64*(4*a^3 + 3*a*b^2) *cos(3*d*x + 3*c)/d - 1/64*(8*a^3 + 9*a*b^2)*cos(d*x + c)/d - 1/128*(6*a^2 *b + b^3)*sin(4*d*x + 4*c)/d
Time = 35.58 (sec) , antiderivative size = 552, normalized size of antiderivative = 2.85 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:
int(cos(c + d*x)^4*sin(c + d*x)*(a + b*sin(c + d*x))^3,x)
Output:
(3*b*atan((3*b*tan(c/2 + (d*x)/2)*(8*a^2 + b^2))/(64*((3*a^2*b)/8 + (3*b^3 )/64)))*(8*a^2 + b^2))/(64*d) - (tan(c/2 + (d*x)/2)*((3*a^2*b)/8 + (3*b^3) /64) + 10*a^3*tan(c/2 + (d*x)/2)^10 + 2*a^3*tan(c/2 + (d*x)/2)^14 + (12*a* b^2)/35 + tan(c/2 + (d*x)/2)^12*(12*a*b^2 + 6*a^3) + tan(c/2 + (d*x)/2)^8* (12*a*b^2 + 14*a^3) - tan(c/2 + (d*x)/2)^4*((12*a*b^2)/5 - (26*a^3)/5) + t an(c/2 + (d*x)/2)^2*((96*a*b^2)/35 + (6*a^3)/5) + tan(c/2 + (d*x)/2)^6*((9 6*a*b^2)/5 + (62*a^3)/5) - tan(c/2 + (d*x)/2)^15*((3*a^2*b)/8 + (3*b^3)/64 ) - tan(c/2 + (d*x)/2)^3*((41*a^2*b)/8 - (23*b^3)/64) + tan(c/2 + (d*x)/2) ^13*((41*a^2*b)/8 - (23*b^3)/64) - tan(c/2 + (d*x)/2)^5*((13*a^2*b)/8 + (3 33*b^3)/64) + tan(c/2 + (d*x)/2)^11*((13*a^2*b)/8 + (333*b^3)/64) + tan(c/ 2 + (d*x)/2)^7*((31*a^2*b)/8 + (671*b^3)/64) - tan(c/2 + (d*x)/2)^9*((31*a ^2*b)/8 + (671*b^3)/64) + (2*a^3)/5)/(d*(8*tan(c/2 + (d*x)/2)^2 + 28*tan(c /2 + (d*x)/2)^4 + 56*tan(c/2 + (d*x)/2)^6 + 70*tan(c/2 + (d*x)/2)^8 + 56*t an(c/2 + (d*x)/2)^10 + 28*tan(c/2 + (d*x)/2)^12 + 8*tan(c/2 + (d*x)/2)^14 + tan(c/2 + (d*x)/2)^16 + 1)) - (3*b*(8*a^2 + b^2)*(atan(tan(c/2 + (d*x)/2 )) - (d*x)/2))/(64*d)
Time = 0.18 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.47 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-560 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} b^{3}-1920 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a \,b^{2}-2240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a^{2} b +840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{3}-896 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+3072 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+3920 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b -70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}+1792 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-384 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-840 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -105 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{3}-896 \cos \left (d x +c \right ) a^{3}-768 \cos \left (d x +c \right ) a \,b^{2}+896 a^{3}+840 a^{2} b d x +768 a \,b^{2}+105 b^{3} d x}{4480 d} \] Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^3,x)
Output:
( - 560*cos(c + d*x)*sin(c + d*x)**7*b**3 - 1920*cos(c + d*x)*sin(c + d*x) **6*a*b**2 - 2240*cos(c + d*x)*sin(c + d*x)**5*a**2*b + 840*cos(c + d*x)*s in(c + d*x)**5*b**3 - 896*cos(c + d*x)*sin(c + d*x)**4*a**3 + 3072*cos(c + d*x)*sin(c + d*x)**4*a*b**2 + 3920*cos(c + d*x)*sin(c + d*x)**3*a**2*b - 70*cos(c + d*x)*sin(c + d*x)**3*b**3 + 1792*cos(c + d*x)*sin(c + d*x)**2*a **3 - 384*cos(c + d*x)*sin(c + d*x)**2*a*b**2 - 840*cos(c + d*x)*sin(c + d *x)*a**2*b - 105*cos(c + d*x)*sin(c + d*x)*b**3 - 896*cos(c + d*x)*a**3 - 768*cos(c + d*x)*a*b**2 + 896*a**3 + 840*a**2*b*d*x + 768*a*b**2 + 105*b** 3*d*x)/(4480*d)