Integrand size = 21, antiderivative size = 247 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {1}{4} a b \left (6 a^2+5 b^2\right ) x+\frac {\left (35 a^4+168 a^2 b^2+24 b^4\right ) \sin (c+d x)}{35 d}+\frac {a b \left (6 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a b \left (6 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{6 d}+\frac {b^2 \left (37 a^2+6 b^2\right ) \cos ^4(c+d x) \sin (c+d x)}{35 d}+\frac {8 a b^3 \cos ^5(c+d x) \sin (c+d x)}{21 d}+\frac {b^2 \cos ^4(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}-\frac {\left (35 a^4+168 a^2 b^2+24 b^4\right ) \sin ^3(c+d x)}{105 d} \] Output:
1/4*a*b*(6*a^2+5*b^2)*x+1/35*(35*a^4+168*a^2*b^2+24*b^4)*sin(d*x+c)/d+1/4* a*b*(6*a^2+5*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/6*a*b*(6*a^2+5*b^2)*cos(d*x+c) ^3*sin(d*x+c)/d+1/35*b^2*(37*a^2+6*b^2)*cos(d*x+c)^4*sin(d*x+c)/d+8/21*a*b ^3*cos(d*x+c)^5*sin(d*x+c)/d+1/7*b^2*cos(d*x+c)^4*(a+b*cos(d*x+c))^2*sin(d *x+c)/d-1/105*(35*a^4+168*a^2*b^2+24*b^4)*sin(d*x+c)^3/d
Time = 1.17 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {1680 a b \left (6 a^2+5 b^2\right ) (c+d x)+105 \left (48 a^4+240 a^2 b^2+35 b^4\right ) \sin (c+d x)+420 a b \left (16 a^2+15 b^2\right ) \sin (2 (c+d x))+35 \left (16 a^4+120 a^2 b^2+21 b^4\right ) \sin (3 (c+d x))+420 a b \left (2 a^2+3 b^2\right ) \sin (4 (c+d x))+21 b^2 \left (24 a^2+7 b^2\right ) \sin (5 (c+d x))+140 a b^3 \sin (6 (c+d x))+15 b^4 \sin (7 (c+d x))}{6720 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + b*Cos[c + d*x])^4,x]
Output:
(1680*a*b*(6*a^2 + 5*b^2)*(c + d*x) + 105*(48*a^4 + 240*a^2*b^2 + 35*b^4)* Sin[c + d*x] + 420*a*b*(16*a^2 + 15*b^2)*Sin[2*(c + d*x)] + 35*(16*a^4 + 1 20*a^2*b^2 + 21*b^4)*Sin[3*(c + d*x)] + 420*a*b*(2*a^2 + 3*b^2)*Sin[4*(c + d*x)] + 21*b^2*(24*a^2 + 7*b^2)*Sin[5*(c + d*x)] + 140*a*b^3*Sin[6*(c + d *x)] + 15*b^4*Sin[7*(c + d*x)])/(6720*d)
Time = 1.24 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.90, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 3272, 3042, 3512, 27, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \frac {1}{7} \int \cos ^3(c+d x) (a+b \cos (c+d x)) \left (16 a b^2 \cos ^2(c+d x)+3 b \left (7 a^2+2 b^2\right ) \cos (c+d x)+a \left (7 a^2+4 b^2\right )\right )dx+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (16 a b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (7 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (7 a^2+4 b^2\right )\right )dx+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{6} \int 2 \cos ^3(c+d x) \left (3 \left (7 a^2+4 b^2\right ) a^2+14 b \left (6 a^2+5 b^2\right ) \cos (c+d x) a+3 b^2 \left (37 a^2+6 b^2\right ) \cos ^2(c+d x)\right )dx+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \cos ^3(c+d x) \left (3 \left (7 a^2+4 b^2\right ) a^2+14 b \left (6 a^2+5 b^2\right ) \cos (c+d x) a+3 b^2 \left (37 a^2+6 b^2\right ) \cos ^2(c+d x)\right )dx+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 \left (7 a^2+4 b^2\right ) a^2+14 b \left (6 a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 b^2 \left (37 a^2+6 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \int \cos ^3(c+d x) \left (3 \left (35 a^4+168 b^2 a^2+24 b^4\right )+70 a b \left (6 a^2+5 b^2\right ) \cos (c+d x)\right )dx+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 \left (35 a^4+168 b^2 a^2+24 b^4\right )+70 a b \left (6 a^2+5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \int \cos ^4(c+d x)dx+3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \int \cos ^3(c+d x)dx\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {3 b^2 \left (37 a^2+6 b^2\right ) \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {1}{5} \left (70 a b \left (6 a^2+5 b^2\right ) \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {3 \left (35 a^4+168 a^2 b^2+24 b^4\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )\right )+\frac {8 a b^3 \sin (c+d x) \cos ^5(c+d x)}{3 d}\right )+\frac {b^2 \sin (c+d x) \cos ^4(c+d x) (a+b \cos (c+d x))^2}{7 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + b*Cos[c + d*x])^4,x]
Output:
(b^2*Cos[c + d*x]^4*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d) + ((8*a*b^3 *Cos[c + d*x]^5*Sin[c + d*x])/(3*d) + ((3*b^2*(37*a^2 + 6*b^2)*Cos[c + d*x ]^4*Sin[c + d*x])/(5*d) + ((-3*(35*a^4 + 168*a^2*b^2 + 24*b^4)*(-Sin[c + d *x] + Sin[c + d*x]^3/3))/d + 70*a*b*(6*a^2 + 5*b^2)*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/5)/3)/7
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 4.15 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.77
\[\frac {\frac {a^{4} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+4 a^{3} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {6 a^{2} b^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+4 a \,b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {b^{4} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\]
Input:
int(cos(d*x+c)^3*(a+cos(d*x+c)*b)^4,x)
Output:
1/d*(1/3*a^4*(cos(d*x+c)^2+2)*sin(d*x+c)+4*a^3*b*(1/4*(cos(d*x+c)^3+3/2*co s(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+6/5*a^2*b^2*(8/3+cos(d*x+c)^4+4/3*cos( d*x+c)^2)*sin(d*x+c)+4*a*b^3*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos( d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+1/7*b^4*(16/5+cos(d*x+c)^6+6/5*cos(d*x +c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))
Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.69 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {105 \, {\left (6 \, a^{3} b + 5 \, a b^{3}\right )} d x + {\left (60 \, b^{4} \cos \left (d x + c\right )^{6} + 280 \, a b^{3} \cos \left (d x + c\right )^{5} + 72 \, {\left (7 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 280 \, a^{4} + 1344 \, a^{2} b^{2} + 192 \, b^{4} + 70 \, {\left (6 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (35 \, a^{4} + 168 \, a^{2} b^{2} + 24 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 105 \, {\left (6 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{420 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^4,x, algorithm="fricas")
Output:
1/420*(105*(6*a^3*b + 5*a*b^3)*d*x + (60*b^4*cos(d*x + c)^6 + 280*a*b^3*co s(d*x + c)^5 + 72*(7*a^2*b^2 + b^4)*cos(d*x + c)^4 + 280*a^4 + 1344*a^2*b^ 2 + 192*b^4 + 70*(6*a^3*b + 5*a*b^3)*cos(d*x + c)^3 + 4*(35*a^4 + 168*a^2* b^2 + 24*b^4)*cos(d*x + c)^2 + 105*(6*a^3*b + 5*a*b^3)*cos(d*x + c))*sin(d *x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (231) = 462\).
Time = 0.49 (sec) , antiderivative size = 495, normalized size of antiderivative = 2.00 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\begin {cases} \frac {2 a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 a^{3} b x \sin ^{4}{\left (c + d x \right )}}{2} + 3 a^{3} b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \frac {3 a^{3} b x \cos ^{4}{\left (c + d x \right )}}{2} + \frac {3 a^{3} b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {5 a^{3} b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {16 a^{2} b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {8 a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {6 a^{2} b^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 a b^{3} x \sin ^{6}{\left (c + d x \right )}}{4} + \frac {15 a b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {15 a b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4} + \frac {5 a b^{3} x \cos ^{6}{\left (c + d x \right )}}{4} + \frac {5 a b^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {10 a b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {11 a b^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{4 d} + \frac {16 b^{4} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 b^{4} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {b^{4} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{4} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*(a+b*cos(d*x+c))**4,x)
Output:
Piecewise((2*a**4*sin(c + d*x)**3/(3*d) + a**4*sin(c + d*x)*cos(c + d*x)** 2/d + 3*a**3*b*x*sin(c + d*x)**4/2 + 3*a**3*b*x*sin(c + d*x)**2*cos(c + d* x)**2 + 3*a**3*b*x*cos(c + d*x)**4/2 + 3*a**3*b*sin(c + d*x)**3*cos(c + d* x)/(2*d) + 5*a**3*b*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 16*a**2*b**2*sin( c + d*x)**5/(5*d) + 8*a**2*b**2*sin(c + d*x)**3*cos(c + d*x)**2/d + 6*a**2 *b**2*sin(c + d*x)*cos(c + d*x)**4/d + 5*a*b**3*x*sin(c + d*x)**6/4 + 15*a *b**3*x*sin(c + d*x)**4*cos(c + d*x)**2/4 + 15*a*b**3*x*sin(c + d*x)**2*co s(c + d*x)**4/4 + 5*a*b**3*x*cos(c + d*x)**6/4 + 5*a*b**3*sin(c + d*x)**5* cos(c + d*x)/(4*d) + 10*a*b**3*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) + 11* a*b**3*sin(c + d*x)*cos(c + d*x)**5/(4*d) + 16*b**4*sin(c + d*x)**7/(35*d) + 8*b**4*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*b**4*sin(c + d*x)**3*c os(c + d*x)**4/d + b**4*sin(c + d*x)*cos(c + d*x)**6/d, Ne(d, 0)), (x*(a + b*cos(c))**4*cos(c)**3, True))
Time = 0.04 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.78 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=-\frac {560 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b - 672 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} b^{2} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{3} + 48 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} b^{4}}{1680 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^4,x, algorithm="maxima")
Output:
-1/1680*(560*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3*b - 672*(3*sin(d*x + c)^5 - 10* sin(d*x + c)^3 + 15*sin(d*x + c))*a^2*b^2 + 35*(4*sin(2*d*x + 2*c)^3 - 60* d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a*b^3 + 48*(5*sin(d *x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*b^4)/ d
Time = 0.50 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.80 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {b^{4} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a b^{3} \sin \left (6 \, d x + 6 \, c\right )}{48 \, d} + \frac {1}{4} \, {\left (6 \, a^{3} b + 5 \, a b^{3}\right )} x + \frac {{\left (24 \, a^{2} b^{2} + 7 \, b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {{\left (16 \, a^{4} + 120 \, a^{2} b^{2} + 21 \, b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {{\left (16 \, a^{3} b + 15 \, a b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{16 \, d} + \frac {{\left (48 \, a^{4} + 240 \, a^{2} b^{2} + 35 \, b^{4}\right )} \sin \left (d x + c\right )}{64 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^4,x, algorithm="giac")
Output:
1/448*b^4*sin(7*d*x + 7*c)/d + 1/48*a*b^3*sin(6*d*x + 6*c)/d + 1/4*(6*a^3* b + 5*a*b^3)*x + 1/320*(24*a^2*b^2 + 7*b^4)*sin(5*d*x + 5*c)/d + 1/16*(2*a ^3*b + 3*a*b^3)*sin(4*d*x + 4*c)/d + 1/192*(16*a^4 + 120*a^2*b^2 + 21*b^4) *sin(3*d*x + 3*c)/d + 1/16*(16*a^3*b + 15*a*b^3)*sin(2*d*x + 2*c)/d + 1/64 *(48*a^4 + 240*a^2*b^2 + 35*b^4)*sin(d*x + c)/d
Time = 42.13 (sec) , antiderivative size = 476, normalized size of antiderivative = 1.93 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {\left (2\,a^4-5\,a^3\,b+12\,a^2\,b^2-\frac {11\,a\,b^3}{2}+2\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (\frac {28\,a^4}{3}-12\,a^3\,b+40\,a^2\,b^2-\frac {14\,a\,b^3}{3}+4\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {58\,a^4}{3}-9\,a^3\,b+\frac {452\,a^2\,b^2}{5}-\frac {85\,a\,b^3}{6}+\frac {86\,b^4}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (24\,a^4+\frac {624\,a^2\,b^2}{5}+\frac {424\,b^4}{35}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {58\,a^4}{3}+9\,a^3\,b+\frac {452\,a^2\,b^2}{5}+\frac {85\,a\,b^3}{6}+\frac {86\,b^4}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {28\,a^4}{3}+12\,a^3\,b+40\,a^2\,b^2+\frac {14\,a\,b^3}{3}+4\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^4+5\,a^3\,b+12\,a^2\,b^2+\frac {11\,a\,b^3}{2}+2\,b^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,b\,\mathrm {atan}\left (\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2+5\,b^2\right )}{2\,\left (3\,a^3\,b+\frac {5\,a\,b^3}{2}\right )}\right )\,\left (6\,a^2+5\,b^2\right )}{2\,d}-\frac {a\,b\,\left (6\,a^2+5\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{2\,d} \] Input:
int(cos(c + d*x)^3*(a + b*cos(c + d*x))^4,x)
Output:
(tan(c/2 + (d*x)/2)^7*(24*a^4 + (424*b^4)/35 + (624*a^2*b^2)/5) + tan(c/2 + (d*x)/2)^13*(2*a^4 - 5*a^3*b - (11*a*b^3)/2 + 2*b^4 + 12*a^2*b^2) + tan( c/2 + (d*x)/2)^3*((14*a*b^3)/3 + 12*a^3*b + (28*a^4)/3 + 4*b^4 + 40*a^2*b^ 2) + tan(c/2 + (d*x)/2)^11*((28*a^4)/3 - 12*a^3*b - (14*a*b^3)/3 + 4*b^4 + 40*a^2*b^2) + tan(c/2 + (d*x)/2)^5*((85*a*b^3)/6 + 9*a^3*b + (58*a^4)/3 + (86*b^4)/5 + (452*a^2*b^2)/5) + tan(c/2 + (d*x)/2)^9*((58*a^4)/3 - 9*a^3* b - (85*a*b^3)/6 + (86*b^4)/5 + (452*a^2*b^2)/5) + tan(c/2 + (d*x)/2)*((11 *a*b^3)/2 + 5*a^3*b + 2*a^4 + 2*b^4 + 12*a^2*b^2))/(d*(7*tan(c/2 + (d*x)/2 )^2 + 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d* x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + ( d*x)/2)^14 + 1)) + (a*b*atan((a*b*tan(c/2 + (d*x)/2)*(6*a^2 + 5*b^2))/(2*( (5*a*b^3)/2 + 3*a^3*b)))*(6*a^2 + 5*b^2))/(2*d) - (a*b*(6*a^2 + 5*b^2)*(at an(tan(c/2 + (d*x)/2)) - (d*x)/2))/(2*d)
Time = 0.17 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a \,b^{3}-420 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b -910 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{3}+1050 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b +1155 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-60 \sin \left (d x +c \right )^{7} b^{4}+504 \sin \left (d x +c \right )^{5} a^{2} b^{2}+252 \sin \left (d x +c \right )^{5} b^{4}-140 \sin \left (d x +c \right )^{3} a^{4}-1680 \sin \left (d x +c \right )^{3} a^{2} b^{2}-420 \sin \left (d x +c \right )^{3} b^{4}+420 \sin \left (d x +c \right ) a^{4}+2520 \sin \left (d x +c \right ) a^{2} b^{2}+420 \sin \left (d x +c \right ) b^{4}+630 a^{3} b d x +525 a \,b^{3} d x}{420 d} \] Input:
int(cos(d*x+c)^3*(a+b*cos(d*x+c))^4,x)
Output:
(280*cos(c + d*x)*sin(c + d*x)**5*a*b**3 - 420*cos(c + d*x)*sin(c + d*x)** 3*a**3*b - 910*cos(c + d*x)*sin(c + d*x)**3*a*b**3 + 1050*cos(c + d*x)*sin (c + d*x)*a**3*b + 1155*cos(c + d*x)*sin(c + d*x)*a*b**3 - 60*sin(c + d*x) **7*b**4 + 504*sin(c + d*x)**5*a**2*b**2 + 252*sin(c + d*x)**5*b**4 - 140* sin(c + d*x)**3*a**4 - 1680*sin(c + d*x)**3*a**2*b**2 - 420*sin(c + d*x)** 3*b**4 + 420*sin(c + d*x)*a**4 + 2520*sin(c + d*x)*a**2*b**2 + 420*sin(c + d*x)*b**4 + 630*a**3*b*d*x + 525*a*b**3*d*x)/(420*d)