Integrand size = 21, antiderivative size = 150 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {1}{16} \left (6 a^2+5 b^2\right ) x+\frac {2 a b \sin (c+d x)}{d}+\frac {\left (6 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (6 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin ^5(c+d x)}{5 d} \] Output:
1/16*(6*a^2+5*b^2)*x+2*a*b*sin(d*x+c)/d+1/16*(6*a^2+5*b^2)*cos(d*x+c)*sin( d*x+c)/d+1/24*(6*a^2+5*b^2)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*b^2*cos(d*x+c)^5 *sin(d*x+c)/d-4/3*a*b*sin(d*x+c)^3/d+2/5*a*b*sin(d*x+c)^5/d
Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {1920 a b \sin (c+d x)-1280 a b \sin ^3(c+d x)+384 a b \sin ^5(c+d x)+5 \left (72 a^2 c+60 b^2 c+72 a^2 d x+60 b^2 d x+\left (48 a^2+45 b^2\right ) \sin (2 (c+d x))+\left (6 a^2+9 b^2\right ) \sin (4 (c+d x))+b^2 \sin (6 (c+d x))\right )}{960 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + b*Cos[c + d*x])^2,x]
Output:
(1920*a*b*Sin[c + d*x] - 1280*a*b*Sin[c + d*x]^3 + 384*a*b*Sin[c + d*x]^5 + 5*(72*a^2*c + 60*b^2*c + 72*a^2*d*x + 60*b^2*d*x + (48*a^2 + 45*b^2)*Sin [2*(c + d*x)] + (6*a^2 + 9*b^2)*Sin[4*(c + d*x)] + b^2*Sin[6*(c + d*x)]))/ (960*d)
Time = 0.56 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3268, 3042, 3113, 2009, 3493, 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.
| \(\displaystyle \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 3268 |
| \(\displaystyle \int \cos ^4(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right )dx+2 a b \int \cos ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3493 |
| \(\displaystyle \frac {1}{6} \left (6 a^2+5 b^2\right ) \int \cos ^4(c+d x)dx-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (6 a^2+5 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{6} \left (6 a^2+5 b^2\right ) \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (6 a^2+5 b^2\right ) \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{6} \left (6 a^2+5 b^2\right ) \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{6} \left (6 a^2+5 b^2\right ) \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {2 a b \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + b*Cos[c + d*x])^2,x]
Output:
(b^2*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*a*b*(-Sin[c + d*x] + (2*Sin[c + d*x]^3)/3 - Sin[c + d*x]^5/5))/d + ((6*a^2 + 5*b^2)*((Cos[c + d*x]^3*Si n[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6
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 _)])^2, x_Symbol] :> Simp[2*c*(d/b) Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
Time = 22.42 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.75
| method | result | size |
| parallelrisch | \(\frac {\left (240 a^{2}+225 b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (30 a^{2}+45 b^{2}\right ) \sin \left (4 d x +4 c \right )+200 a b \sin \left (3 d x +3 c \right )+24 a b \sin \left (5 d x +5 c \right )+5 b^{2} \sin \left (6 d x +6 c \right )+1200 a b \sin \left (d x +c \right )+360 d x \left (a^{2}+\frac {5 b^{2}}{6}\right )}{960 d}\) | \(112\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a b \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(120\) |
| default | \(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a b \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(120\) |
| parts | \(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {2 a b \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}\) | \(125\) |
| risch | \(\frac {3 a^{2} x}{8}+\frac {5 b^{2} x}{16}+\frac {5 a b \sin \left (d x +c \right )}{4 d}+\frac {b^{2} \sin \left (6 d x +6 c \right )}{192 d}+\frac {a b \sin \left (5 d x +5 c \right )}{40 d}+\frac {\sin \left (4 d x +4 c \right ) a^{2}}{32 d}+\frac {3 \sin \left (4 d x +4 c \right ) b^{2}}{64 d}+\frac {5 a b \sin \left (3 d x +3 c \right )}{24 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}+\frac {15 \sin \left (2 d x +2 c \right ) b^{2}}{64 d}\) | \(144\) |
| norman | \(\frac {\left (\frac {3 a^{2}}{8}+\frac {5 b^{2}}{16}\right ) x +\left (\frac {3 a^{2}}{8}+\frac {5 b^{2}}{16}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {9 a^{2}}{4}+\frac {15 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {9 a^{2}}{4}+\frac {15 b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {15 a^{2}}{2}+\frac {25 b^{2}}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {45 a^{2}}{8}+\frac {75 b^{2}}{16}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {45 a^{2}}{8}+\frac {75 b^{2}}{16}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {\left (10 a^{2}-416 a b +75 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}-\frac {\left (10 a^{2}-32 a b +11 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {\left (10 a^{2}+32 a b +11 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (10 a^{2}+416 a b +75 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}-\frac {\left (42 a^{2}-224 a b -5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}+\frac {\left (42 a^{2}+224 a b -5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\) | \(359\) |
| orering | \(\text {Expression too large to display}\) | \(2560\) |
Input:
int(cos(d*x+c)^4*(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
Output:
1/960*((240*a^2+225*b^2)*sin(2*d*x+2*c)+(30*a^2+45*b^2)*sin(4*d*x+4*c)+200 *a*b*sin(3*d*x+3*c)+24*a*b*sin(5*d*x+5*c)+5*b^2*sin(6*d*x+6*c)+1200*a*b*si n(d*x+c)+360*d*x*(a^2+5/6*b^2))/d
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.73 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {15 \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} d x + {\left (40 \, b^{2} \cos \left (d x + c\right )^{5} + 96 \, a b \cos \left (d x + c\right )^{4} + 128 \, a b \cos \left (d x + c\right )^{2} + 10 \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 256 \, a b + 15 \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/240*(15*(6*a^2 + 5*b^2)*d*x + (40*b^2*cos(d*x + c)^5 + 96*a*b*cos(d*x + c)^4 + 128*a*b*cos(d*x + c)^2 + 10*(6*a^2 + 5*b^2)*cos(d*x + c)^3 + 256*a* b + 15*(6*a^2 + 5*b^2)*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (141) = 282\).
Time = 0.35 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.29 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {16 a b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {8 a b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 a b \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{2} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4*(a+b*cos(d*x+c))**2,x)
Output:
Piecewise((3*a**2*x*sin(c + d*x)**4/8 + 3*a**2*x*sin(c + d*x)**2*cos(c + d *x)**2/4 + 3*a**2*x*cos(c + d*x)**4/8 + 3*a**2*sin(c + d*x)**3*cos(c + d*x )/(8*d) + 5*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 16*a*b*sin(c + d*x)* *5/(15*d) + 8*a*b*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 2*a*b*sin(c + d* x)*cos(c + d*x)**4/d + 5*b**2*x*sin(c + d*x)**6/16 + 15*b**2*x*sin(c + d*x )**4*cos(c + d*x)**2/16 + 15*b**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5 *b**2*x*cos(c + d*x)**6/16 + 5*b**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*b**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*cos(c))**2*cos(c)**4, True))
Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a b - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{960 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
1/960*(30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2 + 12 8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a*b - 5*(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)) *b^2)/d
Time = 0.52 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {1}{16} \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} x + \frac {b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a b \sin \left (5 \, d x + 5 \, c\right )}{40 \, d} + \frac {5 \, a b \sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac {5 \, a b \sin \left (d x + c\right )}{4 \, d} + \frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, a^{2} + 15 \, b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
1/16*(6*a^2 + 5*b^2)*x + 1/192*b^2*sin(6*d*x + 6*c)/d + 1/40*a*b*sin(5*d*x + 5*c)/d + 5/24*a*b*sin(3*d*x + 3*c)/d + 5/4*a*b*sin(d*x + c)/d + 1/64*(2 *a^2 + 3*b^2)*sin(4*d*x + 4*c)/d + 1/64*(16*a^2 + 15*b^2)*sin(2*d*x + 2*c) /d
Time = 39.97 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3\,a^2\,x}{8}+\frac {5\,b^2\,x}{16}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {15\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {b^2\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {5\,a\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {5\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{24\,d}+\frac {a\,b\,\sin \left (5\,c+5\,d\,x\right )}{40\,d} \] Input:
int(cos(c + d*x)^4*(a + b*cos(c + d*x))^2,x)
Output:
(3*a^2*x)/8 + (5*b^2*x)/16 + (a^2*sin(2*c + 2*d*x))/(4*d) + (a^2*sin(4*c + 4*d*x))/(32*d) + (15*b^2*sin(2*c + 2*d*x))/(64*d) + (3*b^2*sin(4*c + 4*d* x))/(64*d) + (b^2*sin(6*c + 6*d*x))/(192*d) + (5*a*b*sin(c + d*x))/(4*d) + (5*a*b*sin(3*c + 3*d*x))/(24*d) + (a*b*sin(5*c + 5*d*x))/(40*d)
Time = 0.17 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{2}-60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2}-130 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}+150 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}+165 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}+96 \sin \left (d x +c \right )^{5} a b -320 \sin \left (d x +c \right )^{3} a b +480 \sin \left (d x +c \right ) a b +90 a^{2} d x +75 b^{2} d x}{240 d} \] Input:
int(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x)
Output:
(40*cos(c + d*x)*sin(c + d*x)**5*b**2 - 60*cos(c + d*x)*sin(c + d*x)**3*a* *2 - 130*cos(c + d*x)*sin(c + d*x)**3*b**2 + 150*cos(c + d*x)*sin(c + d*x) *a**2 + 165*cos(c + d*x)*sin(c + d*x)*b**2 + 96*sin(c + d*x)**5*a*b - 320* sin(c + d*x)**3*a*b + 480*sin(c + d*x)*a*b + 90*a**2*d*x + 75*b**2*d*x)/(2 40*d)