Integrand size = 21, antiderivative size = 101 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {1}{8} \left (4 a^2+3 b^2\right ) x+\frac {2 a b \sin (c+d x)}{d}+\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a b \sin ^3(c+d x)}{3 d} \] Output:
1/8*(4*a^2+3*b^2)*x+2*a*b*sin(d*x+c)/d+1/8*(4*a^2+3*b^2)*cos(d*x+c)*sin(d* x+c)/d+1/4*b^2*cos(d*x+c)^3*sin(d*x+c)/d-2/3*a*b*sin(d*x+c)^3/d
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {48 a^2 c+36 b^2 c+48 a^2 d x+36 b^2 d x+192 a b \sin (c+d x)-64 a b \sin ^3(c+d x)+24 \left (a^2+b^2\right ) \sin (2 (c+d x))+3 b^2 \sin (4 (c+d x))}{96 d} \] Input:
Integrate[Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2,x]
Output:
(48*a^2*c + 36*b^2*c + 48*a^2*d*x + 36*b^2*d*x + 192*a*b*Sin[c + d*x] - 64 *a*b*Sin[c + d*x]^3 + 24*(a^2 + b^2)*Sin[2*(c + d*x)] + 3*b^2*Sin[4*(c + d *x)])/(96*d)
Time = 0.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3268, 3042, 3113, 2009, 3493, 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.
\(\displaystyle \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 3268 |
\(\displaystyle \int \cos ^2(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right )dx+2 a b \int \cos ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \frac {1}{4} \left (4 a^2+3 b^2\right ) \int \cos ^2(c+d x)dx-\frac {2 a b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (4 a^2+3 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {2 a b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{4} \left (4 a^2+3 b^2\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {2 a b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} \left (4 a^2+3 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {2 a b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2,x]
Output:
(b^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((4*a^2 + 3*b^2)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4 - (2*a*b*(-Sin[c + d*x] + Sin[c + d*x]^3/3)) /d
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 = 7.48 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(\frac {24 \left (a^{2}+b^{2}\right ) \sin \left (2 d x +2 c \right )+16 a b \sin \left (3 d x +3 c \right )+3 \sin \left (4 d x +4 c \right ) b^{2}+144 a b \sin \left (d x +c \right )+48 d x \left (a^{2}+\frac {3 b^{2}}{4}\right )}{96 d}\) | \(75\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(89\) |
default | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(89\) |
risch | \(\frac {3 b^{2} x}{8}+\frac {a^{2} x}{2}+\frac {3 a b \sin \left (d x +c \right )}{2 d}+\frac {\sin \left (4 d x +4 c \right ) b^{2}}{32 d}+\frac {a b \sin \left (3 d x +3 c \right )}{6 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) b^{2}}{4 d}\) | \(94\) |
parts | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {2 a b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}\) | \(94\) |
norman | \(\frac {\left (\frac {3 b^{2}}{8}+\frac {a^{2}}{2}\right ) x +\left (\frac {3 b^{2}}{2}+2 a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3 b^{2}}{2}+2 a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 b^{2}}{8}+\frac {a^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {9 b^{2}}{4}+3 a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {\left (4 a^{2}-16 a b +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (4 a^{2}+16 a b +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a^{2}-80 a b -9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {\left (12 a^{2}+80 a b -9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(249\) |
orering | \(\text {Expression too large to display}\) | \(1120\) |
Input:
int(cos(d*x+c)^2*(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
Output:
1/96*(24*(a^2+b^2)*sin(2*d*x+2*c)+16*a*b*sin(3*d*x+3*c)+3*sin(4*d*x+4*c)*b ^2+144*a*b*sin(d*x+c)+48*d*x*(a^2+3/4*b^2))/d
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} d x + {\left (6 \, b^{2} \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right )^{2} + 32 \, a b + 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/24*(3*(4*a^2 + 3*b^2)*d*x + (6*b^2*cos(d*x + c)^3 + 16*a*b*cos(d*x + c)^ 2 + 32*a*b + 3*(4*a^2 + 3*b^2)*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (92) = 184\).
Time = 0.16 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.09 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 a b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**2*(a+b*cos(d*x+c))**2,x)
Output:
Piecewise((a**2*x*sin(c + d*x)**2/2 + a**2*x*cos(c + d*x)**2/2 + a**2*sin( c + d*x)*cos(c + d*x)/(2*d) + 4*a*b*sin(c + d*x)**3/(3*d) + 2*a*b*sin(c + d*x)*cos(c + d*x)**2/d + 3*b**2*x*sin(c + d*x)**4/8 + 3*b**2*x*sin(c + d*x )**2*cos(c + d*x)**2/4 + 3*b**2*x*cos(c + d*x)**4/8 + 3*b**2*sin(c + d*x)* *3*cos(c + d*x)/(8*d) + 5*b**2*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0 )), (x*(a + b*cos(c))**2*cos(c)**2, True))
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{96 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 - 64*(sin(d*x + c)^3 - 3*sin (d*x + c))*a*b + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c)) *b^2)/d
Time = 0.50 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {1}{8} \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} x + \frac {b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a b \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} + \frac {3 \, a b \sin \left (d x + c\right )}{2 \, d} + \frac {{\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
1/8*(4*a^2 + 3*b^2)*x + 1/32*b^2*sin(4*d*x + 4*c)/d + 1/6*a*b*sin(3*d*x + 3*c)/d + 3/2*a*b*sin(d*x + c)/d + 1/4*(a^2 + b^2)*sin(2*d*x + 2*c)/d
Time = 39.82 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {a^2\,x}{2}+\frac {3\,b^2\,x}{8}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d} \] Input:
int(cos(c + d*x)^2*(a + b*cos(c + d*x))^2,x)
Output:
(a^2*x)/2 + (3*b^2*x)/8 + (a^2*sin(2*c + 2*d*x))/(4*d) + (b^2*sin(2*c + 2* d*x))/(4*d) + (b^2*sin(4*c + 4*d*x))/(32*d) + (3*a*b*sin(c + d*x))/(2*d) + (a*b*sin(3*c + 3*d*x))/(6*d)
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}+12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-16 \sin \left (d x +c \right )^{3} a b +48 \sin \left (d x +c \right ) a b +12 a^{2} d x +9 b^{2} d x}{24 d} \] Input:
int(cos(d*x+c)^2*(a+b*cos(d*x+c))^2,x)
Output:
( - 6*cos(c + d*x)*sin(c + d*x)**3*b**2 + 12*cos(c + d*x)*sin(c + d*x)*a** 2 + 15*cos(c + d*x)*sin(c + d*x)*b**2 - 16*sin(c + d*x)**3*a*b + 48*sin(c + d*x)*a*b + 12*a**2*d*x + 9*b**2*d*x)/(24*d)