Integrand size = 21, antiderivative size = 87 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {7 a^2 x}{8}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d} \]
7/8*a^2*x+2*a^2*sin(d*x+c)/d+7/8*a^2*cos(d*x+c)*sin(d*x+c)/d+1/4*a^2*cos(d *x+c)^3*sin(d*x+c)/d-2/3*a^2*sin(d*x+c)^3/d
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.61 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {a^2 (84 d x+144 \sin (c+d x)+48 \sin (2 (c+d x))+16 \sin (3 (c+d x))+3 \sin (4 (c+d x)))}{96 d} \]
(a^2*(84*d*x + 144*Sin[c + d*x] + 48*Sin[2*(c + d*x)] + 16*Sin[3*(c + d*x) ] + 3*Sin[4*(c + d*x)]))/(96*d)
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3236, 2009}
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.
\(\displaystyle \int \cos ^2(c+d x) (a \cos (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^2 \cos ^4(c+d x)+2 a^2 \cos ^3(c+d x)+a^2 \cos ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^2 x}{8}\) |
(7*a^2*x)/8 + (2*a^2*Sin[c + d*x])/d + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/( 8*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (2*a^2*Sin[c + d*x]^3)/(3 *d)
3.1.15.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Time = 2.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {a^{2} \left (84 d x +144 \sin \left (d x +c \right )+3 \sin \left (4 d x +4 c \right )+16 \sin \left (3 d x +3 c \right )+48 \sin \left (2 d x +2 c \right )\right )}{96 d}\) | \(55\) |
risch | \(\frac {7 a^{2} x}{8}+\frac {3 a^{2} \sin \left (d x +c \right )}{2 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{2} \sin \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{2 d}\) | \(73\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(90\) |
default | \(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(90\) |
parts | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {2 a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\) | \(95\) |
norman | \(\frac {\frac {7 a^{2} x}{8}+\frac {25 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {83 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {77 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {7 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {21 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {7 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {7 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(166\) |
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {21 \, a^{2} d x + {\left (6 \, a^{2} \cos \left (d x + c\right )^{3} + 16 \, a^{2} \cos \left (d x + c\right )^{2} + 21 \, a^{2} \cos \left (d x + c\right ) + 32 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]
1/24*(21*a^2*d*x + (6*a^2*cos(d*x + c)^3 + 16*a^2*cos(d*x + c)^2 + 21*a^2* cos(d*x + c) + 32*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (82) = 164\).
Time = 0.18 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.43 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
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 + a**2*x*sin(c + d*x)**2/2 + 3*a**2*x*cos(c + d*x)**4/8 + a**2*x* cos(c + d*x)**2/2 + 3*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 4*a**2*sin (c + d*x)**3/(3*d) + 5*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 2*a**2*si n(c + d*x)*cos(c + d*x)**2/d + a**2*sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(a*cos(c) + a)**2*cos(c)**2, True))
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=-\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \]
-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 3*(12*d*x + 12*c + sin(4 *d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c) )*a^2)/d
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {7}{8} \, a^{2} x + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {3 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]
7/8*a^2*x + 1/32*a^2*sin(4*d*x + 4*c)/d + 1/6*a^2*sin(3*d*x + 3*c)/d + 1/2 *a^2*sin(2*d*x + 2*c)/d + 3/2*a^2*sin(d*x + c)/d
Time = 17.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {7\,a^2\,x}{8}+\frac {\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {77\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]