Integrand size = 21, antiderivative size = 103 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {3 a^2 x}{4}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {a^2 \sin ^5(c+d x)}{5 d} \]
3/4*a^2*x+2*a^2*sin(d*x+c)/d+3/4*a^2*cos(d*x+c)*sin(d*x+c)/d+1/2*a^2*cos(d *x+c)^3*sin(d*x+c)/d-a^2*sin(d*x+c)^3/d+1/5*a^2*sin(d*x+c)^5/d
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.59 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {a^2 (60 d x+110 \sin (c+d x)+40 \sin (2 (c+d x))+15 \sin (3 (c+d x))+5 \sin (4 (c+d x))+\sin (5 (c+d x)))}{80 d} \]
(a^2*(60*d*x + 110*Sin[c + d*x] + 40*Sin[2*(c + d*x)] + 15*Sin[3*(c + d*x) ] + 5*Sin[4*(c + d*x)] + Sin[5*(c + d*x)]))/(80*d)
Time = 0.31 (sec) , antiderivative size = 103, 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.
\(\displaystyle \int \cos ^3(c+d x) (a \cos (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^2 \cos ^5(c+d x)+2 a^2 \cos ^4(c+d x)+a^2 \cos ^3(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a^2 x}{4}\) |
(3*a^2*x)/4 + (2*a^2*Sin[c + d*x])/d + (3*a^2*Cos[c + d*x]*Sin[c + d*x])/( 4*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(2*d) - (a^2*Sin[c + d*x]^3)/d + (a^2*Sin[c + d*x]^5)/(5*d)
3.1.14.3.1 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 = 2.76 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62
method | result | size |
parallelrisch | \(\frac {a^{2} \left (60 d x +110 \sin \left (d x +c \right )+\sin \left (5 d x +5 c \right )+5 \sin \left (4 d x +4 c \right )+15 \sin \left (3 d x +3 c \right )+40 \sin \left (2 d x +2 c \right )\right )}{80 d}\) | \(64\) |
risch | \(\frac {3 a^{2} x}{4}+\frac {11 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{16 d}+\frac {3 a^{2} \sin \left (3 d x +3 c \right )}{16 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{2 d}\) | \(90\) |
derivativedivides | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(96\) |
default | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(96\) |
parts | \(\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {2 a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(101\) |
norman | \(\frac {\frac {3 a^{2} x}{4}+\frac {13 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {9 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {72 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {15 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 a^{2} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(202\) |
1/80*a^2*(60*d*x+110*sin(d*x+c)+sin(5*d*x+5*c)+5*sin(4*d*x+4*c)+15*sin(3*d *x+3*c)+40*sin(2*d*x+2*c))/d
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {15 \, a^{2} d x + {\left (4 \, a^{2} \cos \left (d x + c\right )^{4} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 15 \, a^{2} \cos \left (d x + c\right ) + 24 \, a^{2}\right )} \sin \left (d x + c\right )}{20 \, d} \]
1/20*(15*a^2*d*x + (4*a^2*cos(d*x + c)^4 + 10*a^2*cos(d*x + c)^3 + 12*a^2* cos(d*x + c)^2 + 15*a^2*cos(d*x + c) + 24*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (94) = 188\).
Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((3*a**2*x*sin(c + d*x)**4/4 + 3*a**2*x*sin(c + d*x)**2*cos(c + d *x)**2/2 + 3*a**2*x*cos(c + d*x)**4/4 + 8*a**2*sin(c + d*x)**5/(15*d) + 4* a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 3*a**2*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 2*a**2*sin(c + d*x)**3/(3*d) + a**2*sin(c + d*x)*cos(c + d* x)**4/d + 5*a**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) + a**2*sin(c + d*x)*co s(c + d*x)**2/d, Ne(d, 0)), (x*(a*cos(c) + a)**2*cos(c)**3, True))
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{240 \, d} \]
1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 - 8 0*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2)/d
Time = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {3}{4} \, a^{2} x + \frac {a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {3 \, a^{2} \sin \left (3 \, d x + 3 \, c\right )}{16 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {11 \, a^{2} \sin \left (d x + c\right )}{8 \, d} \]
3/4*a^2*x + 1/80*a^2*sin(5*d*x + 5*c)/d + 1/16*a^2*sin(4*d*x + 4*c)/d + 3/ 16*a^2*sin(3*d*x + 3*c)/d + 1/2*a^2*sin(2*d*x + 2*c)/d + 11/8*a^2*sin(d*x + c)/d
Time = 17.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {3\,a^2\,x}{4}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]