Integrand size = 21, antiderivative size = 105 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {13 a^3 x}{8}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {a^3 \sin ^5(c+d x)}{5 d} \]
13/8*a^3*x+4*a^3*sin(d*x+c)/d+13/8*a^3*cos(d*x+c)*sin(d*x+c)/d+3/4*a^3*cos (d*x+c)^3*sin(d*x+c)/d-5/3*a^3*sin(d*x+c)^3/d+1/5*a^3*sin(d*x+c)^5/d
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.60 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {a^3 (780 d x+1380 \sin (c+d x)+480 \sin (2 (c+d x))+170 \sin (3 (c+d x))+45 \sin (4 (c+d x))+6 \sin (5 (c+d x)))}{480 d} \]
(a^3*(780*d*x + 1380*Sin[c + d*x] + 480*Sin[2*(c + d*x)] + 170*Sin[3*(c + d*x)] + 45*Sin[4*(c + d*x)] + 6*Sin[5*(c + d*x)]))/(480*d)
Time = 0.32 (sec) , antiderivative size = 105, 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.
\(\displaystyle \int \cos ^2(c+d x) (a \cos (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^3 \cos ^5(c+d x)+3 a^3 \cos ^4(c+d x)+3 a^3 \cos ^3(c+d x)+a^3 \cos ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \sin ^5(c+d x)}{5 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {13 a^3 x}{8}\) |
(13*a^3*x)/8 + (4*a^3*Sin[c + d*x])/d + (13*a^3*Cos[c + d*x]*Sin[c + d*x]) /(8*d) + (3*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (5*a^3*Sin[c + d*x]^3 )/(3*d) + (a^3*Sin[c + d*x]^5)/(5*d)
3.1.24.3.1 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 = 2.47 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {a^{3} \left (780 d x +1380 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )+45 \sin \left (4 d x +4 c \right )+170 \sin \left (3 d x +3 c \right )+480 \sin \left (2 d x +2 c \right )\right )}{480 d}\) | \(66\) |
risch | \(\frac {13 a^{3} x}{8}+\frac {23 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {a^{3} \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a^{3} \sin \left (4 d x +4 c \right )}{32 d}+\frac {17 a^{3} \sin \left (3 d x +3 c \right )}{48 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{d}\) | \(89\) |
derivativedivides | \(\frac {\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(121\) |
default | \(\frac {\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(121\) |
parts | \(\frac {a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}+\frac {3 a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(129\) |
norman | \(\frac {\frac {13 a^{3} x}{8}+\frac {51 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {133 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {416 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {91 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {13 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {65 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {65 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {65 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {65 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {13 a^{3} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(202\) |
1/480*a^3*(780*d*x+1380*sin(d*x+c)+6*sin(5*d*x+5*c)+45*sin(4*d*x+4*c)+170* sin(3*d*x+3*c)+480*sin(2*d*x+2*c))/d
Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {195 \, a^{3} d x + {\left (24 \, a^{3} \cos \left (d x + c\right )^{4} + 90 \, a^{3} \cos \left (d x + c\right )^{3} + 152 \, a^{3} \cos \left (d x + c\right )^{2} + 195 \, a^{3} \cos \left (d x + c\right ) + 304 \, a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(195*a^3*d*x + (24*a^3*cos(d*x + c)^4 + 90*a^3*cos(d*x + c)^3 + 152* a^3*cos(d*x + c)^2 + 195*a^3*cos(d*x + c) + 304*a^3)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (99) = 198\).
Time = 0.27 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.59 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\begin {cases} \frac {9 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {9 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{3} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((9*a**3*x*sin(c + d*x)**4/8 + 9*a**3*x*sin(c + d*x)**2*cos(c + d *x)**2/4 + a**3*x*sin(c + d*x)**2/2 + 9*a**3*x*cos(c + d*x)**4/8 + a**3*x* cos(c + d*x)**2/2 + 8*a**3*sin(c + d*x)**5/(15*d) + 4*a**3*sin(c + d*x)**3 *cos(c + d*x)**2/(3*d) + 9*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*a** 3*sin(c + d*x)**3/d + a**3*sin(c + d*x)*cos(c + d*x)**4/d + 15*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*a**3*sin(c + d*x)*cos(c + d*x)**2/d + a** 3*sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(a*cos(c) + a)**3*cos(c)* *2, True))
Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{480 \, d} \]
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 - 4 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 )/d
Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {13}{8} \, a^{3} x + \frac {a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {17 \, a^{3} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {23 \, a^{3} \sin \left (d x + c\right )}{8 \, d} \]
13/8*a^3*x + 1/80*a^3*sin(5*d*x + 5*c)/d + 3/32*a^3*sin(4*d*x + 4*c)/d + 1 7/48*a^3*sin(3*d*x + 3*c)/d + a^3*sin(2*d*x + 2*c)/d + 23/8*a^3*sin(d*x + c)/d
Time = 17.43 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \cos ^2(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {13\,a^3\,x}{8}+\frac {\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {91\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {416\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\frac {133\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {51\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]