Integrand size = 29, antiderivative size = 103 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 a^2 x}{16}-\frac {a^2 \cos ^5(c+d x)}{10 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 a d} \]
3/16*a^2*x-1/10*a^2*cos(d*x+c)^5/d+3/16*a^2*cos(d*x+c)*sin(d*x+c)/d+1/8*a^ 2*cos(d*x+c)^3*sin(d*x+c)/d-1/6*cos(d*x+c)^3*(a+a*sin(d*x+c))^3/a/d
Time = 5.74 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (180 c+180 d x-240 \cos (c+d x)-40 \cos (3 (c+d x))+24 \cos (5 (c+d x))-15 \sin (2 (c+d x))-45 \sin (4 (c+d x))+5 \sin (6 (c+d x)))}{960 d} \]
(a^2*(180*c + 180*d*x - 240*Cos[c + d*x] - 40*Cos[3*(c + d*x)] + 24*Cos[5* (c + d*x)] - 15*Sin[2*(c + d*x)] - 45*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x) ]))/(960*d)
Time = 0.52 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3349, 3042, 3148, 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.
| \(\displaystyle \int \sin ^2(c+d x) \cos ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^2 (a \sin (c+d x)+a)^2dx\) |
| \(\Big \downarrow \) 3349 |
| \(\displaystyle \frac {1}{2} a \int \cos ^4(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \int \cos (c+d x)^4 (\sin (c+d x) a+a)dx-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{2} a \left (a \int \cos ^4(c+d x)dx-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{2} a \left (a \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (a \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{2} a \left (a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{2} a \left (a \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 a d}\) |
-1/6*(Cos[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(a*d) + (a*(-1/5*(a*Cos[c + d *x]^5)/d + a*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x ]*Sin[c + d*x])/(2*d)))/4)))/2
3.3.76.3.1 Defintions of rubi rules used
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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^( p + 1))*((a + b*Sin[e + f*x])^(m + 1)/(2*b*f*g*(m + 1))), x] + Simp[a/(2*g^ 2) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; F reeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[m - p, 0]
Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\frac {\left (36 d x +\sin \left (6 d x +6 c \right )-48 \cos \left (d x +c \right )-8 \cos \left (3 d x +3 c \right )+\frac {24 \cos \left (5 d x +5 c \right )}{5}-3 \sin \left (2 d x +2 c \right )-9 \sin \left (4 d x +4 c \right )-\frac {256}{5}\right ) a^{2}}{192 d}\) | \(76\) |
| risch | \(\frac {3 a^{2} x}{16}-\frac {a^{2} \cos \left (d x +c \right )}{4 d}+\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}+\frac {a^{2} \cos \left (5 d x +5 c \right )}{40 d}-\frac {3 a^{2} \sin \left (4 d x +4 c \right )}{64 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{24 d}-\frac {a^{2} \sin \left (2 d x +2 c \right )}{64 d}\) | \(107\) |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+2 a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\) | \(142\) |
| default | \(\frac {a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+2 a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\) | \(142\) |
| norman | \(\frac {\frac {3 a^{2} x}{16}-\frac {8 a^{2}}{15 d}-\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {13 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {25 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {25 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {13 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {3 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {9 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {45 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {15 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {45 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {9 a^{2} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a^{2} x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {16 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(303\) |
1/192*(36*d*x+sin(6*d*x+6*c)-48*cos(d*x+c)-8*cos(3*d*x+3*c)+24/5*cos(5*d*x +5*c)-3*sin(2*d*x+2*c)-9*sin(4*d*x+4*c)-256/5)*a^2/d
Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {96 \, a^{2} \cos \left (d x + c\right )^{5} - 160 \, a^{2} \cos \left (d x + c\right )^{3} + 45 \, a^{2} d x + 5 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 26 \, a^{2} \cos \left (d x + c\right )^{3} + 9 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]
1/240*(96*a^2*cos(d*x + c)^5 - 160*a^2*cos(d*x + c)^3 + 45*a^2*d*x + 5*(8* a^2*cos(d*x + c)^5 - 26*a^2*cos(d*x + c)^3 + 9*a^2*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (92) = 184\).
Time = 0.37 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.00 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {2 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {4 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a**2*x*sin(c + d*x)**6/16 + 3*a**2*x*sin(c + d*x)**4*cos(c + d* x)**2/16 + a**2*x*sin(c + d*x)**4/8 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x )**4/16 + a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**2*x*cos(c + d*x)** 6/16 + a**2*x*cos(c + d*x)**4/8 + a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 2*a**2*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) - a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 4*a**2*cos(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**2*c os(c)**2, True))
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {128 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 30 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{960 \, d} \]
1/960*(128*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2 - 5*(4*sin(2*d*x + 2* c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*a^2 + 30*(4*d*x + 4*c - sin(4*d *x + 4*c))*a^2)/d
Time = 0.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3}{16} \, a^{2} x + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{4 \, d} + \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {3 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]
3/16*a^2*x + 1/40*a^2*cos(5*d*x + 5*c)/d - 1/24*a^2*cos(3*d*x + 3*c)/d - 1 /4*a^2*cos(d*x + c)/d + 1/192*a^2*sin(6*d*x + 6*c)/d - 3/64*a^2*sin(4*d*x + 4*c)/d - 1/64*a^2*sin(2*d*x + 2*c)/d
Time = 13.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.50 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3\,a^2\,x}{16}-\frac {\frac {3\,a^2\,\left (c+d\,x\right )}{16}-\frac {13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}-\frac {25\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {25\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}-\frac {a^2\,\left (45\,c+45\,d\,x-128\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {9\,a^2\,\left (c+d\,x\right )}{8}-\frac {a^2\,\left (270\,c+270\,d\,x-768\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {15\,a^2\,\left (c+d\,x\right )}{4}-\frac {a^2\,\left (900\,c+900\,d\,x-1280\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {45\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (675\,c+675\,d\,x-1920\right )}{240}\right )+\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]
(3*a^2*x)/16 - ((3*a^2*(c + d*x))/16 - (13*a^2*tan(c/2 + (d*x)/2)^3)/24 - (25*a^2*tan(c/2 + (d*x)/2)^5)/4 + (25*a^2*tan(c/2 + (d*x)/2)^7)/4 + (13*a^ 2*tan(c/2 + (d*x)/2)^9)/24 - (3*a^2*tan(c/2 + (d*x)/2)^11)/8 - (a^2*(45*c + 45*d*x - 128))/240 + tan(c/2 + (d*x)/2)^2*((9*a^2*(c + d*x))/8 - (a^2*(2 70*c + 270*d*x - 768))/240) + tan(c/2 + (d*x)/2)^6*((15*a^2*(c + d*x))/4 - (a^2*(900*c + 900*d*x - 1280))/240) + tan(c/2 + (d*x)/2)^8*((45*a^2*(c + d*x))/16 - (a^2*(675*c + 675*d*x - 1920))/240) + (3*a^2*tan(c/2 + (d*x)/2) )/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)