Integrand size = 21, antiderivative size = 102 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {7 a^2 x}{16}-\frac {7 a^2 \cos ^5(c+d x)}{30 d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d} \] Output:
7/16*a^2*x-7/30*a^2*cos(d*x+c)^5/d+7/16*a^2*cos(d*x+c)*sin(d*x+c)/d+7/24*a ^2*cos(d*x+c)^3*sin(d*x+c)/d-1/6*cos(d*x+c)^5*(a^2+a^2*sin(d*x+c))/d
Time = 0.59 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.48 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \cos ^5(c+d x) \left (-210 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (-96+231 \sin (c+d x)+57 \sin ^2(c+d x)-182 \sin ^3(c+d x)-106 \sin ^4(c+d x)+56 \sin ^5(c+d x)+40 \sin ^6(c+d x)\right )\right )}{240 d (-1+\sin (c+d x))^3 (1+\sin (c+d x))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
Output:
-1/240*(a^2*Cos[c + d*x]^5*(-210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sq rt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]]*(-96 + 231*Sin[c + d*x] + 57 *Sin[c + d*x]^2 - 182*Sin[c + d*x]^3 - 106*Sin[c + d*x]^4 + 56*Sin[c + d*x ]^5 + 40*Sin[c + d*x]^6)))/(d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/ 2))
Time = 0.47 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3157, 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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^4 (a \sin (c+d x)+a)^2dx\) |
| \(\Big \downarrow \) 3157 |
| \(\displaystyle \frac {7}{6} a \int \cos ^4(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \int \cos (c+d x)^4 (\sin (c+d x) a+a)dx-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {7}{6} a \left (a \int \cos ^4(c+d x)dx-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {7}{6} a \left (a \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \left (a \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {7}{6} a \left (a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {7}{6} a \left (a \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
Output:
-1/6*(Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/d + (7*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)))/6
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_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers Q[2*m, 2*p]
Time = 13.48 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75
| method | result | size |
| parallelrisch | \(-\frac {\left (-84 d x +\sin \left (6 d x +6 c \right )+48 \cos \left (d x +c \right )+24 \cos \left (3 d x +3 c \right )+\frac {24 \cos \left (5 d x +5 c \right )}{5}-51 \sin \left (2 d x +2 c \right )-3 \sin \left (4 d x +4 c \right )+\frac {384}{5}\right ) a^{2}}{192 d}\) | \(76\) |
| risch | \(\frac {7 a^{2} x}{16}-\frac {a^{2} \cos \left (d x +c \right )}{4 d}-\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{64 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{8 d}+\frac {17 a^{2} \sin \left (2 d x +2 c \right )}{64 d}\) | \(107\) |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a^{2} \cos \left (d x +c \right )^{5}}{5}+a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(109\) |
| default | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a^{2} \cos \left (d x +c \right )^{5}}{5}+a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(109\) |
| norman | \(\frac {\frac {7 a^{2} x}{16}-\frac {4 a^{2}}{5 d}+\frac {9 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {89 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {89 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {9 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {21 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {105 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {35 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {105 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {21 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {7 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\) | \(341\) |
| orering | \(\text {Expression too large to display}\) | \(2404\) |
Input:
int(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/192*(-84*d*x+sin(6*d*x+6*c)+48*cos(d*x+c)+24*cos(3*d*x+3*c)+24/5*cos(5* d*x+5*c)-51*sin(2*d*x+2*c)-3*sin(4*d*x+4*c)+384/5)*a^2/d
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {96 \, a^{2} \cos \left (d x + c\right )^{5} - 105 \, a^{2} d x + 5 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 14 \, a^{2} \cos \left (d x + c\right )^{3} - 21 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/240*(96*a^2*cos(d*x + c)^5 - 105*a^2*d*x + 5*(8*a^2*cos(d*x + c)^5 - 14 *a^2*cos(d*x + c)^3 - 21*a^2*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (95) = 190\).
Time = 0.39 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.81 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))**2,x)
Output:
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 + 3*a**2*x*sin(c + d*x)**4/8 + 3*a**2*x*sin(c + d*x)**2*cos(c + d *x)**4/16 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**2*x*cos(c + d* x)**6/16 + 3*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) + 3*a**2*sin(c + d*x)* *3*cos(c + d*x)/(8*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 5*a**2* sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*a**2*cos(c + d*x)**5/(5*d), Ne(d, 0 )), (x*(a*sin(c) + a)**2*cos(c)**4, True))
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {384 \, a^{2} \cos \left (d x + c\right )^{5} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{960 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/960*(384*a^2*cos(d*x + c)^5 - 5*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2 - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2* d*x + 2*c))*a^2)/d
Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {7}{16} \, a^{2} x - \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{4 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {17 \, a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
7/16*a^2*x - 1/40*a^2*cos(5*d*x + 5*c)/d - 1/8*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 + 1/64*a^2*sin(4*d*x + 4*c)/d + 17/64*a^2*sin(2*d*x + 2*c)/d
Time = 28.11 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.42 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {7\,a^2\,x}{16}-\frac {\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {a^2\,\left (105\,c+105\,d\,x\right )}{240}-\frac {a^2\,\left (105\,c+105\,d\,x-192\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^2\,\left (630\,c+630\,d\,x-192\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^2\,\left (630\,c+630\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^2\,\left (1575\,c+1575\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^2\,\left (1575\,c+1575\,d\,x-1920\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{12}-\frac {a^2\,\left (2100\,c+2100\,d\,x-1920\right )}{240}\right )-\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \] Input:
int(cos(c + d*x)^4*(a + a*sin(c + d*x))^2,x)
Output:
(7*a^2*x)/16 - ((11*a^2*tan(c/2 + (d*x)/2)^5)/4 - (89*a^2*tan(c/2 + (d*x)/ 2)^3)/24 - (11*a^2*tan(c/2 + (d*x)/2)^7)/4 + (89*a^2*tan(c/2 + (d*x)/2)^9) /24 + (9*a^2*tan(c/2 + (d*x)/2)^11)/8 + (a^2*(105*c + 105*d*x))/240 - (a^2 *(105*c + 105*d*x - 192))/240 + tan(c/2 + (d*x)/2)^2*((a^2*(105*c + 105*d* x))/40 - (a^2*(630*c + 630*d*x - 192))/240) + tan(c/2 + (d*x)/2)^10*((a^2* (105*c + 105*d*x))/40 - (a^2*(630*c + 630*d*x - 960))/240) + tan(c/2 + (d* x)/2)^8*((a^2*(105*c + 105*d*x))/16 - (a^2*(1575*c + 1575*d*x - 960))/240) + tan(c/2 + (d*x)/2)^4*((a^2*(105*c + 105*d*x))/16 - (a^2*(1575*c + 1575* d*x - 1920))/240) + tan(c/2 + (d*x)/2)^6*((a^2*(105*c + 105*d*x))/12 - (a^ 2*(2100*c + 2100*d*x - 1920))/240) - (9*a^2*tan(c/2 + (d*x)/2))/8)/(d*(tan (c/2 + (d*x)/2)^2 + 1)^6)
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.98 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+135 \cos \left (d x +c \right ) \sin \left (d x +c \right )-96 \cos \left (d x +c \right )+105 d x +96\right )}{240 d} \] Input:
int(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 40*cos(c + d*x)*sin(c + d*x)**5 - 96*cos(c + d*x)*sin(c + d*x)** 4 + 10*cos(c + d*x)*sin(c + d*x)**3 + 192*cos(c + d*x)*sin(c + d*x)**2 + 1 35*cos(c + d*x)*sin(c + d*x) - 96*cos(c + d*x) + 105*d*x + 96))/(240*d)