Integrand size = 27, antiderivative size = 129 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 x}{8}-\frac {a^2 \cos ^5(c+d x)}{15 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d} \] Output:
1/8*a^2*x-1/15*a^2*cos(d*x+c)^5/d+1/8*a^2*cos(d*x+c)*sin(d*x+c)/d+1/12*a^2 *cos(d*x+c)^3*sin(d*x+c)/d-1/7*cos(d*x+c)^5*(a+a*sin(d*x+c))^2/d-1/21*cos( d*x+c)^5*(a^2+a^2*sin(d*x+c))/d
Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.67 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (840 c+840 d x-1155 \cos (c+d x)-525 \cos (3 (c+d x))-63 \cos (5 (c+d x))+15 \cos (7 (c+d x))+210 \sin (2 (c+d x))-210 \sin (4 (c+d x))-70 \sin (6 (c+d x)))}{6720 d} \] Input:
Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(840*c + 840*d*x - 1155*Cos[c + d*x] - 525*Cos[3*(c + d*x)] - 63*Cos[ 5*(c + d*x)] + 15*Cos[7*(c + d*x)] + 210*Sin[2*(c + d*x)] - 210*Sin[4*(c + d*x)] - 70*Sin[6*(c + d*x)]))/(6720*d)
Time = 0.63 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3339, 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.
| \(\displaystyle \int \sin (c+d x) \cos ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \cos (c+d x)^4 (a \sin (c+d x)+a)^2dx\) |
| \(\Big \downarrow \) 3339 |
| \(\displaystyle \frac {2}{7} \int \cos ^4(c+d x) (\sin (c+d x) a+a)^2dx-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{7} \int \cos (c+d x)^4 (\sin (c+d x) a+a)^2dx-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3157 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \int \cos ^4(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \int \cos (c+d x)^4 (\sin (c+d x) a+a)dx-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \left (a \int \cos ^4(c+d x)dx-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \left (a \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \left (a \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \left (a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {2}{7} \left (\frac {7}{6} a \left (a \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {a \cos ^5(c+d x)}{5 d}\right )-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}\right )-\frac {\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d}\) |
Input:
Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
-1/7*(Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/d + (2*(-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))/7
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && NeQ[m + p + 1, 0]
Time = 85.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.69
| method | result | size |
| parallelrisch | \(\frac {a^{2} \left (840 d x +210 \sin \left (2 d x +2 c \right )-70 \sin \left (6 d x +6 c \right )-210 \sin \left (4 d x +4 c \right )-63 \cos \left (5 d x +5 c \right )-525 \cos \left (3 d x +3 c \right )-1155 \cos \left (d x +c \right )+15 \cos \left (7 d x +7 c \right )-1728\right )}{6720 d}\) | \(89\) |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5}}{d}\) | \(106\) |
| default | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5}}{d}\) | \(106\) |
| risch | \(\frac {a^{2} x}{8}-\frac {11 a^{2} \cos \left (d x +c \right )}{64 d}+\frac {a^{2} \cos \left (7 d x +7 c \right )}{448 d}-\frac {a^{2} \sin \left (6 d x +6 c \right )}{96 d}-\frac {3 a^{2} \cos \left (5 d x +5 c \right )}{320 d}-\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}-\frac {5 a^{2} \cos \left (3 d x +3 c \right )}{64 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{32 d}\) | \(124\) |
| norman | \(\frac {\frac {a^{2} x}{8}-\frac {18 a^{2}}{35 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {31 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {31 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {7 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {21 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {35 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}+\frac {35 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {21 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {7 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {14 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 d}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}-\frac {8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {16 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(377\) |
| orering | \(\text {Expression too large to display}\) | \(3490\) |
Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/6720*a^2*(840*d*x+210*sin(2*d*x+2*c)-70*sin(6*d*x+6*c)-210*sin(4*d*x+4*c )-63*cos(5*d*x+5*c)-525*cos(3*d*x+3*c)-1155*cos(d*x+c)+15*cos(7*d*x+7*c)-1 728)/d
Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.66 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {120 \, a^{2} \cos \left (d x + c\right )^{7} - 336 \, a^{2} \cos \left (d x + c\right )^{5} + 105 \, a^{2} d x - 35 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/840*(120*a^2*cos(d*x + c)^7 - 336*a^2*cos(d*x + c)^5 + 105*a^2*d*x - 35* (8*a^2*cos(d*x + c)^5 - 2*a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/d
Time = 0.49 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.73 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin {\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)*(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((a**2*x*sin(c + d*x)**6/8 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x )**2/8 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/8 + a**2*x*cos(c + d*x)* *6/8 + a**2*sin(c + d*x)**5*cos(c + d*x)/(8*d) + a**2*sin(c + d*x)**3*cos( c + d*x)**3/(3*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - a**2*sin( c + d*x)*cos(c + d*x)**5/(8*d) - 2*a**2*cos(c + d*x)**7/(35*d) - a**2*cos( c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)*cos(c)**4, True) )
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.64 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {672 \, a^{2} \cos \left (d x + c\right )^{5} - 96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{3360 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-1/3360*(672*a^2*cos(d*x + c)^5 - 96*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5) *a^2 - 35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2) /d
Time = 0.17 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {1}{8} \, a^{2} x + \frac {a^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {3 \, a^{2} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {11 \, a^{2} \cos \left (d x + c\right )}{64 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/8*a^2*x + 1/448*a^2*cos(7*d*x + 7*c)/d - 3/320*a^2*cos(5*d*x + 5*c)/d - 5/64*a^2*cos(3*d*x + 3*c)/d - 11/64*a^2*cos(d*x + c)/d - 1/96*a^2*sin(6*d* x + 6*c)/d - 1/32*a^2*sin(4*d*x + 4*c)/d + 1/32*a^2*sin(2*d*x + 2*c)/d
Time = 19.12 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.01 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,x}{8}-\frac {\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}-\frac {18}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (7\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {7\,c}{8}+\frac {7\,d\,x}{8}-2\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (7\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {7\,c}{8}+\frac {7\,d\,x}{8}-\frac {8}{5}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (21\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {21\,c}{8}+\frac {21\,d\,x}{8}-8\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (21\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {21\,c}{8}+\frac {21\,d\,x}{8}-\frac {14}{5}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (35\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {35\,c}{8}+\frac {35\,d\,x}{8}-2\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (35\,a^2\,\left (\frac {c}{8}+\frac {d\,x}{8}\right )-a^2\,\left (\frac {35\,c}{8}+\frac {35\,d\,x}{8}-16\right )\right )+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \] Input:
int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^2,x)
Output:
(a^2*x)/8 - ((31*a^2*tan(c/2 + (d*x)/2)^5)/12 - (11*a^2*tan(c/2 + (d*x)/2) ^3)/3 - (31*a^2*tan(c/2 + (d*x)/2)^9)/12 + (11*a^2*tan(c/2 + (d*x)/2)^11)/ 3 - (a^2*tan(c/2 + (d*x)/2)^13)/4 + a^2*(c/8 + (d*x)/8) - a^2*(c/8 + (d*x) /8 - 18/35) + tan(c/2 + (d*x)/2)^12*(7*a^2*(c/8 + (d*x)/8) - a^2*((7*c)/8 + (7*d*x)/8 - 2)) + tan(c/2 + (d*x)/2)^2*(7*a^2*(c/8 + (d*x)/8) - a^2*((7* c)/8 + (7*d*x)/8 - 8/5)) + tan(c/2 + (d*x)/2)^10*(21*a^2*(c/8 + (d*x)/8) - a^2*((21*c)/8 + (21*d*x)/8 - 8)) + tan(c/2 + (d*x)/2)^4*(21*a^2*(c/8 + (d *x)/8) - a^2*((21*c)/8 + (21*d*x)/8 - 14/5)) + tan(c/2 + (d*x)/2)^8*(35*a^ 2*(c/8 + (d*x)/8) - a^2*((35*c)/8 + (35*d*x)/8 - 2)) + tan(c/2 + (d*x)/2)^ 6*(35*a^2*(c/8 + (d*x)/8) - a^2*((35*c)/8 + (35*d*x)/8 - 16)) + (a^2*tan(c /2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)
Time = 0.16 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+490 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+312 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right )-216 \cos \left (d x +c \right )+105 d x +216\right )}{840 d} \] Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 120*cos(c + d*x)*sin(c + d*x)**6 - 280*cos(c + d*x)*sin(c + d*x) **5 + 24*cos(c + d*x)*sin(c + d*x)**4 + 490*cos(c + d*x)*sin(c + d*x)**3 + 312*cos(c + d*x)*sin(c + d*x)**2 - 105*cos(c + d*x)*sin(c + d*x) - 216*co s(c + d*x) + 105*d*x + 216))/(840*d)