Integrand size = 27, antiderivative size = 134 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)}-\frac {4 (a+a \sin (c+d x))^{2+m}}{a^2 d (2+m)}+\frac {6 (a+a \sin (c+d x))^{3+m}}{a^3 d (3+m)}-\frac {4 (a+a \sin (c+d x))^{4+m}}{a^4 d (4+m)}+\frac {(a+a \sin (c+d x))^{5+m}}{a^5 d (5+m)} \] Output:
(a+a*sin(d*x+c))^(1+m)/a/d/(1+m)-4*(a+a*sin(d*x+c))^(2+m)/a^2/d/(2+m)+6*(a +a*sin(d*x+c))^(3+m)/a^3/d/(3+m)-4*(a+a*sin(d*x+c))^(4+m)/a^4/d/(4+m)+(a+a *sin(d*x+c))^(5+m)/a^5/d/(5+m)
Time = 1.81 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a (1+\sin (c+d x)))^{1+m} \left (\frac {7}{1+m}-\frac {40 (1+\sin (c+d x))}{2+m}+\frac {84 (1+\sin (c+d x))^2}{3+m}-\frac {64 (1+\sin (c+d x))^3}{4+m}+\frac {16 (1+\sin (c+d x))^4}{5+m}+\frac {3 \left (6+m+m^2-2 \left (2+3 m+m^2\right ) \cos (2 (c+d x))-8 (1+m) \sin (c+d x)\right )}{(1+m) (2+m) (3+m)}\right )}{16 a d} \] Input:
Integrate[Cos[c + d*x]*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]
Output:
((a*(1 + Sin[c + d*x]))^(1 + m)*(7/(1 + m) - (40*(1 + Sin[c + d*x]))/(2 + m) + (84*(1 + Sin[c + d*x])^2)/(3 + m) - (64*(1 + Sin[c + d*x])^3)/(4 + m) + (16*(1 + Sin[c + d*x])^4)/(5 + m) + (3*(6 + m + m^2 - 2*(2 + 3*m + m^2) *Cos[2*(c + d*x)] - 8*(1 + m)*Sin[c + d*x]))/((1 + m)*(2 + m)*(3 + m))))/( 16*a*d)
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 53, 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.
| \(\displaystyle \int \sin ^4(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^4 \cos (c+d x) (a \sin (c+d x)+a)^mdx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \sin ^4(c+d x) (\sin (c+d x) a+a)^md(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int a^4 \sin ^4(c+d x) (\sin (c+d x) a+a)^md(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\int \left (a^4 (\sin (c+d x) a+a)^m-4 a^3 (\sin (c+d x) a+a)^{m+1}+6 a^2 (\sin (c+d x) a+a)^{m+2}-4 a (\sin (c+d x) a+a)^{m+3}+(\sin (c+d x) a+a)^{m+4}\right )d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^4 (a \sin (c+d x)+a)^{m+1}}{m+1}-\frac {4 a^3 (a \sin (c+d x)+a)^{m+2}}{m+2}+\frac {6 a^2 (a \sin (c+d x)+a)^{m+3}}{m+3}-\frac {4 a (a \sin (c+d x)+a)^{m+4}}{m+4}+\frac {(a \sin (c+d x)+a)^{m+5}}{m+5}}{a^5 d}\) |
Input:
Int[Cos[c + d*x]*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]
Output:
((a^4*(a + a*Sin[c + d*x])^(1 + m))/(1 + m) - (4*a^3*(a + a*Sin[c + d*x])^ (2 + m))/(2 + m) + (6*a^2*(a + a*Sin[c + d*x])^(3 + m))/(3 + m) - (4*a*(a + a*Sin[c + d*x])^(4 + m))/(4 + m) + (a + a*Sin[c + d*x])^(5 + m)/(5 + m)) /(a^5*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 2.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.84
| method | result | size |
| parallelrisch | \(\frac {16 \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{m} \left (\frac {3}{2}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}+3 \left (-1-m \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+3 \left (m^{2}+3 m +4\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-2 m^{3}-12 m^{2}-31 m -21\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (m^{4}+10 m^{3}+41 m^{2}+68 m +45\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-2 m^{3}-12 m^{2}-31 m -21\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+3 \left (m^{2}+3 m +4\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \left (-1-m \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}{d \left (5+m \right ) \left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(246\) |
| derivativedivides | \(\frac {\sin \left (d x +c \right )^{5} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (5+m \right )}+\frac {m \sin \left (d x +c \right )^{4} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{2}+9 m +20\right ) d}+\frac {24 \,{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{5}+15 m^{4}+85 m^{3}+225 m^{2}+274 m +120\right )}-\frac {4 m \sin \left (d x +c \right )^{3} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}-\frac {24 m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{5}+15 m^{4}+85 m^{3}+225 m^{2}+274 m +120\right )}+\frac {12 m \sin \left (d x +c \right )^{2} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{4}+14 m^{3}+71 m^{2}+154 m +120\right ) d}\) | \(256\) |
| default | \(\frac {\sin \left (d x +c \right )^{5} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (5+m \right )}+\frac {m \sin \left (d x +c \right )^{4} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{2}+9 m +20\right ) d}+\frac {24 \,{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{5}+15 m^{4}+85 m^{3}+225 m^{2}+274 m +120\right )}-\frac {4 m \sin \left (d x +c \right )^{3} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}-\frac {24 m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{5}+15 m^{4}+85 m^{3}+225 m^{2}+274 m +120\right )}+\frac {12 m \sin \left (d x +c \right )^{2} {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{4}+14 m^{3}+71 m^{2}+154 m +120\right ) d}\) | \(256\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^m,x,method=_RETURNVERBOSE)
Output:
16*(a*(1+sin(d*x+c)))^m*(3/2+3/2*tan(1/2*d*x+1/2*c)^8+3*(-1-m)*tan(1/2*d*x +1/2*c)^7+3*(m^2+3*m+4)*tan(1/2*d*x+1/2*c)^6+(-2*m^3-12*m^2-31*m-21)*tan(1 /2*d*x+1/2*c)^5+(m^4+10*m^3+41*m^2+68*m+45)*tan(1/2*d*x+1/2*c)^4+(-2*m^3-1 2*m^2-31*m-21)*tan(1/2*d*x+1/2*c)^3+3*(m^2+3*m+4)*tan(1/2*d*x+1/2*c)^2+3*( -1-m)*tan(1/2*d*x+1/2*c))*(tan(1/2*d*x+1/2*c)+1)^2/d/(5+m)/(4+m)/(3+m)/(2+ m)/(1+m)/(1+tan(1/2*d*x+1/2*c)^2)^5
Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.47 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m^{4} + 6 \, m^{3} + 11 \, m^{2} + 6 \, m\right )} \cos \left (d x + c\right )^{4} + m^{4} + 6 \, m^{3} - 2 \, {\left (m^{4} + 6 \, m^{3} + 17 \, m^{2} + 12 \, m\right )} \cos \left (d x + c\right )^{2} + 23 \, m^{2} + {\left ({\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} \cos \left (d x + c\right )^{4} + m^{4} + 6 \, m^{3} - 2 \, {\left (m^{4} + 8 \, m^{3} + 29 \, m^{2} + 46 \, m + 24\right )} \cos \left (d x + c\right )^{2} + 23 \, m^{2} + 18 \, m + 24\right )} \sin \left (d x + c\right ) + 18 \, m + 24\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{5} + 15 \, d m^{4} + 85 \, d m^{3} + 225 \, d m^{2} + 274 \, d m + 120 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="fricas" )
Output:
((m^4 + 6*m^3 + 11*m^2 + 6*m)*cos(d*x + c)^4 + m^4 + 6*m^3 - 2*(m^4 + 6*m^ 3 + 17*m^2 + 12*m)*cos(d*x + c)^2 + 23*m^2 + ((m^4 + 10*m^3 + 35*m^2 + 50* m + 24)*cos(d*x + c)^4 + m^4 + 6*m^3 - 2*(m^4 + 8*m^3 + 29*m^2 + 46*m + 24 )*cos(d*x + c)^2 + 23*m^2 + 18*m + 24)*sin(d*x + c) + 18*m + 24)*(a*sin(d* x + c) + a)^m/(d*m^5 + 15*d*m^4 + 85*d*m^3 + 225*d*m^2 + 274*d*m + 120*d)
Leaf count of result is larger than twice the leaf count of optimal. 2747 vs. \(2 (112) = 224\).
Time = 11.25 (sec) , antiderivative size = 2747, normalized size of antiderivative = 20.50 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**4*(a+a*sin(d*x+c))**m,x)
Output:
Piecewise((x*(a*sin(c) + a)**m*sin(c)**4*cos(c), Eq(d, 0)), (12*log(sin(c + d*x) + 1)*sin(c + d*x)**4/(12*a**5*d*sin(c + d*x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48*a**5*d*sin(c + d*x) + 12*a**5*d) + 48*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(12*a**5*d*sin(c + d*x)**4 + 4 8*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48*a**5*d*sin(c + d *x) + 12*a**5*d) + 72*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(12*a**5*d*sin (c + d*x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48* a**5*d*sin(c + d*x) + 12*a**5*d) + 48*log(sin(c + d*x) + 1)*sin(c + d*x)/( 12*a**5*d*sin(c + d*x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48*a**5*d*sin(c + d*x) + 12*a**5*d) + 12*log(sin(c + d*x) + 1)/( 12*a**5*d*sin(c + d*x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48*a**5*d*sin(c + d*x) + 12*a**5*d) + 48*sin(c + d*x)**3/(12*a** 5*d*sin(c + d*x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)** 2 + 48*a**5*d*sin(c + d*x) + 12*a**5*d) + 108*sin(c + d*x)**2/(12*a**5*d*s in(c + d*x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 4 8*a**5*d*sin(c + d*x) + 12*a**5*d) + 88*sin(c + d*x)/(12*a**5*d*sin(c + d* x)**4 + 48*a**5*d*sin(c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48*a**5*d* sin(c + d*x) + 12*a**5*d) + 25/(12*a**5*d*sin(c + d*x)**4 + 48*a**5*d*sin( c + d*x)**3 + 72*a**5*d*sin(c + d*x)**2 + 48*a**5*d*sin(c + d*x) + 12*a**5 *d), Eq(m, -5)), (-12*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(3*a**4*d*s...
Time = 0.04 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.19 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} a^{m} \sin \left (d x + c\right )^{5} + {\left (m^{4} + 6 \, m^{3} + 11 \, m^{2} + 6 \, m\right )} a^{m} \sin \left (d x + c\right )^{4} - 4 \, {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (d x + c\right )^{3} + 12 \, {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} - 24 \, a^{m} m \sin \left (d x + c\right ) + 24 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{5} + 15 \, m^{4} + 85 \, m^{3} + 225 \, m^{2} + 274 \, m + 120\right )} d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="maxima" )
Output:
((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*a^m*sin(d*x + c)^5 + (m^4 + 6*m^3 + 1 1*m^2 + 6*m)*a^m*sin(d*x + c)^4 - 4*(m^3 + 3*m^2 + 2*m)*a^m*sin(d*x + c)^3 + 12*(m^2 + m)*a^m*sin(d*x + c)^2 - 24*a^m*m*sin(d*x + c) + 24*a^m)*(sin( d*x + c) + 1)^m/((m^5 + 15*m^4 + 85*m^3 + 225*m^2 + 274*m + 120)*d)
Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (134) = 268\).
Time = 0.13 (sec) , antiderivative size = 791, normalized size of antiderivative = 5.90 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="giac")
Output:
((a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^m*m^4 - 4*(a*sin(d*x + c) + a )^4*(a*sin(d*x + c) + a)^m*a*m^4 + 6*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c ) + a)^m*a^2*m^4 - 4*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^3*m^4 + (a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^4*m^4 + 10*(a*sin(d*x + c ) + a)^5*(a*sin(d*x + c) + a)^m*m^3 - 44*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a*m^3 + 72*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2* m^3 - 52*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^3*m^3 + 14*(a*sin (d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^4*m^3 + 35*(a*sin(d*x + c) + a)^5* (a*sin(d*x + c) + a)^m*m^2 - 164*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a*m^2 + 294*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2*m^2 - 2 36*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^3*m^2 + 71*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^4*m^2 + 50*(a*sin(d*x + c) + a)^5*(a*sin (d*x + c) + a)^m*m - 244*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a*m + 468*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2*m - 428*(a*sin(d* x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^3*m + 154*(a*sin(d*x + c) + a)*(a*s in(d*x + c) + a)^m*a^4*m + 24*(a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^ m - 120*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a + 240*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2 - 240*(a*sin(d*x + c) + a)^2*(a*sin( d*x + c) + a)^m*a^3 + 120*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^4) /((a^4*m^5 + 15*a^4*m^4 + 85*a^4*m^3 + 225*a^4*m^2 + 274*a^4*m + 120*a^...
Time = 36.20 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.60 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (132\,m+240\,\sin \left (c+d\,x\right )-120\,\sin \left (3\,c+3\,d\,x\right )+24\,\sin \left (5\,c+5\,d\,x\right )+20\,m\,\sin \left (c+d\,x\right )-144\,m\,\cos \left (2\,c+2\,d\,x\right )+12\,m\,\cos \left (4\,c+4\,d\,x\right )-218\,m\,\sin \left (3\,c+3\,d\,x\right )+50\,m\,\sin \left (5\,c+5\,d\,x\right )+206\,m^2\,\sin \left (c+d\,x\right )+52\,m^3\,\sin \left (c+d\,x\right )+10\,m^4\,\sin \left (c+d\,x\right )+162\,m^2+36\,m^3+6\,m^4-184\,m^2\,\cos \left (2\,c+2\,d\,x\right )-48\,m^3\,\cos \left (2\,c+2\,d\,x\right )-8\,m^4\,\cos \left (2\,c+2\,d\,x\right )+22\,m^2\,\cos \left (4\,c+4\,d\,x\right )+12\,m^3\,\cos \left (4\,c+4\,d\,x\right )+2\,m^4\,\cos \left (4\,c+4\,d\,x\right )-127\,m^2\,\sin \left (3\,c+3\,d\,x\right )-34\,m^3\,\sin \left (3\,c+3\,d\,x\right )-5\,m^4\,\sin \left (3\,c+3\,d\,x\right )+35\,m^2\,\sin \left (5\,c+5\,d\,x\right )+10\,m^3\,\sin \left (5\,c+5\,d\,x\right )+m^4\,\sin \left (5\,c+5\,d\,x\right )+384\right )}{16\,d\,\left (m^5+15\,m^4+85\,m^3+225\,m^2+274\,m+120\right )} \] Input:
int(cos(c + d*x)*sin(c + d*x)^4*(a + a*sin(c + d*x))^m,x)
Output:
((a*(sin(c + d*x) + 1))^m*(132*m + 240*sin(c + d*x) - 120*sin(3*c + 3*d*x) + 24*sin(5*c + 5*d*x) + 20*m*sin(c + d*x) - 144*m*cos(2*c + 2*d*x) + 12*m *cos(4*c + 4*d*x) - 218*m*sin(3*c + 3*d*x) + 50*m*sin(5*c + 5*d*x) + 206*m ^2*sin(c + d*x) + 52*m^3*sin(c + d*x) + 10*m^4*sin(c + d*x) + 162*m^2 + 36 *m^3 + 6*m^4 - 184*m^2*cos(2*c + 2*d*x) - 48*m^3*cos(2*c + 2*d*x) - 8*m^4* cos(2*c + 2*d*x) + 22*m^2*cos(4*c + 4*d*x) + 12*m^3*cos(4*c + 4*d*x) + 2*m ^4*cos(4*c + 4*d*x) - 127*m^2*sin(3*c + 3*d*x) - 34*m^3*sin(3*c + 3*d*x) - 5*m^4*sin(3*c + 3*d*x) + 35*m^2*sin(5*c + 5*d*x) + 10*m^3*sin(5*c + 5*d*x ) + m^4*sin(5*c + 5*d*x) + 384))/(16*d*(274*m + 225*m^2 + 85*m^3 + 15*m^4 + m^5 + 120))
Time = 0.17 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.65 \[ \int \cos (c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {\left (\sin \left (d x +c \right ) a +a \right )^{m} \left (\sin \left (d x +c \right )^{5} m^{4}+10 \sin \left (d x +c \right )^{5} m^{3}+35 \sin \left (d x +c \right )^{5} m^{2}+50 \sin \left (d x +c \right )^{5} m +24 \sin \left (d x +c \right )^{5}+\sin \left (d x +c \right )^{4} m^{4}+6 \sin \left (d x +c \right )^{4} m^{3}+11 \sin \left (d x +c \right )^{4} m^{2}+6 \sin \left (d x +c \right )^{4} m -4 \sin \left (d x +c \right )^{3} m^{3}-12 \sin \left (d x +c \right )^{3} m^{2}-8 \sin \left (d x +c \right )^{3} m +12 \sin \left (d x +c \right )^{2} m^{2}+12 \sin \left (d x +c \right )^{2} m -24 \sin \left (d x +c \right ) m +24\right )}{d \left (m^{5}+15 m^{4}+85 m^{3}+225 m^{2}+274 m +120\right )} \] Input:
int(cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^m,x)
Output:
((sin(c + d*x)*a + a)**m*(sin(c + d*x)**5*m**4 + 10*sin(c + d*x)**5*m**3 + 35*sin(c + d*x)**5*m**2 + 50*sin(c + d*x)**5*m + 24*sin(c + d*x)**5 + sin (c + d*x)**4*m**4 + 6*sin(c + d*x)**4*m**3 + 11*sin(c + d*x)**4*m**2 + 6*s in(c + d*x)**4*m - 4*sin(c + d*x)**3*m**3 - 12*sin(c + d*x)**3*m**2 - 8*si n(c + d*x)**3*m + 12*sin(c + d*x)**2*m**2 + 12*sin(c + d*x)**2*m - 24*sin( c + d*x)*m + 24))/(d*(m**5 + 15*m**4 + 85*m**3 + 225*m**2 + 274*m + 120))