Integrand size = 21, antiderivative size = 81 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {4 (a+a \sin (c+d x))^{3+m}}{a^3 d (3+m)}-\frac {4 (a+a \sin (c+d x))^{4+m}}{a^4 d (4+m)}+\frac {(a+a \sin (c+d x))^{5+m}}{a^5 d (5+m)} \] Output:
4*(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 = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a (1+\sin (c+d x)))^{3+m} \left (\frac {4 a^2}{3+m}-\frac {4 a^2 (1+\sin (c+d x))}{4+m}+\frac {(a+a \sin (c+d x))^2}{5+m}\right )}{a^5 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^m,x]
Output:
((a*(1 + Sin[c + d*x]))^(3 + m)*((4*a^2)/(3 + m) - (4*a^2*(1 + Sin[c + d*x ]))/(4 + m) + (a + a*Sin[c + d*x])^2/(5 + m)))/(a^5*d)
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a)^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)^mdx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^{m+2}d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\int \left (4 a^2 (\sin (c+d x) a+a)^{m+2}-4 a (\sin (c+d x) a+a)^{m+3}+(\sin (c+d x) a+a)^{m+4}\right )d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {4 a^2 (a \sin (c+d x)+a)^{m+3}}{m+3}-\frac {4 a (a \sin (c+d x)+a)^{m+4}}{m+4}+\frac {(a \sin (c+d x)+a)^{m+5}}{m+5}}{a^5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^m,x]
Output:
((4*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_.) + (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_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 2.06 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.68
| method | result | size |
| parallelrisch | \(\frac {\left (\left (3 m^{2}+53 m +100\right ) \sin \left (3 d x +3 c \right )+\left (m^{2}+7 m +12\right ) \sin \left (5 d x +5 c \right )+\left (8 m^{2}+88 m \right ) \cos \left (2 d x +2 c \right )+\left (2 m^{2}+6 m \right ) \cos \left (4 d x +4 c \right )+\left (2 m^{2}+46 m +600\right ) \sin \left (d x +c \right )+6 m^{2}+82 m +512\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{m}}{16 d \left (4+m \right ) \left (3+m \right ) \left (5+m \right )}\) | \(136\) |
| derivativedivides | \(\frac {\sin \left (d x +c \right )^{5} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (5+m \right )}+\frac {\left (m^{2}+11 m +32\right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}+\frac {m \sin \left (d x +c \right )^{4} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{2}+9 m +20\right )}+\frac {\left (m^{2}+15 m +60\right ) \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}-\frac {2 \left (m^{2}+11 m +20\right ) \sin \left (d x +c \right )^{3} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}-\frac {2 m \left (7+m \right ) \sin \left (d x +c \right )^{2} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}\) | \(254\) |
| default | \(\frac {\sin \left (d x +c \right )^{5} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (5+m \right )}+\frac {\left (m^{2}+11 m +32\right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}+\frac {m \sin \left (d x +c \right )^{4} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{2}+9 m +20\right )}+\frac {\left (m^{2}+15 m +60\right ) \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}-\frac {2 \left (m^{2}+11 m +20\right ) \sin \left (d x +c \right )^{3} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}-\frac {2 m \left (7+m \right ) \sin \left (d x +c \right )^{2} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (d x +c \right )\right )\right )}}{d \left (m^{3}+12 m^{2}+47 m +60\right )}\) | \(254\) |
Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^m,x,method=_RETURNVERBOSE)
Output:
1/16*((3*m^2+53*m+100)*sin(3*d*x+3*c)+(m^2+7*m+12)*sin(5*d*x+5*c)+(8*m^2+8 8*m)*cos(2*d*x+2*c)+(2*m^2+6*m)*cos(4*d*x+4*c)+(2*m^2+46*m+600)*sin(d*x+c) +6*m^2+82*m+512)*(a*(1+sin(d*x+c)))^m/d/(4+m)/(3+m)/(5+m)
Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.26 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m^{2} + 3 \, m\right )} \cos \left (d x + c\right )^{4} + 8 \, m \cos \left (d x + c\right )^{2} + {\left ({\left (m^{2} + 7 \, m + 12\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (m + 2\right )} \cos \left (d x + c\right )^{2} + 32\right )} \sin \left (d x + c\right ) + 32\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} + 12 \, d m^{2} + 47 \, d m + 60 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
Output:
((m^2 + 3*m)*cos(d*x + c)^4 + 8*m*cos(d*x + c)^2 + ((m^2 + 7*m + 12)*cos(d *x + c)^4 + 8*(m + 2)*cos(d*x + c)^2 + 32)*sin(d*x + c) + 32)*(a*sin(d*x + c) + a)^m/(d*m^3 + 12*d*m^2 + 47*d*m + 60*d)
Leaf count of result is larger than twice the leaf count of optimal. 5534 vs. \(2 (68) = 136\).
Time = 38.10 (sec) , antiderivative size = 5534, normalized size of antiderivative = 68.32 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**m,x)
Output:
Piecewise((x*(a*sin(c) + a)**m*cos(c)**5, 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*l og(sin(c + d*x) + 1)*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) + 1 2*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) + 20*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) + 6*sin(c + d*x)**2*cos(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) + 56*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) + 8*sin(c + d*x)*cos(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) + 52*sin(c + d*x)/(12*a**5*d*...
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (81) = 162\).
Time = 0.04 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.28 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {\frac {{\left ({\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} a^{m} \sin \left (d x + c\right )^{5} + {\left (m^{4} + 6 \, m^{3} + 11 \, m^{2} + 6 \, m\right )} a^{m} \sin \left (d x + c\right )^{4} - 4 \, {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (d x + c\right )^{3} + 12 \, {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} - 24 \, a^{m} m \sin \left (d x + c\right ) + 24 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{m^{5} + 15 \, m^{4} + 85 \, m^{3} + 225 \, m^{2} + 274 \, m + 120} - \frac {2 \, {\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (d x + c\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} - 2 \, a^{m} m \sin \left (d x + c\right ) + 2 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} + \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m + 1}}{a {\left (m + 1\right )}}}{d} \] Input:
integrate(cos(d*x+c)^5*(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 + 11*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) - 2*((m^2 + 3*m + 2)*a^m*sin(d*x + c)^3 + (m^2 + m)*a^m*sin(d*x + c)^2 - 2*a^m*m*si n(d*x + c) + 2*a^m)*(sin(d*x + c) + 1)^m/(m^3 + 6*m^2 + 11*m + 6) + (a*sin (d*x + c) + a)^(m + 1)/(a*(m + 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (81) = 162\).
Time = 0.15 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.63 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{5} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} + 7 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{5} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 32 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + 36 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m + 12 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{5} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 60 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a + 80 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2}}{{\left (a^{4} m^{3} + 12 \, a^{4} m^{2} + 47 \, a^{4} m + 60 \, a^{4}\right )} a d} \] Input:
integrate(cos(d*x+c)^5*(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^2 - 4*(a*sin(d*x + c) + a )^4*(a*sin(d*x + c) + a)^m*a*m^2 + 4*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c ) + a)^m*a^2*m^2 + 7*(a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^m*m - 32* (a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a*m + 36*(a*sin(d*x + c) + a )^3*(a*sin(d*x + c) + a)^m*a^2*m + 12*(a*sin(d*x + c) + a)^5*(a*sin(d*x + c) + a)^m - 60*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*a + 80*(a*sin (d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a^2)/((a^4*m^3 + 12*a^4*m^2 + 47*a ^4*m + 60*a^4)*a*d)
Time = 1.22 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.41 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (82\,m+600\,\sin \left (c+d\,x\right )+100\,\sin \left (3\,c+3\,d\,x\right )+12\,\sin \left (5\,c+5\,d\,x\right )+46\,m\,\sin \left (c+d\,x\right )+88\,m\,\cos \left (2\,c+2\,d\,x\right )+6\,m\,\cos \left (4\,c+4\,d\,x\right )+53\,m\,\sin \left (3\,c+3\,d\,x\right )+7\,m\,\sin \left (5\,c+5\,d\,x\right )+2\,m^2\,\sin \left (c+d\,x\right )+6\,m^2+8\,m^2\,\cos \left (2\,c+2\,d\,x\right )+2\,m^2\,\cos \left (4\,c+4\,d\,x\right )+3\,m^2\,\sin \left (3\,c+3\,d\,x\right )+m^2\,\sin \left (5\,c+5\,d\,x\right )+512\right )}{16\,d\,\left (m^3+12\,m^2+47\,m+60\right )} \] Input:
int(cos(c + d*x)^5*(a + a*sin(c + d*x))^m,x)
Output:
((a*(sin(c + d*x) + 1))^m*(82*m + 600*sin(c + d*x) + 100*sin(3*c + 3*d*x) + 12*sin(5*c + 5*d*x) + 46*m*sin(c + d*x) + 88*m*cos(2*c + 2*d*x) + 6*m*co s(4*c + 4*d*x) + 53*m*sin(3*c + 3*d*x) + 7*m*sin(5*c + 5*d*x) + 2*m^2*sin( c + d*x) + 6*m^2 + 8*m^2*cos(2*c + 2*d*x) + 2*m^2*cos(4*c + 4*d*x) + 3*m^2 *sin(3*c + 3*d*x) + m^2*sin(5*c + 5*d*x) + 512))/(16*d*(47*m + 12*m^2 + m^ 3 + 60))
\[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\int \left (\sin \left (d x +c \right ) a +a \right )^{m} \cos \left (d x +c \right )^{5}d x \] Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^m,x)
Output:
int((sin(c + d*x)*a + a)**m*cos(c + d*x)**5,x)