Integrand size = 29, antiderivative size = 156 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{5005 d (a+a \sin (c+d x))^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1001 d (a+a \sin (c+d x))^{3/2}}-\frac {8 a^2 \cos ^5(c+d x)}{143 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d} \] Output:
-256/5005*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-64/1001*a^3*cos(d*x+c) ^5/d/(a+a*sin(d*x+c))^(3/2)-8/143*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2 )-6/143*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d-2/13*cos(d*x+c)^5*(a+a*sin (d*x+c))^(3/2)/d
Time = 4.64 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (19559-12600 \cos (2 (c+d x))+385 \cos (4 (c+d x))+28230 \sin (c+d x)-3290 \sin (3 (c+d x)))}{20020 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
Output:
-1/20020*(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d* x])]*(19559 - 12600*Cos[2*(c + d*x)] + 385*Cos[4*(c + d*x)] + 28230*Sin[c + d*x] - 3290*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.87 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3335, 3042, 3153, 3042, 3153, 3042, 3153, 3042, 3152}
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.
| \(\displaystyle \int \sin (c+d x) \cos ^4(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \cos (c+d x)^4 (a \sin (c+d x)+a)^{3/2}dx\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle \frac {3}{13} \int \cos ^4(c+d x) (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{13} \int \cos (c+d x)^4 (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\) |
Input:
Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(-2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d) + (3*((-2*a*Cos[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])/(11*d) + (12*a*((-2*a*Cos[c + d*x]^5)/(9* d*Sqrt[a + a*Sin[c + d*x]]) + (8*a*((-8*a^2*Cos[c + d*x]^5)/(35*d*(a + a*S in[c + d*x])^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2) )))/9))/11))/13
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 - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
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] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
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] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1, 0]
Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{3} \left (385 \sin \left (d x +c \right )^{4}+1645 \sin \left (d x +c \right )^{3}+2765 \sin \left (d x +c \right )^{2}+2295 \sin \left (d x +c \right )+918\right )}{5005 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(87\) |
Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE )
Output:
2/5005*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(385*sin(d*x+c)^4+1645*sin(d*x+ c)^3+2765*sin(d*x+c)^2+2295*sin(d*x+c)+918)/cos(d*x+c)/(a+a*sin(d*x+c))^(1 /2)/d
Time = 0.11 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.21 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (385 \, a \cos \left (d x + c\right )^{7} - 490 \, a \cos \left (d x + c\right )^{6} - 1015 \, a \cos \left (d x + c\right )^{5} + 20 \, a \cos \left (d x + c\right )^{4} - 32 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - {\left (385 \, a \cos \left (d x + c\right )^{6} + 875 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 160 \, a \cos \left (d x + c\right )^{3} - 192 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - 512 \, a\right )} \sin \left (d x + c\right ) - 512 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5005 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fri cas")
Output:
2/5005*(385*a*cos(d*x + c)^7 - 490*a*cos(d*x + c)^6 - 1015*a*cos(d*x + c)^ 5 + 20*a*cos(d*x + c)^4 - 32*a*cos(d*x + c)^3 + 64*a*cos(d*x + c)^2 - 256* a*cos(d*x + c) - (385*a*cos(d*x + c)^6 + 875*a*cos(d*x + c)^5 - 140*a*cos( d*x + c)^4 - 160*a*cos(d*x + c)^3 - 192*a*cos(d*x + c)^2 - 256*a*cos(d*x + c) - 512*a)*sin(d*x + c) - 512*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c ) + d*sin(d*x + c) + d)
Timed out. \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="max ima")
Output:
integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^4*sin(d*x + c), x)
Time = 0.15 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {64 \, \sqrt {2} {\left (770 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 3185 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 5005 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 3575 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1001 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{5005 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="gia c")
Output:
64/5005*sqrt(2)*(770*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1 /2*d*x + 1/2*c)^13 - 3185*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*p i + 1/2*d*x + 1/2*c)^11 + 5005*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(- 1/4*pi + 1/2*d*x + 1/2*c)^9 - 3575*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s in(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 1001*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c ))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d
Timed out. \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^(3/2),x)
Output:
int(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^(3/2), x)
\[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )d x \right ) \] Input:
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4*sin(c + d*x)**2,x) + int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**4*sin(c + d*x),x))