Integrand size = 31, antiderivative size = 124 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {152 a^2 \cos ^5(c+d x)}{3465 d (a+a \sin (c+d x))^{5/2}}-\frac {38 a \cos ^5(c+d x)}{693 d (a+a \sin (c+d x))^{3/2}}+\frac {20 \cos ^5(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a d} \]
-152/3465*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-38/693*a*cos(d*x+c)^5/ d/(a+a*sin(d*x+c))^(3/2)+20/99*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)-2/11* cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/a/d
Time = 2.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (5773 \cos \left (\frac {1}{2} (c+d x)\right )-3495 \cos \left (\frac {3}{2} (c+d x)\right )-1505 \cos \left (\frac {5}{2} (c+d x)\right )+315 \cos \left (\frac {7}{2} (c+d x)\right )+5773 \sin \left (\frac {1}{2} (c+d x)\right )+3495 \sin \left (\frac {3}{2} (c+d x)\right )-1505 \sin \left (\frac {5}{2} (c+d x)\right )-315 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{13860 d \sqrt {a (1+\sin (c+d x))}} \]
-1/13860*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(5773*Cos[(c + d*x)/2] - 3495*Cos[(3*(c + d*x))/2] - 1505*Cos[(5*(c + d*x))/2] + 315*Cos[(7*(c + d *x))/2] + 5773*Sin[(c + d*x)/2] + 3495*Sin[(3*(c + d*x))/2] - 1505*Sin[(5* (c + d*x))/2] - 315*Sin[(7*(c + d*x))/2]))/(d*Sqrt[a*(1 + Sin[c + d*x])])
Time = 0.87 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.39, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3356, 27, 3042, 3335, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^4(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^4}{\sqrt {a \sin (c+d x)+a}}dx\) |
| \(\Big \downarrow \) 3356 |
| \(\displaystyle \frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}-\frac {\int -\frac {1}{2} \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a} (8 \sin (c+d x) a+a)dx}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a} (8 \sin (c+d x) a+a)dx}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a} (8 \sin (c+d x) a+a)dx}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle \frac {\frac {19}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {19}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {19}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {\frac {19}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {16 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{8 a^2}+\frac {\cos ^5(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
Cos[c + d*x]^5/(4*d*Sqrt[a + a*Sin[c + d*x]]) + ((-16*a*Cos[c + d*x]^5*Sqr t[a + a*Sin[c + d*x]])/(11*d) + (19*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*Sin[c + d*x ])^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))))/9))/11 )/(8*a^2)
3.5.64.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((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/(a*f*g*(2*m + p + 1))), x] - Simp[1/(a^2*(2* m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*m - b* (2*m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[ a^2 - b^2, 0] && LeQ[m, -2^(-1)] && NeQ[2*m + p + 1, 0]
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.60
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (315 \left (\sin ^{3}\left (d x +c \right )\right )+595 \left (\sin ^{2}\left (d x +c \right )\right )+340 \sin \left (d x +c \right )+136\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(74\) |
2/3465*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(315*sin(d*x+c)^3+595*sin(d*x+c)^2+ 340*sin(d*x+c)+136)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} - 35 \, \cos \left (d x + c\right )^{5} - 445 \, \cos \left (d x + c\right )^{4} + 19 \, \cos \left (d x + c\right )^{3} - 38 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} + 350 \, \cos \left (d x + c\right )^{4} - 95 \, \cos \left (d x + c\right )^{3} - 114 \, \cos \left (d x + c\right )^{2} - 152 \, \cos \left (d x + c\right ) - 304\right )} \sin \left (d x + c\right ) + 152 \, \cos \left (d x + c\right ) + 304\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]
-2/3465*(315*cos(d*x + c)^6 - 35*cos(d*x + c)^5 - 445*cos(d*x + c)^4 + 19* cos(d*x + c)^3 - 38*cos(d*x + c)^2 + (315*cos(d*x + c)^5 + 350*cos(d*x + c )^4 - 95*cos(d*x + c)^3 - 114*cos(d*x + c)^2 - 152*cos(d*x + c) - 304)*sin (d*x + c) + 152*cos(d*x + c) + 304)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)
\[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {16 \, \sqrt {2} {\left (1260 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 3080 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2475 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 693 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{3465 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]
-16/3465*sqrt(2)*(1260*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 3080*sq rt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 + 2475*sqrt(a)*sin(-1/4*pi + 1/2*d* x + 1/2*c)^7 - 693*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)/(a*d*sgn(cos( -1/4*pi + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]