Integrand size = 31, antiderivative size = 156 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {1472 a^3 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac {368 a^2 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {46 a \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}+\frac {20 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d} \]
-1472/45045*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-368/9009*a^2*cos(d*x +c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/13*cos(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/a/ d-46/1287*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)+20/143*cos(d*x+c)^5*(a+a *sin(d*x+c))^(1/2)/d
Time = 2.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (81183-62440 \cos (2 (c+d x))+3465 \cos (4 (c+d x))+119780 \sin (c+d x)-21420 \sin (3 (c+d x)))}{180180 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
-1/180180*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x ])]*(81183 - 62440*Cos[2*(c + d*x)] + 3465*Cos[4*(c + d*x)] + 119780*Sin[c + d*x] - 21420*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2] ))
Time = 0.90 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3357, 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.
| \(\displaystyle \int \sin ^2(c+d x) \cos ^4(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^4 \sqrt {a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3357 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \cos ^4(c+d x) (3 a-10 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^4(c+d x) (3 a-10 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos (c+d x)^4 (3 a-10 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle \frac {\frac {23}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {23}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {23}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {23}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {23}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {23}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {\frac {23}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {20 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{13 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}\) |
(-2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*a*d) + ((20*a*Cos[c + d *x]^5*Sqrt[a + a*Sin[c + d*x]])/(11*d) + (23*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*Si n[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*a)
3.5.43.3.1 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[(-(g*Cos[e + f*x])^( p + 1))*((a + b*Sin[e + f*x])^(m + 1)/(b*f*g*(m + p + 2))), x] + Simp[1/(b* (m + p + 2)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a *(p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^ 2 - b^2, 0] && NeQ[m + p + 2, 0]
Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.54
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (3465 \left (\sin ^{4}\left (d x +c \right )\right )+10710 \left (\sin ^{3}\left (d x +c \right )\right )+12145 \left (\sin ^{2}\left (d x +c \right )\right )+6940 \sin \left (d x +c \right )+2776\right )}{45045 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(85\) |
2/45045*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^3*(3465*sin(d*x+c)^4+10710*sin(d*x +c)^3+12145*sin(d*x+c)^2+6940*sin(d*x+c)+2776)/cos(d*x+c)/(a+a*sin(d*x+c)) ^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (3465 \, \cos \left (d x + c\right )^{7} - 315 \, \cos \left (d x + c\right )^{6} - 4585 \, \cos \left (d x + c\right )^{5} + 115 \, \cos \left (d x + c\right )^{4} - 184 \, \cos \left (d x + c\right )^{3} + 368 \, \cos \left (d x + c\right )^{2} - {\left (3465 \, \cos \left (d x + c\right )^{6} + 3780 \, \cos \left (d x + c\right )^{5} - 805 \, \cos \left (d x + c\right )^{4} - 920 \, \cos \left (d x + c\right )^{3} - 1104 \, \cos \left (d x + c\right )^{2} - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sin \left (d x + c\right ) - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45045 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]
2/45045*(3465*cos(d*x + c)^7 - 315*cos(d*x + c)^6 - 4585*cos(d*x + c)^5 + 115*cos(d*x + c)^4 - 184*cos(d*x + c)^3 + 368*cos(d*x + c)^2 - (3465*cos(d *x + c)^6 + 3780*cos(d*x + c)^5 - 805*cos(d*x + c)^4 - 920*cos(d*x + c)^3 - 1104*cos(d*x + c)^2 - 1472*cos(d*x + c) - 2944)*sin(d*x + c) - 1472*cos( d*x + c) - 2944)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}\, dx \]
\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {32 \, \sqrt {2} {\left (13860 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 49140 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 65065 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 38610 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9009 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{45045 \, d} \]
32/45045*sqrt(2)*(13860*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^13 - 49140*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 + 65065*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1 /4*pi + 1/2*d*x + 1/2*c)^9 - 38610*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin (-1/4*pi + 1/2*d*x + 1/2*c)^7 + 9009*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s in(-1/4*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d
Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]