Integrand size = 31, antiderivative size = 92 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {46 a \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac {20 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 a d \sqrt {a+a \sin (c+d x)}} \]
-46/315*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)+20/63*cos(d*x+c)^5/d/(a+a* sin(d*x+c))^(3/2)-2/9*cos(d*x+c)^5/a/d/(a+a*sin(d*x+c))^(1/2)
Time = 4.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (51-35 \cos (2 (c+d x))+40 \sin (c+d x))}{315 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
-1/315*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])] *(51 - 35*Cos[2*(c + d*x)] + 40*Sin[c + d*x]))/(a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.75 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.48, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3356, 27, 3042, 3335, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^4(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^4}{(a \sin (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3356 |
| \(\displaystyle \frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int -\frac {\cos ^4(c+d x) (4 \sin (c+d x) a+3 a)}{2 \sqrt {\sin (c+d x) a+a}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\cos ^4(c+d x) (4 \sin (c+d x) a+3 a)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x)^4 (4 \sin (c+d x) a+3 a)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle \frac {\frac {23}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {8 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {23}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {8 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {23}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {8 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {23}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {8 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {\frac {23}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {8 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cos ^5(c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
Cos[c + d*x]^5/(2*d*(a + a*Sin[c + d*x])^(3/2)) + ((-8*a*Cos[c + d*x]^5)/( 9*d*Sqrt[a + a*Sin[c + d*x]]) + (23*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)/(4*a^2)
3.5.73.3.1 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 = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.73
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (35 \left (\sin ^{2}\left (d x +c \right )\right )+20 \sin \left (d x +c \right )+8\right )}{315 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(67\) |
2/315/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+20*sin(d*x+c)+8)/ cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} + 85 \, \cos \left (d x + c\right )^{4} - 73 \, \cos \left (d x + c\right )^{3} - 169 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{3} - 123 \, \cos \left (d x + c\right )^{2} + 46 \, \cos \left (d x + c\right ) + 92\right )} \sin \left (d x + c\right ) + 46 \, \cos \left (d x + c\right ) + 92\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]
-2/315*(35*cos(d*x + c)^5 + 85*cos(d*x + c)^4 - 73*cos(d*x + c)^3 - 169*co s(d*x + c)^2 - (35*cos(d*x + c)^4 - 50*cos(d*x + c)^3 - 123*cos(d*x + c)^2 + 46*cos(d*x + c) + 92)*sin(d*x + c) + 46*cos(d*x + c) + 92)*sqrt(a*sin(d *x + c) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {8 \, \sqrt {2} {\left (140 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{315 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]
8/315*sqrt(2)*(140*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 180*sqrt(a)* sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 63*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2* c)^5)/(a^2*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]