Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {8 a^2 \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac {2 a \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}} \]
8/315*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)+2/63*a*cos(d*x+c)^5/d/(a+a *sin(d*x+c))^(3/2)-2/9*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)
Time = 1.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (87-35 \cos (2 (c+d x))+130 \sin (c+d x))}{315 d \sqrt {a (1+\sin (c+d x))}} \]
-1/315*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(87 - 35*Cos[2*(c + d*x)] + 130*Sin[c + d*x]))/(d*Sqrt[a*(1 + Sin[c + d*x])])
Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {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.
\(\displaystyle \int \frac {\sin (c+d x) \cos ^4(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^4}{\sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3335 |
\(\displaystyle -\frac {1}{9} \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{9} \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {1}{9} \left (\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}-\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx\right )-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{9} \left (\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}-\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx\right )-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {1}{9} \left (\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}+\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\) |
(-2*Cos[c + d*x]^5)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + ((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*Si n[c + d*x])^(3/2)))/9
3.5.65.3.1 Defintions of rubi rules used
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 = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (35 \left (\sin ^{2}\left (d x +c \right )\right )+65 \sin \left (d x +c \right )+26\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(64\) |
2/315*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+65*sin(d*x+c)+26)/c os(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) - 16\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]
-2/315*(35*cos(d*x + c)^5 + 40*cos(d*x + c)^4 - cos(d*x + c)^3 + 2*cos(d*x + c)^2 - (35*cos(d*x + c)^4 - 5*cos(d*x + c)^3 - 6*cos(d*x + c)^2 - 8*cos (d*x + c) - 16)*sin(d*x + c) - 8*cos(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {16 \, \sqrt {2} {\left (70 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 135 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{315 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]
16/315*sqrt(2)*(70*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 135*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*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]