Integrand size = 23, antiderivative size = 127 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{1155 d (a+a \sin (c+d x))^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{231 d (a+a \sin (c+d x))^{3/2}}-\frac {8 a^2 \cos ^5(c+d x)}{33 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d} \]
-256/1155*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-64/231*a^3*cos(d*x+c)^ 5/d/(a+a*sin(d*x+c))^(3/2)-8/33*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)- 2/11*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.54 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \cos ^5(c+d x) (a (1+\sin (c+d x)))^{3/2} \left (533+755 \sin (c+d x)+455 \sin ^2(c+d x)+105 \sin ^3(c+d x)\right )}{1155 d (1+\sin (c+d x))^4} \]
(-2*Cos[c + d*x]^5*(a*(1 + Sin[c + d*x]))^(3/2)*(533 + 755*Sin[c + d*x] + 455*Sin[c + d*x]^2 + 105*Sin[c + d*x]^3))/(1155*d*(1 + Sin[c + d*x])^4)
Time = 0.62 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {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.
\(\displaystyle \int \cos ^4(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^4 (a \sin (c+d x)+a)^{3/2}dx\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {12}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {12}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
(-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*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
3.2.17.3.1 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]
Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{3} \left (105 \left (\sin ^{3}\left (d x +c \right )\right )+455 \left (\sin ^{2}\left (d x +c \right )\right )+755 \sin \left (d x +c \right )+533\right )}{1155 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(77\) |
2/1155*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(105*sin(d*x+c)^3+455*sin(d*x+c )^2+755*sin(d*x+c)+533)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.31 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (105 \, a \cos \left (d x + c\right )^{6} + 245 \, a \cos \left (d x + c\right )^{5} - 20 \, a \cos \left (d x + c\right )^{4} + 32 \, a \cos \left (d x + c\right )^{3} - 64 \, a \cos \left (d x + c\right )^{2} + 256 \, a \cos \left (d x + c\right ) + {\left (105 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 160 \, a \cos \left (d x + c\right )^{3} - 192 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - 512 \, a\right )} \sin \left (d x + c\right ) + 512 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1155 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]
-2/1155*(105*a*cos(d*x + c)^6 + 245*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) + (105*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)
\[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \cos ^{4}{\left (c + d x \right )}\, dx \]
\[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 \, \sqrt {2} {\left (105 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 385 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 495 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 231 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{1155 \, d} \]
-64/1155*sqrt(2)*(105*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 385*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)^9 + 495*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/ 4*pi + 1/2*d*x + 1/2*c)^7 - 231*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. \[ \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]