Integrand size = 29, antiderivative size = 124 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 a^3 \cos ^3(c+d x)}{315 d (a+a \sin (c+d x))^{3/2}}-\frac {16 a^2 \cos ^3(c+d x)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d} \] Output:
-64/315*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-16/105*a^2*cos(d*x+c)^3/ d/(a+a*sin(d*x+c))^(1/2)-2/21*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d-2/9* cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d
Time = 0.79 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))} (-664+240 \cos (2 (c+d x))-741 \sin (c+d x)+35 \sin (3 (c+d x)))}{630 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[Cos[c + d*x]^2*Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(-66 4 + 240*Cos[2*(c + d*x)] - 741*Sin[c + d*x] + 35*Sin[3*(c + d*x)]))/(630*d *(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.68 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {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.
| \(\displaystyle \int \sin (c+d x) \cos ^2(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \cos (c+d x)^2 (a \sin (c+d x)+a)^{3/2}dx\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \cos (c+d x)^2 (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {1}{3} \left (\frac {8}{7} a \int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {8}{7} a \int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {1}{3} \left (\frac {8}{7} a \left (\frac {4}{5} a \int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {8}{7} a \left (\frac {4}{5} a \int \frac {\cos (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {1}{3} \left (\frac {8}{7} a \left (-\frac {8 a^2 \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}\) |
Input:
Int[Cos[c + d*x]^2*Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(-2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(9*d) + ((-2*a*Cos[c + d*x] ^3*Sqrt[a + a*Sin[c + d*x]])/(7*d) + (8*a*((-8*a^2*Cos[c + d*x]^3)/(15*d*( a + a*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^3)/(5*d*Sqrt[a + a*Sin[c + d*x]])))/7)/3
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 = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62
| method | result | size |
| default | \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} \left (35 \sin \left (d x +c \right )^{3}+120 \sin \left (d x +c \right )^{2}+159 \sin \left (d x +c \right )+106\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(77\) |
Input:
int(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE )
Output:
-2/315*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(35*sin(d*x+c)^3+120*sin(d*x+c) ^2+159*sin(d*x+c)+106)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.07 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 109 \, a \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} - 32 \, a \cos \left (d x + c\right ) - {\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 32 \, a \cos \left (d x + c\right ) - 64 \, a\right )} \sin \left (d x + c\right ) - 64 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fri cas")
Output:
2/315*(35*a*cos(d*x + c)^5 - 50*a*cos(d*x + c)^4 - 109*a*cos(d*x + c)^3 + 8*a*cos(d*x + c)^2 - 32*a*cos(d*x + c) - (35*a*cos(d*x + c)^4 + 85*a*cos(d *x + c)^3 - 24*a*cos(d*x + c)^2 - 32*a*cos(d*x + c) - 64*a)*sin(d*x + c) - 64*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
\[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**2*sin(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)
Output:
Integral((a*(sin(c + d*x) + 1))**(3/2)*sin(c + d*x)*cos(c + d*x)**2, x)
\[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="max ima")
Output:
integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^2*sin(d*x + c), x)
Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {16 \, \sqrt {2} {\left (70 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 252 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{315 \, d} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="gia c")
Output:
-16/315*sqrt(2)*(70*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/ 2*d*x + 1/2*c)^9 - 225*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 + 252*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)^5 - 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)^3)*sqrt(a)/d
Timed out. \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(cos(c + d*x)^2*sin(c + d*x)*(a + a*sin(c + d*x))^(3/2),x)
Output:
int(cos(c + d*x)^2*sin(c + d*x)*(a + a*sin(c + d*x))^(3/2), x)
\[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )d x \right ) \] Input:
int(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**2,x) + int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x),x))