Integrand size = 21, antiderivative size = 86 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {8 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d} \] Output:
-8/5*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/5*a*cos(d*x+c)*(a+a*sin(d*x +c))^(1/2)/d-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d
Time = 1.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.83 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\left (-4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-8+5 \sqrt {2} \sqrt {1+\cos (c+d x)}\right )+\sqrt {2} \sqrt {1+\cos (c+d x)} \left (-5 \cos \left (\frac {3}{2} (c+d x)\right )+\cos \left (\frac {5}{2} (c+d x)\right )+8 (4+\cos (c+d x)) \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )\right ) (a (1+\sin (c+d x)))^{3/2}}{20 d \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \] Input:
Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
Output:
((-4*Cos[(c + d*x)/2]*(-8 + 5*Sqrt[2]*Sqrt[1 + Cos[c + d*x]]) + Sqrt[2]*Sq rt[1 + Cos[c + d*x]]*(-5*Cos[(3*(c + d*x))/2] + Cos[(5*(c + d*x))/2] + 8*( 4 + Cos[c + d*x])*Sin[(c + d*x)/2]^3))*(a*(1 + Sin[c + d*x]))^(3/2))/(20*d *Sqrt[Cos[(c + d*x)/2]^2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)
Time = 0.38 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3230, 3042, 3126, 3042, 3125}
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.
| \(\displaystyle \int \sin (c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) (a \sin (c+d x)+a)^{3/2}dx\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {3}{5} \int (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{5} \int (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {3}{5} \left (\frac {4}{3} a \int \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{5} \left (\frac {4}{3} a \int \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {3}{5} \left (-\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
Input:
Int[Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(-2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*d) + (3*((-8*a^2*Cos[c + d *x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d)))/5
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+6\right )}{5 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(63\) |
Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(sin(d*x+c)^2+3*sin(d*x+c)+6)/cos(d* x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} - 7 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) - 4 \, a\right )} \sin \left (d x + c\right ) - 4 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
2/5*(a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 - 7*a*cos(d*x + c) - (a*cos(d*x + c)^2 + 3*a*cos(d*x + c) - 4*a)*sin(d*x + c) - 4*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
\[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin {\left (c + d x \right )}\, dx \] Input:
integrate(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), x)
\[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(3/2)*sin(d*x + c), x)
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (20 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )} \sqrt {a}}{10 \, d} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
1/10*sqrt(2)*(20*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*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d* x + 3/2*c) + a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d*x + 5/2*c))*sqrt(a)/d
Timed out. \[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(sin(c + d*x)*(a + a*sin(c + d*x))^(3/2),x)
Output:
int(sin(c + d*x)*(a + a*sin(c + d*x))^(3/2), x)
\[ \int \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{2}d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )d x \right ) \] Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*sin(c + d*x)**2,x) + int(sqrt(sin(c + d*x) + 1)*sin(c + d*x),x))