Integrand size = 21, antiderivative size = 116 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {64 a^3 \cos (c+d x)}{21 d \sqrt {a+a \sin (c+d x)}}-\frac {16 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{7 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 d} \] Output:
-64/21*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-16/21*a^2*cos(d*x+c)*(a+a*s in(d*x+c))^(1/2)/d-2/7*a*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/7*cos(d*x+c )*(a+a*sin(d*x+c))^(5/2)/d
Leaf count is larger than twice the leaf count of optimal. \(271\) vs. \(2(116)=232\).
Time = 3.40 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.34 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a (1+\sin (c+d x)))^{5/2} \left (6 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}+10 \sqrt {2} \sqrt {1+\cos (c+d x)} \sin ^2\left (\frac {1}{2} (c+d x)\right )-45 \sqrt {2} \sqrt {1+\cos (c+d x)} \sin ^4\left (\frac {1}{2} (c+d x)\right )+3 \sin ^6\left (\frac {1}{2} (c+d x)\right ) \left (5 \sqrt {2} \sqrt {1+\cos (c+d x)}+8 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} \tan \left (\frac {1}{2} (c+d x)\right )\right )+10 \left (-4+2 \sqrt {2} \sqrt {1+\cos (c+d x)}-7 \cos ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan ^3\left (\frac {1}{2} (c+d x)\right )\right )\right )}{21 d \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:
Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-2*Sec[(c + d*x)/2]^4*(a*(1 + Sin[c + d*x]))^(5/2)*(6*Cos[(c + d*x)/2]^6* Sqrt[Cos[(c + d*x)/2]^2] + 10*Sqrt[2]*Sqrt[1 + Cos[c + d*x]]*Sin[(c + d*x) /2]^2 - 45*Sqrt[2]*Sqrt[1 + Cos[c + d*x]]*Sin[(c + d*x)/2]^4 + 3*Sin[(c + d*x)/2]^6*(5*Sqrt[2]*Sqrt[1 + Cos[c + d*x]] + 8*Sqrt[Cos[(c + d*x)/2]^2]*T an[(c + d*x)/2]) + 10*(-4 + 2*Sqrt[2]*Sqrt[1 + Cos[c + d*x]] - 7*(Cos[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]^3)))/(21*d*Sqrt[Cos[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2])^5)
Time = 0.48 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3230, 3042, 3126, 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.
| \(\displaystyle \int \sin (c+d x) (a \sin (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) (a \sin (c+d x)+a)^{5/2}dx\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {5}{7} \int (\sin (c+d x) a+a)^{5/2}dx-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{7} \int (\sin (c+d x) a+a)^{5/2}dx-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {5}{7} \left (\frac {8}{5} a \int (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{7} \left (\frac {8}{5} a \int (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {5}{7} \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{7} \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {5}{7} \left (\frac {8}{5} a \left (-\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\right )-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 d}\) |
Input:
Int[Sin[c + d*x]*(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*d) + (5*((-2*a*Cos[c + d*x ]*(a + a*Sin[c + d*x])^(3/2))/(5*d) + (8*a*((-8*a^2*Cos[c + d*x])/(3*d*Sqr t[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) ))/5))/7
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 = 0.44 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (3 \sin \left (d x +c \right )^{3}+12 \sin \left (d x +c \right )^{2}+23 \sin \left (d x +c \right )+46\right )}{21 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(75\) |
Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/21*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(3*sin(d*x+c)^3+12*sin(d*x+c)^2+23* sin(d*x+c)+46)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.08 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} + 12 \, a^{2} \cos \left (d x + c\right )^{3} - 17 \, a^{2} \cos \left (d x + c\right )^{2} - 58 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2} + {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 9 \, a^{2} \cos \left (d x + c\right )^{2} - 26 \, a^{2} \cos \left (d x + c\right ) + 32 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{21 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
2/21*(3*a^2*cos(d*x + c)^4 + 12*a^2*cos(d*x + c)^3 - 17*a^2*cos(d*x + c)^2 - 58*a^2*cos(d*x + c) - 32*a^2 + (3*a^2*cos(d*x + c)^3 - 9*a^2*cos(d*x + c)^2 - 26*a^2*cos(d*x + c) + 32*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a )/(d*cos(d*x + c) + d*sin(d*x + c) + d)
\[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \sin {\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))**(5/2),x)
Output:
Integral((a*(sin(c + d*x) + 1))**(5/2)*sin(c + d*x), x)
\[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(5/2)*sin(d*x + c), x)
Time = 0.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14 \[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (315 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 77 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 21 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 3 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )} \sqrt {a}}{84 \, d} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
1/84*sqrt(2)*(315*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/ 2*d*x + 1/2*c) + 77*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 21*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d*x + 5/2*c) + 3*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c))*sqrt(a)/d
Timed out. \[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int(sin(c + d*x)*(a + a*sin(c + d*x))^(5/2),x)
Output:
int(sin(c + d*x)*(a + a*sin(c + d*x))^(5/2), x)
\[ \int \sin (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{3}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{2}d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )d x \right ) \] Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(c + d*x) + 1)*sin(c + d*x)**3,x) + 2*int(sqrt(s in(c + d*x) + 1)*sin(c + d*x)**2,x) + int(sqrt(sin(c + d*x) + 1)*sin(c + d *x),x))