Integrand size = 14, antiderivative size = 89 \[ \int (a+a \sin (c+d x))^{5/2} \, dx=-\frac {64 a^3 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {16 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {2 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d} \] Output:
-64/15*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-16/15*a^2*cos(d*x+c)*(a+a*s in(d*x+c))^(1/2)/d-2/5*a*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d
Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.31 \[ \int (a+a \sin (c+d x))^{5/2} \, dx=-\frac {(a (1+\sin (c+d x)))^{5/2} \left (150 \cos \left (\frac {1}{2} (c+d x)\right )+25 \cos \left (\frac {3}{2} (c+d x)\right )-3 \cos \left (\frac {5}{2} (c+d x)\right )-150 \sin \left (\frac {1}{2} (c+d x)\right )+25 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:
Integrate[(a + a*Sin[c + d*x])^(5/2),x]
Output:
-1/30*((a*(1 + Sin[c + d*x]))^(5/2)*(150*Cos[(c + d*x)/2] + 25*Cos[(3*(c + d*x))/2] - 3*Cos[(5*(c + d*x))/2] - 150*Sin[(c + d*x)/2] + 25*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2])^5)
Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {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.
\(\displaystyle \int (a \sin (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (c+d x)+a)^{5/2}dx\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {8}{5} a \int (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} a \int (\sin (c+d x) a+a)^{3/2}dx-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {8}{5} a \left (-\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )-\frac {2 a \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
Input:
Int[(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-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*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]
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (3 \sin \left (d x +c \right )^{2}+14 \sin \left (d x +c \right )+43\right )}{15 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(65\) |
Input:
int((a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/15*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(3*sin(d*x+c)^2+14*sin(d*x+c)+43)/c os(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.29 \[ \int (a+a \sin (c+d x))^{5/2} \, dx=\frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 11 \, a^{2} \cos \left (d x + c\right )^{2} - 46 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2} - {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) - 32 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate((a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
2/15*(3*a^2*cos(d*x + c)^3 - 11*a^2*cos(d*x + c)^2 - 46*a^2*cos(d*x + c) - 32*a^2 - (3*a^2*cos(d*x + c)^2 + 14*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)
\[ \int (a+a \sin (c+d x))^{5/2} \, dx=\int \left (a \sin {\left (c + d x \right )} + a\right )^{\frac {5}{2}}\, dx \] Input:
integrate((a+a*sin(d*x+c))**(5/2),x)
Output:
Integral((a*sin(c + d*x) + a)**(5/2), x)
\[ \int (a+a \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(5/2), x)
Time = 0.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15 \[ \int (a+a \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (150 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \] Input:
integrate((a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
1/30*sqrt(2)*(150*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) + 25*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) + 3*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))*sqrt(a)/d
Timed out. \[ \int (a+a \sin (c+d x))^{5/2} \, dx=\int {\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int((a + a*sin(c + d*x))^(5/2),x)
Output:
int((a + a*sin(c + d*x))^(5/2), x)
\[ \int (a+a \sin (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )+1}d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )d x \right )\right ) \] Input:
int((a+a*sin(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(c + d*x) + 1),x) + int(sqrt(sin(c + d*x) + 1)*s in(c + d*x)**2,x) + 2*int(sqrt(sin(c + d*x) + 1)*sin(c + d*x),x))