\(\int (a-a \sin (c+d x))^{7/2} \, dx\) [20]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 123 \[ \int (a-a \sin (c+d x))^{7/2} \, dx=\frac {256 a^4 \cos (c+d x)}{35 d \sqrt {a-a \sin (c+d x)}}+\frac {64 a^3 \cos (c+d x) \sqrt {a-a \sin (c+d x)}}{35 d}+\frac {24 a^2 \cos (c+d x) (a-a \sin (c+d x))^{3/2}}{35 d}+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d} \] Output:

256/35*a^4*cos(d*x+c)/d/(a-a*sin(d*x+c))^(1/2)+64/35*a^3*cos(d*x+c)*(a-a*s 
in(d*x+c))^(1/2)/d+24/35*a^2*cos(d*x+c)*(a-a*sin(d*x+c))^(3/2)/d+2/7*a*cos 
(d*x+c)*(a-a*sin(d*x+c))^(5/2)/d
 

Mathematica [A] (verified)

Time = 3.50 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.28 \[ \int (a-a \sin (c+d x))^{7/2} \, dx=-\frac {a^3 (-1+\sin (c+d x))^3 \sqrt {a-a \sin (c+d x)} \left (1225 \cos \left (\frac {1}{2} (c+d x)\right )+245 \cos \left (\frac {3}{2} (c+d x)\right )-49 \cos \left (\frac {5}{2} (c+d x)\right )-5 \cos \left (\frac {7}{2} (c+d x)\right )+1225 \sin \left (\frac {1}{2} (c+d x)\right )-245 \sin \left (\frac {3}{2} (c+d x)\right )-49 \sin \left (\frac {5}{2} (c+d x)\right )+5 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{140 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^7} \] Input:

Integrate[(a - a*Sin[c + d*x])^(7/2),x]
 

Output:

-1/140*(a^3*(-1 + Sin[c + d*x])^3*Sqrt[a - a*Sin[c + d*x]]*(1225*Cos[(c + 
d*x)/2] + 245*Cos[(3*(c + d*x))/2] - 49*Cos[(5*(c + d*x))/2] - 5*Cos[(7*(c 
 + d*x))/2] + 1225*Sin[(c + d*x)/2] - 245*Sin[(3*(c + d*x))/2] - 49*Sin[(5 
*(c + d*x))/2] + 5*Sin[(7*(c + d*x))/2]))/(d*(Cos[(c + d*x)/2] - Sin[(c + 
d*x)/2])^7)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3126, 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.

\(\displaystyle \int (a-a \sin (c+d x))^{7/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a-a \sin (c+d x))^{7/2}dx\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {12}{7} a \int (a-a \sin (c+d x))^{5/2}dx+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{7} a \int (a-a \sin (c+d x))^{5/2}dx+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \int (a-a \sin (c+d x))^{3/2}dx+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \int (a-a \sin (c+d x))^{3/2}dx+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {a-a \sin (c+d x)}dx+\frac {2 a \cos (c+d x) \sqrt {a-a \sin (c+d x)}}{3 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {a-a \sin (c+d x)}dx+\frac {2 a \cos (c+d x) \sqrt {a-a \sin (c+d x)}}{3 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {12}{7} a \left (\frac {8}{5} a \left (\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a-a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a-a \sin (c+d x)}}{3 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 a \cos (c+d x) (a-a \sin (c+d x))^{5/2}}{7 d}\)

Input:

Int[(a - a*Sin[c + d*x])^(7/2),x]
 

Output:

(2*a*Cos[c + d*x]*(a - a*Sin[c + d*x])^(5/2))/(7*d) + (12*a*((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*S 
qrt[a - a*Sin[c + d*x]]) + (2*a*Cos[c + d*x]*Sqrt[a - a*Sin[c + d*x]])/(3* 
d)))/5))/7
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3125
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]
 

rule 3126
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]
 
Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.62

method result size
default \(\frac {2 \left (\sin \left (d x +c \right )-1\right ) a^{4} \left (1+\sin \left (d x +c \right )\right ) \left (5 \sin \left (d x +c \right )^{3}-27 \sin \left (d x +c \right )^{2}+71 \sin \left (d x +c \right )-177\right )}{35 \cos \left (d x +c \right ) \sqrt {a -a \sin \left (d x +c \right )}\, d}\) \(76\)

Input:

int((a-a*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/35*(sin(d*x+c)-1)*a^4*(1+sin(d*x+c))*(5*sin(d*x+c)^3-27*sin(d*x+c)^2+71* 
sin(d*x+c)-177)/cos(d*x+c)/(a-a*sin(d*x+c))^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.16 \[ \int (a-a \sin (c+d x))^{7/2} \, dx=-\frac {2 \, {\left (5 \, a^{3} \cos \left (d x + c\right )^{4} + 27 \, a^{3} \cos \left (d x + c\right )^{3} - 54 \, a^{3} \cos \left (d x + c\right )^{2} - 204 \, a^{3} \cos \left (d x + c\right ) - 128 \, a^{3} - {\left (5 \, a^{3} \cos \left (d x + c\right )^{3} - 22 \, a^{3} \cos \left (d x + c\right )^{2} - 76 \, a^{3} \cos \left (d x + c\right ) + 128 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a \sin \left (d x + c\right ) + a}}{35 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate((a-a*sin(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-2/35*(5*a^3*cos(d*x + c)^4 + 27*a^3*cos(d*x + c)^3 - 54*a^3*cos(d*x + c)^ 
2 - 204*a^3*cos(d*x + c) - 128*a^3 - (5*a^3*cos(d*x + c)^3 - 22*a^3*cos(d* 
x + c)^2 - 76*a^3*cos(d*x + c) + 128*a^3)*sin(d*x + c))*sqrt(-a*sin(d*x + 
c) + a)/(d*cos(d*x + c) - d*sin(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int (a-a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate((a-a*sin(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a-a \sin (c+d x))^{7/2} \, dx=\int { {\left (-a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \,d x } \] Input:

integrate((a-a*sin(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate((-a*sin(d*x + c) + a)^(7/2), x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.07 \[ \int (a-a \sin (c+d x))^{7/2} \, dx=-\frac {\sqrt {2} {\left (1225 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 245 \, a^{3} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 49 \, a^{3} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 5 \, a^{3} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{140 \, d} \] Input:

integrate((a-a*sin(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

-1/140*sqrt(2)*(1225*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)*sgn(sin(-1/4*pi + 
1/2*d*x + 1/2*c)) - 245*a^3*cos(-3/4*pi + 3/2*d*x + 3/2*c)*sgn(sin(-1/4*pi 
 + 1/2*d*x + 1/2*c)) + 49*a^3*cos(-5/4*pi + 5/2*d*x + 5/2*c)*sgn(sin(-1/4* 
pi + 1/2*d*x + 1/2*c)) - 5*a^3*cos(-7/4*pi + 7/2*d*x + 7/2*c)*sgn(sin(-1/4 
*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int (a-a \sin (c+d x))^{7/2} \, dx=\int {\left (a-a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \] Input:

int((a - a*sin(c + d*x))^(7/2),x)
 

Output:

int((a - a*sin(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int (a-a \sin (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (\int \sqrt {-\sin \left (d x +c \right )+1}d x -\left (\int \sqrt {-\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{3}d x \right )+3 \left (\int \sqrt {-\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{2}d x \right )-3 \left (\int \sqrt {-\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )d x \right )\right ) \] Input:

int((a-a*sin(d*x+c))^(7/2),x)
                                                                                    
                                                                                    
 

Output:

sqrt(a)*a**3*(int(sqrt( - sin(c + d*x) + 1),x) - int(sqrt( - sin(c + d*x) 
+ 1)*sin(c + d*x)**3,x) + 3*int(sqrt( - sin(c + d*x) + 1)*sin(c + d*x)**2, 
x) - 3*int(sqrt( - sin(c + d*x) + 1)*sin(c + d*x),x))