\(\int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx\) [33]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {-a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cos (c+d x)}{4 d (-a+a \sin (c+d x))^{5/2}}-\frac {3 \cos (c+d x)}{16 a d (-a+a \sin (c+d x))^{3/2}} \] Output:

3/32*arctan(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(-a+a*sin(d*x+c))^(1/2))*2^(1/2 
)/a^(5/2)/d+1/4*cos(d*x+c)/d/(-a+a*sin(d*x+c))^(5/2)-3/16*cos(d*x+c)/a/d/( 
-a+a*sin(d*x+c))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (11 \cos \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {3}{2} (c+d x)\right )+11 \sin \left (\frac {1}{2} (c+d x)\right )+(3+3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) (-3+\cos (2 (c+d x))+4 \sin (c+d x))-3 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{32 d (a (-1+\sin (c+d x)))^{5/2}} \] Input:

Integrate[(-a + a*Sin[c + d*x])^(-5/2),x]
 

Output:

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(11*Cos[(c + d*x)/2] + 3*Cos[(3*(c 
+ d*x))/2] + 11*Sin[(c + d*x)/2] + (3 + 3*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2) 
*(-1)^(1/4)*(1 + Tan[(c + d*x)/4])]*(-3 + Cos[2*(c + d*x)] + 4*Sin[c + d*x 
]) - 3*Sin[(3*(c + d*x))/2]))/(32*d*(a*(-1 + Sin[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 3129, 3042, 3129, 3042, 3128, 217}

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.

\(\displaystyle \int \frac {1}{(a \sin (c+d x)-a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (c+d x)-a)^{5/2}}dx\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {\cos (c+d x)}{4 d (a \sin (c+d x)-a)^{5/2}}-\frac {3 \int \frac {1}{(a \sin (c+d x)-a)^{3/2}}dx}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos (c+d x)}{4 d (a \sin (c+d x)-a)^{5/2}}-\frac {3 \int \frac {1}{(a \sin (c+d x)-a)^{3/2}}dx}{8 a}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {\cos (c+d x)}{4 d (a \sin (c+d x)-a)^{5/2}}-\frac {3 \left (\frac {\cos (c+d x)}{2 d (a \sin (c+d x)-a)^{3/2}}-\frac {\int \frac {1}{\sqrt {a \sin (c+d x)-a}}dx}{4 a}\right )}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos (c+d x)}{4 d (a \sin (c+d x)-a)^{5/2}}-\frac {3 \left (\frac {\cos (c+d x)}{2 d (a \sin (c+d x)-a)^{3/2}}-\frac {\int \frac {1}{\sqrt {a \sin (c+d x)-a}}dx}{4 a}\right )}{8 a}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\cos (c+d x)}{4 d (a \sin (c+d x)-a)^{5/2}}-\frac {3 \left (\frac {\int \frac {1}{-\frac {a^2 \cos ^2(c+d x)}{a \sin (c+d x)-a}-2 a}d\frac {a \cos (c+d x)}{\sqrt {a \sin (c+d x)-a}}}{2 a d}+\frac {\cos (c+d x)}{2 d (a \sin (c+d x)-a)^{3/2}}\right )}{8 a}\)

\(\Big \downarrow \) 217

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

Input:

Int[(-a + a*Sin[c + d*x])^(-5/2),x]
 

Output:

Cos[c + d*x]/(4*d*(-a + a*Sin[c + d*x])^(5/2)) - (3*(-1/2*ArcTan[(Sqrt[a]* 
Cos[c + d*x])/(Sqrt[2]*Sqrt[-a + a*Sin[c + d*x]])]/(Sqrt[2]*a^(3/2)*d) + C 
os[c + d*x]/(2*d*(-a + a*Sin[c + d*x])^(3/2))))/(8*a)
 

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3129

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c 
+ d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n 
+ 1))   Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] 
&& EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(210\) vs. \(2(94)=188\).

Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.87

method	result	size
default	\(-\frac {\left (3 \sqrt {2}\, \arctan \left (\frac {\sqrt {-a \sin \left (d x +c \right )-a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \cos \left (d x +c \right )^{2}+6 \sqrt {2}\, \arctan \left (\frac {\sqrt {-a \sin \left (d x +c \right )-a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )+6 \sqrt {-a \sin \left (d x +c \right )-a}\, a^{\frac {3}{2}} \sin \left (d x +c \right )-6 \sqrt {2}\, \arctan \left (\frac {\sqrt {-a \sin \left (d x +c \right )-a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-14 \sqrt {-a \sin \left (d x +c \right )-a}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (1+\sin \left (d x +c \right )\right )}}{32 a^{\frac {9}{2}} \left (\sin \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \sqrt {-a +a \sin \left (d x +c \right )}\, d}\)	\(211\)

Input:

int(1/(-a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32*(3*2^(1/2)*arctan(1/2*(-a*sin(d*x+c)-a)^(1/2)*2^(1/2)/a^(1/2))*a^2*c 
os(d*x+c)^2+6*2^(1/2)*arctan(1/2*(-a*sin(d*x+c)-a)^(1/2)*2^(1/2)/a^(1/2))* 
a^2*sin(d*x+c)+6*(-a*sin(d*x+c)-a)^(1/2)*a^(3/2)*sin(d*x+c)-6*2^(1/2)*arct 
an(1/2*(-a*sin(d*x+c)-a)^(1/2)*2^(1/2)/a^(1/2))*a^2-14*(-a*sin(d*x+c)-a)^( 
1/2)*a^(3/2))*(-a*(1+sin(d*x+c)))^(1/2)/a^(9/2)/(sin(d*x+c)-1)/cos(d*x+c)/ 
(-a+a*sin(d*x+c))^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (94) = 188\).

Time = 0.11 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.89 \[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {-a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) - a} \sqrt {-a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 7 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {a \sin \left (d x + c\right ) - a}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

integrate(1/(-a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-1/64*(3*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 - (cos(d*x + c)^2 - 2* 
cos(d*x + c) - 4)*sin(d*x + c) - 2*cos(d*x + c) - 4)*sqrt(-a)*log(-(a*cos( 
d*x + c)^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c) - a)*sqrt(-a)*(cos(d*x + c) + s 
in(d*x + c) + 1) + 3*a*cos(d*x + c) + (a*cos(d*x + c) - 2*a)*sin(d*x + c) 
+ 2*a)/(cos(d*x + c)^2 + (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 
2)) - 4*(3*cos(d*x + c)^2 - (3*cos(d*x + c) - 4)*sin(d*x + c) + 7*cos(d*x 
+ c) + 4)*sqrt(a*sin(d*x + c) - a))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d* 
x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d - (a^3*d*cos(d*x + c)^2 - 2*a^3* 
d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a \sin {\left (c + d x \right )} - a\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(-a+a*sin(d*x+c))**(5/2),x)
 

Output:

Integral((a*sin(c + d*x) - a)**(-5/2), x)
 

Maxima [F]

\[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (d x + c\right ) - a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(-a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(d*x + c) - a)^(-5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (94) = 188\).

Time = 0.20 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.77 \[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (\frac {8 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} - \frac {18 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}}{\sqrt {-a} a^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {12 \, \sqrt {2} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {\frac {8 \, \sqrt {2} \sqrt {-a} a^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} - \frac {\sqrt {2} \sqrt {-a} a^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}}}{a^{5}}}{256 \, d} \] Input:

integrate(1/(-a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

1/256*(sqrt(2)*(8*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)/(cos(-1/4*pi + 1/2* 
d*x + 1/2*c) + 1) - 18*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)^2/(cos(-1/4*pi 
 + 1/2*d*x + 1/2*c) + 1)^2 - 1)*(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)^2/(sq 
rt(-a)*a^2*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)^2*sgn(sin(-1/4*pi + 1/2*d* 
x + 1/2*c))) + 12*sqrt(2)*log(-(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)/(cos(- 
1/4*pi + 1/2*d*x + 1/2*c) + 1))/(sqrt(-a)*a^2*sgn(sin(-1/4*pi + 1/2*d*x + 
1/2*c))) + (8*sqrt(2)*sqrt(-a)*a^2*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)*sg 
n(sin(-1/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) - s 
qrt(2)*sqrt(-a)*a^2*(cos(-1/4*pi + 1/2*d*x + 1/2*c) - 1)^2*sgn(sin(-1/4*pi 
 + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)^2)/a^5)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a\,\sin \left (c+d\,x\right )-a\right )}^{5/2}} \,d x \] Input:

int(1/(a*sin(c + d*x) - a)^(5/2),x)
 

Output:

int(1/(a*sin(c + d*x) - a)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{(-a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )-1}}{\sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )-1}d x \right )}{a^{3}} \] Input:

int(1/(-a+a*sin(d*x+c))^(5/2),x)
 

Output:

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