\(\int \frac {\csc (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [82]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int \frac {\csc (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {43 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {11 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}} \] Output:

-2*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d+43/32*arct 
anh(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)+11/16*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 = 1.81 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.06 \[ \int \frac {\csc (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-8 \sin \left (\frac {1}{2} (c+d x)\right )+4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-22 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+11 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3-(43+43 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4-16 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+16 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4\right )}{16 d (a (1+\sin (c+d x)))^{5/2}} \] Input:

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

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8*Sin[(c + d*x)/2] + 4*(Cos[(c + 
d*x)/2] + Sin[(c + d*x)/2]) - 22*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[ 
(c + d*x)/2])^2 + 11*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (43 + 43*I) 
*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[( 
c + d*x)/2] + Sin[(c + d*x)/2])^4 - 16*Log[1 + Cos[(c + d*x)/2] - Sin[(c + 
 d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 16*Log[1 - Cos[(c + d* 
x)/2] + Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4))/(16*d* 
(a*(1 + Sin[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3245, 27, 3042, 3457, 27, 3042, 3464, 3042, 3128, 219, 3252, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3245

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {\csc (c+d x) \left (32 a^2-11 a^2 \sin (c+d x)\right )}{2 \sqrt {\sin (c+d x) a+a}}dx}{2 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\csc (c+d x) \left (32 a^2-11 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {32 a^2-11 a^2 \sin (c+d x)}{\sin (c+d x) \sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {32 a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-43 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {32 a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-43 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {86 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}+32 a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {32 a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx+\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {64 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {64 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {11 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}+\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\)

Input:

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

Output:

Cos[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) + (((-64*a^(3/2)*ArcTanh[(Sq 
rt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (43*Sqrt[2]*a^(3/2)*Arc 
Tanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d)/(4*a^2 
) + (11*a*Cos[c + d*x])/(2*d*(a + a*Sin[c + d*x])^(3/2)))/(8*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[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 3245

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3252

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

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(261\) vs. \(2(119)=238\).

Time = 0.33 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {\left (-a^{5} \left (43 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-64 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )\right ) \cos \left (d x +c \right )^{2}+2 \sin \left (d x +c \right ) a^{5} \left (43 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-64 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )\right )+52 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {9}{2}}-22 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{\frac {7}{2}}+86 \sqrt {2}\, a^{5} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-128 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{5}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {15}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(262\)

Input:

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

Output:

1/32/a^(15/2)*(-a^5*(43*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2) 
/a^(1/2))-64*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2)))*cos(d*x+c)^2+2*sin(d 
*x+c)*a^5*(43*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))- 
64*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2)))+52*(a-a*sin(d*x+c))^(1/2)*a^(9 
/2)-22*(a-a*sin(d*x+c))^(3/2)*a^(7/2)+86*2^(1/2)*a^5*arctanh(1/2*(a-a*sin( 
d*x+c))^(1/2)*2^(1/2)/a^(1/2))-128*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2)) 
*a^5)*(-a*(sin(d*x+c)-1))^(1/2)/(1+sin(d*x+c))/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. 539 vs. \(2 (119) = 238\).

Time = 0.12 (sec) , antiderivative size = 539, normalized size of antiderivative = 3.74 \[ \int \frac {\csc (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/64*(43*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) - sin 
(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) 
) + 32*(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)^3 - 
 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) 
- 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) 
+ (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c 
)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 
1)) - 4*(11*cos(d*x + c)^2 + (11*cos(d*x + c) - 4)*sin(d*x + c) + 15*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]

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

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

Output:

Integral(csc(c + d*x)/(a*(sin(c + d*x) + 1))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.13 \[ \int \frac {\csc (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\frac {32 \, \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {\sqrt {2} {\left (11 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{32 \, d} \] Input:

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

Output:

1/32*(32*log(abs(1/2*sqrt(2) + sin(-1/4*pi + 1/2*d*x + 1/2*c)))/(a^(5/2)*s 
gn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 32*log(abs(-1/2*sqrt(2) + sin(-1/4*p 
i + 1/2*d*x + 1/2*c)))/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + sqr 
t(2)*(11*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 13*sqrt(a)*sin(-1/4*pi 
 + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos 
(-1/4*pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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