\(\int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [75]

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

Optimal result

Integrand size = 23, antiderivative size = 186 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{3/2} d}+\frac {13 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac {7 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{a d \sqrt {a+a \sin (c+d x)}} \] Output:

-19/4*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d+13/4*ar 
ctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/a^(3/ 
2)/d+1/2*cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)+7/4*cot(d*x+c)/a/d 
/(a+a*sin(d*x+c))^(1/2)-cot(d*x+c)*csc(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.08 (sec) , antiderivative size = 620, normalized size of antiderivative = 3.33 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[Csc[c + d*x]^3/(a + a*Sin[c + d*x])^(3/2),x]
 

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-32*Sin[(c + d*x)/2] + 16*(Cos[(c 
+ d*x)/2] + Sin[(c + d*x)/2]) - 24*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 
 - (208 + 208*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])^2 + 12*Cot[(c + d*x)/4]*(C 
os[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - Csc[(c + d*x)/4]^2*(Cos[(c + d*x)/ 
2] + Sin[(c + d*x)/2])^2 - 76*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] 
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 76*Log[1 - Cos[(c + d*x)/2] + S 
in[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + Sec[(c + d*x)/4 
]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*(Cos[(c + d*x)/2] + Sin[( 
c + d*x)/2])^2)/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^2 - (24*Sin[(c + d*x 
)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/4] - Sin[(c + 
 d*x)/4]) - (2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/4] 
+ Sin[(c + d*x)/4])^2 + (24*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + 
d*x)/2])^2)/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 12*(Cos[(c + d*x)/2] + 
 Sin[(c + d*x)/2])^2*Tan[(c + d*x)/4]))/(32*d*(a*(1 + Sin[c + d*x]))^(3/2) 
)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 3245, 27, 3042, 3463, 27, 3042, 3463, 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.

\(\displaystyle \int \frac {\csc ^3(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3245

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {\int -\frac {2 \csc ^2(c+d x) \left (7 a^2-6 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{2 a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\csc ^2(c+d x) \left (7 a^2-6 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {7 a^2-6 a^2 \sin (c+d x)}{\sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}}dx}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {-\frac {\frac {\int -\frac {\csc (c+d x) \left (19 a^3-7 a^3 \sin (c+d x)\right )}{2 \sqrt {\sin (c+d x) a+a}}dx}{a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\csc (c+d x) \left (19 a^3-7 a^3 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {\int \frac {19 a^3-7 a^3 \sin (c+d x)}{\sin (c+d x) \sqrt {\sin (c+d x) a+a}}dx}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {-\frac {-\frac {19 a^2 \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-26 a^3 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {19 a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-26 a^3 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {-\frac {19 a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx+\frac {52 a^3 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {-\frac {19 a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx+\frac {26 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {-\frac {-\frac {\frac {26 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {38 a^3 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {-\frac {\frac {26 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {38 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}}{2 a}-\frac {7 a^2 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a}-\frac {4 a \cot (c+d x) \csc (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}+\frac {\cot (c+d x) \csc (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\)

Input:

Int[Csc[c + d*x]^3/(a + a*Sin[c + d*x])^(3/2),x]
 

Output:

(Cot[c + d*x]*Csc[c + d*x])/(2*d*(a + a*Sin[c + d*x])^(3/2)) + ((-4*a*Cot[ 
c + d*x]*Csc[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]) - (-1/2*((-38*a^(5/2)* 
ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (26*Sqrt[2]* 
a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])] 
)/d)/a - (7*a^2*Cot[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/a)/(4*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 3463
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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 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 [A] (verified)

Time = 0.36 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.61

method result size
default \(-\frac {\left (-13 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right )^{3} a^{2}-2 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \sin \left (d x +c \right )^{2}-13 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right )^{2} a^{2}+19 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \sin \left (d x +c \right )^{3} a^{2}+5 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}\, \sin \left (d x +c \right )-3 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \sin \left (d x +c \right )+19 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \sin \left (d x +c \right )^{2} a^{2}+5 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}-3 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{4 a^{\frac {7}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) \(299\)

Input:

int(csc(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4*(-13*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*s 
in(d*x+c)^3*a^2-2*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)^2-13*2^(1/2 
)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+ 
19*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+5*(-a*(sin( 
d*x+c)-1))^(3/2)*a^(1/2)*sin(d*x+c)-3*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*si 
n(d*x+c)+19*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+5* 
(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)-3*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2))*(-a 
*(sin(d*x+c)-1))^(1/2)/a^(7/2)/sin(d*x+c)^2/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. 626 vs. \(2 (157) = 314\).

Time = 0.12 (sec) , antiderivative size = 626, normalized size of antiderivative = 3.37 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

1/16*(26*sqrt(2)*(cos(d*x + c)^4 - cos(d*x + c)^3 - 3*cos(d*x + c)^2 - (co 
s(d*x + c)^3 + 2*cos(d*x + c)^2 - cos(d*x + c) - 2)*sin(d*x + c) + cos(d*x 
 + c) + 2)*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)) + 19*(cos(d*x + c)^4 - cos(d*x + c)^3 - 
3*cos(d*x + c)^2 - (cos(d*x + c)^3 + 2*cos(d*x + c)^2 - cos(d*x + c) - 2)* 
sin(d*x + c) + cos(d*x + c) + 2)*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*(7 
*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - (7*cos(d*x + c)^2 + 3*cos(d*x + c) - 
2)*sin(d*x + c) - 5*cos(d*x + c) - 2)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos 
(d*x + c)^4 - a^2*d*cos(d*x + c)^3 - 3*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d* 
x + c) + 2*a^2*d - (a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 - a^2*d* 
cos(d*x + c) - 2*a^2*d)*sin(d*x + c))
 

Sympy [F]

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

integrate(csc(d*x+c)**3/(a+a*sin(d*x+c))**(3/2),x)
 

Output:

Integral(csc(c + d*x)**3/(a*(sin(c + d*x) + 1))**(3/2), x)
 

Maxima [F]

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

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(csc(d*x + c)^3/(a*sin(d*x + c) + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {19 \, \log \left (\frac {{\left | -16 \, \sqrt {2} - 32 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 16 \, \sqrt {2} - 32 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (10 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \] Input:

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

1/8*(19*log(abs(-16*sqrt(2) - 32*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(16*sq 
rt(2) - 32*sin(-1/4*pi + 1/2*d*x + 1/2*c)))/(a^(3/2)*sgn(cos(-1/4*pi + 1/2 
*d*x + 1/2*c))) + 2*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)/((sin(-1/4*pi + 
 1/2*d*x + 1/2*c)^2 - 1)*a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2* 
sqrt(2)*(10*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 3*sqrt(a)*sin(-1/4* 
pi + 1/2*d*x + 1/2*c))/((2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^2*sgn 
(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(csc(d*x+c)^3/(a+a*sin(d*x+c))^(3/2),x)
 

Output:

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