\(\int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx\) [183]

Optimal result

Integrand size = 13, antiderivative size = 84 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx=-\frac {2 b^3 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \sqrt {a^2-b^2}}-\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^3}+\frac {b \cot (x)}{a^2}-\frac {\cot (x) \csc (x)}{2 a} \] Output:

-2*b^3*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/a^3/(a^2-b^2)^(1/2)-1/2*(a 
^2+2*b^2)*arctanh(cos(x))/a^3+b*cot(x)/a^2-1/2*cot(x)*csc(x)/a
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.71 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx=\frac {-\frac {16 b^3 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+4 a b \cot \left (\frac {x}{2}\right )-a^2 \csc ^2\left (\frac {x}{2}\right )-4 a^2 \log \left (\cos \left (\frac {x}{2}\right )\right )-8 b^2 \log \left (\cos \left (\frac {x}{2}\right )\right )+4 a^2 \log \left (\sin \left (\frac {x}{2}\right )\right )+8 b^2 \log \left (\sin \left (\frac {x}{2}\right )\right )+a^2 \sec ^2\left (\frac {x}{2}\right )-4 a b \tan \left (\frac {x}{2}\right )}{8 a^3} \] Input:

Integrate[Csc[x]^3/(a + b*Sin[x]),x]
 

Output:

((-16*b^3*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 4*a* 
b*Cot[x/2] - a^2*Csc[x/2]^2 - 4*a^2*Log[Cos[x/2]] - 8*b^2*Log[Cos[x/2]] + 
4*a^2*Log[Sin[x/2]] + 8*b^2*Log[Sin[x/2]] + a^2*Sec[x/2]^2 - 4*a*b*Tan[x/2 
])/(8*a^3)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.23, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3281, 25, 3042, 3534, 25, 3042, 3480, 3042, 3139, 1083, 217, 4257}

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.

\(\displaystyle \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (x)^3 (a+b \sin (x))}dx\)

\(\Big \downarrow \) 3281

\(\displaystyle \frac {\int -\frac {\csc ^2(x) \left (-b \sin ^2(x)-a \sin (x)+2 b\right )}{a+b \sin (x)}dx}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\csc ^2(x) \left (-b \sin ^2(x)-a \sin (x)+2 b\right )}{a+b \sin (x)}dx}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {-b \sin (x)^2-a \sin (x)+2 b}{\sin (x)^2 (a+b \sin (x))}dx}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 3534

\(\displaystyle -\frac {\frac {\int -\frac {\csc (x) \left (a^2+b \sin (x) a+2 b^2\right )}{a+b \sin (x)}dx}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {\int \frac {\csc (x) \left (a^2+b \sin (x) a+2 b^2\right )}{a+b \sin (x)}dx}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {a^2+b \sin (x) a+2 b^2}{\sin (x) (a+b \sin (x))}dx}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 3480

\(\displaystyle -\frac {-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc (x)dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \sin (x)}dx}{a}}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc (x)dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \sin (x)}dx}{a}}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 3139

\(\displaystyle -\frac {-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc (x)dx}{a}-\frac {4 b^3 \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{a}}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc (x)dx}{a}+\frac {8 b^3 \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a}}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc (x)dx}{a}-\frac {4 b^3 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {-\frac {-\frac {4 b^3 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{a}}{a}-\frac {2 b \cot (x)}{a}}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\)

Input:

Int[Csc[x]^3/(a + b*Sin[x]),x]
 

Output:

-1/2*(-(((-4*b^3*ArcTan[(2*b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(a*Sqrt 
[a^2 - b^2]) - ((a^2 + 2*b^2)*ArcTanh[Cos[x]])/a)/a) - (2*b*Cot[x])/a)/a - 
 (Cot[x]*Csc[x])/(2*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3281

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 + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

rule 3480

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

rule 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {\frac {a \tan \left (\frac {x}{2}\right )^{2}}{2}-2 b \tan \left (\frac {x}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {x}{2}\right )^{2}}+\frac {\left (2 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {x}{2}\right )}-\frac {2 b^{3} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \sqrt {a^{2}-b^{2}}}\)	\(112\)
risch	\(\frac {i \left (-i a \,{\mathrm e}^{3 i x}-i a \,{\mathrm e}^{i x}+2 b \,{\mathrm e}^{2 i x}-2 b \right )}{\left ({\mathrm e}^{2 i x}-1\right )^{2} a^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{2 a}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) b^{2}}{a^{3}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{2 a}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) b^{2}}{a^{3}}-\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{3}}+\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{3}}\)	\(236\)

Input:

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

Output:

1/4/a^2*(1/2*a*tan(1/2*x)^2-2*b*tan(1/2*x))-1/8/a/tan(1/2*x)^2+1/4/a^3*(2* 
a^2+4*b^2)*ln(tan(1/2*x))+1/2/a^2*b/tan(1/2*x)-2/a^3*b^3/(a^2-b^2)^(1/2)*a 
rctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (74) = 148\).

Time = 0.19 (sec) , antiderivative size = 490, normalized size of antiderivative = 5.83 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx=\left [\frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) + 2 \, {\left (b^{3} \cos \left (x\right )^{2} - b^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (x\right ) - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{5} - a^{3} b^{2} - {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )^{2}\right )}}, \frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) - 4 \, {\left (b^{3} \cos \left (x\right )^{2} - b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (x\right ) - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{5} - a^{3} b^{2} - {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/4*(4*(a^3*b - a*b^3)*cos(x)*sin(x) + 2*(b^3*cos(x)^2 - b^3)*sqrt(-a^2 + 
 b^2)*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 - 2*(a*cos(x 
)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 
- b^2)) - 2*(a^4 - a^2*b^2)*cos(x) - (a^4 + a^2*b^2 - 2*b^4 - (a^4 + a^2*b 
^2 - 2*b^4)*cos(x)^2)*log(1/2*cos(x) + 1/2) + (a^4 + a^2*b^2 - 2*b^4 - (a^ 
4 + a^2*b^2 - 2*b^4)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a^5 - a^3*b^2 - (a 
^5 - a^3*b^2)*cos(x)^2), 1/4*(4*(a^3*b - a*b^3)*cos(x)*sin(x) - 4*(b^3*cos 
(x)^2 - b^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x 
))) - 2*(a^4 - a^2*b^2)*cos(x) - (a^4 + a^2*b^2 - 2*b^4 - (a^4 + a^2*b^2 - 
 2*b^4)*cos(x)^2)*log(1/2*cos(x) + 1/2) + (a^4 + a^2*b^2 - 2*b^4 - (a^4 + 
a^2*b^2 - 2*b^4)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a^5 - a^3*b^2 - (a^5 - 
 a^3*b^2)*cos(x)^2)]
 

Sympy [F]

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

integrate(csc(x)**3/(a+b*sin(x)),x)
 

Output:

Integral(csc(x)**3/(a + b*sin(x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{3}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {a \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, x\right )}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a^{3}} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, x\right ) + a^{2}}{8 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2}} \] Input:

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

Output:

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - 
 b^2)))*b^3/(sqrt(a^2 - b^2)*a^3) + 1/8*(a*tan(1/2*x)^2 - 4*b*tan(1/2*x))/ 
a^2 + 1/2*(a^2 + 2*b^2)*log(abs(tan(1/2*x)))/a^3 - 1/8*(6*a^2*tan(1/2*x)^2 
 + 12*b^2*tan(1/2*x)^2 - 4*a*b*tan(1/2*x) + a^2)/(a^3*tan(1/2*x)^2)
 

Mupad [B] (verification not implemented)

Time = 17.27 (sec) , antiderivative size = 531, normalized size of antiderivative = 6.32 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx =\text {Too large to display} \] Input:

int(1/(sin(x)^3*(a + b*sin(x))),x)
 

Output:

(a^4*(cos(x)/2 - log(sin(x/2)/cos(x/2))/4 + (cos(2*x)*log(sin(x/2)/cos(x/2 
)))/4) - a^2*((b^2*log(sin(x/2)/cos(x/2)))/4 + (b^2*cos(x))/2 - (b^2*cos(2 
*x)*log(sin(x/2)/cos(x/2)))/4) + (b^4*log(sin(x/2)/cos(x/2)))/2 - b^3*atan 
((b^4*sin(x/2)*(b^2 - a^2)^(1/2)*8i - a^4*sin(x/2)*(b^2 - a^2)^(1/2)*1i + 
a*b^3*cos(x/2)*(b^2 - a^2)^(1/2)*4i + a^3*b*cos(x/2)*(b^2 - a^2)^(1/2)*1i) 
/(a^5*cos(x/2) - 8*b^5*sin(x/2) + a^3*b^2*cos(x/2) + 4*a^2*b^3*sin(x/2) - 
4*a*b^4*cos(x/2) + 2*a^4*b*sin(x/2)))*(b^2 - a^2)^(1/2)*1i - (b^4*cos(2*x) 
*log(sin(x/2)/cos(x/2)))/2 + (a*b^3*sin(2*x))/2 - (a^3*b*sin(2*x))/2 + b^3 
*cos(2*x)*atan((b^4*sin(x/2)*(b^2 - a^2)^(1/2)*8i - a^4*sin(x/2)*(b^2 - a^ 
2)^(1/2)*1i + a*b^3*cos(x/2)*(b^2 - a^2)^(1/2)*4i + a^3*b*cos(x/2)*(b^2 - 
a^2)^(1/2)*1i)/(a^5*cos(x/2) - 8*b^5*sin(x/2) + a^3*b^2*cos(x/2) + 4*a^2*b 
^3*sin(x/2) - 4*a*b^4*cos(x/2) + 2*a^4*b*sin(x/2)))*(b^2 - a^2)^(1/2)*1i)/ 
((a^5*cos(2*x))/2 - a^5/2 + (a^3*b^2)/2 - (a^3*b^2*cos(2*x))/2)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{2} b^{3}+2 \cos \left (x \right ) \sin \left (x \right ) a^{3} b -2 \cos \left (x \right ) \sin \left (x \right ) a \,b^{3}-\cos \left (x \right ) a^{4}+\cos \left (x \right ) a^{2} b^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2} a^{4}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2} a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2} b^{4}}{2 \sin \left (x \right )^{2} a^{3} \left (a^{2}-b^{2}\right )} \] Input:

int(csc(x)^3/(a+b*sin(x)),x)
 

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)**2 
*b**3 + 2*cos(x)*sin(x)*a**3*b - 2*cos(x)*sin(x)*a*b**3 - cos(x)*a**4 + co 
s(x)*a**2*b**2 + log(tan(x/2))*sin(x)**2*a**4 + log(tan(x/2))*sin(x)**2*a* 
*2*b**2 - 2*log(tan(x/2))*sin(x)**2*b**4)/(2*sin(x)**2*a**3*(a**2 - b**2))