\(\int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx\) [184]

Optimal result

Integrand size = 13, antiderivative size = 112 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}+\frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a} \] Output:

2*b^4*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(1/2)+1/2*b*( 
a^2+2*b^2)*arctanh(cos(x))/a^4-1/3*(2*a^2+3*b^2)*cot(x)/a^3+1/2*b*cot(x)*c 
sc(x)/a^2-1/3*cot(x)*csc(x)^2/a
 

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {\frac {24 b^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a \left (2 a^2+3 b^2\right ) \cos (3 x) \csc ^3(x)-3 a \cot (x) \csc (x) \left (-2 a b+\left (2 a^2+b^2\right ) \csc (x)\right )+6 b \left (a^2+2 b^2\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{12 a^4} \] Input:

Integrate[Csc[x]^4/(a + b*Sin[x]),x]
 

Output:

((24*b^4*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + a*(2* 
a^2 + 3*b^2)*Cos[3*x]*Csc[x]^3 - 3*a*Cot[x]*Csc[x]*(-2*a*b + (2*a^2 + b^2) 
*Csc[x]) + 6*b*(a^2 + 2*b^2)*(Log[Cos[x/2]] - Log[Sin[x/2]]))/(12*a^4)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.231, Rules used = {3042, 3281, 25, 3042, 3534, 25, 3042, 3534, 27, 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.

\(\displaystyle \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (x)^4 (a+b \sin (x))}dx\)

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

Int[Csc[x]^4/(a + b*Sin[x]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

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 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 = 0.49 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {\frac {\tan \left (\frac {x}{2}\right )^{3} a^{2}}{3}-a b \tan \left (\frac {x}{2}\right )^{2}+3 \tan \left (\frac {x}{2}\right ) a^{2}+4 b^{2} \tan \left (\frac {x}{2}\right )}{8 a^{3}}+\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} \sqrt {a^{2}-b^{2}}}-\frac {1}{24 a \tan \left (\frac {x}{2}\right )^{3}}-\frac {3 a^{2}+4 b^{2}}{8 a^{3} \tan \left (\frac {x}{2}\right )}+\frac {b}{8 a^{2} \tan \left (\frac {x}{2}\right )^{2}}-\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{4}}\)	\(156\)
risch	\(-\frac {6 i b^{2} {\mathrm e}^{4 i x}+3 a b \,{\mathrm e}^{5 i x}-12 i a^{2} {\mathrm e}^{2 i x}-12 i b^{2} {\mathrm e}^{2 i x}+4 i a^{2}+6 i b^{2}-3 a b \,{\mathrm e}^{i x}}{3 a^{3} \left ({\mathrm e}^{2 i x}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i x}+1\right )}{2 a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i x}+1\right )}{a^{4}}-\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, a^{4}}+\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, a^{4}}-\frac {b \ln \left ({\mathrm e}^{i x}-1\right )}{2 a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i x}-1\right )}{a^{4}}\)	\(268\)

Input:

int(csc(x)^4/(a+b*sin(x)),x,method=_RETURNVERBOSE)
 

Output:

1/8/a^3*(1/3*tan(1/2*x)^3*a^2-a*b*tan(1/2*x)^2+3*tan(1/2*x)*a^2+4*b^2*tan( 
1/2*x))+2/a^4*b^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2 
)^(1/2))-1/24/a/tan(1/2*x)^3-1/8*(3*a^2+4*b^2)/a^3/tan(1/2*x)+1/8/a^2*b/ta 
n(1/2*x)^2-1/2/a^4*b*(a^2+2*b^2)*ln(tan(1/2*x))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (98) = 196\).

Time = 0.20 (sec) , antiderivative size = 577, normalized size of antiderivative = 5.15 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx =\text {Too large to display} \] Input:

integrate(csc(x)^4/(a+b*sin(x)),x, algorithm="fricas")
 

Output:

[1/12*(4*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(x)^3 + 6*(b^4*cos(x)^2 - b^4)*sqr 
t(-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))*sin(x) + 6*(a^4*b - a^2*b^3)*cos(x)*sin(x) + 3*(a^4*b + a^2 
*b^3 - 2*b^5 - (a^4*b + a^2*b^3 - 2*b^5)*cos(x)^2)*log(1/2*cos(x) + 1/2)*s 
in(x) - 3*(a^4*b + a^2*b^3 - 2*b^5 - (a^4*b + a^2*b^3 - 2*b^5)*cos(x)^2)*l 
og(-1/2*cos(x) + 1/2)*sin(x) - 12*(a^5 - a*b^4)*cos(x))/((a^6 - a^4*b^2 - 
(a^6 - a^4*b^2)*cos(x)^2)*sin(x)), 1/12*(4*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos 
(x)^3 + 12*(b^4*cos(x)^2 - b^4)*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sq 
rt(a^2 - b^2)*cos(x)))*sin(x) + 6*(a^4*b - a^2*b^3)*cos(x)*sin(x) + 3*(a^4 
*b + a^2*b^3 - 2*b^5 - (a^4*b + a^2*b^3 - 2*b^5)*cos(x)^2)*log(1/2*cos(x) 
+ 1/2)*sin(x) - 3*(a^4*b + a^2*b^3 - 2*b^5 - (a^4*b + a^2*b^3 - 2*b^5)*cos 
(x)^2)*log(-1/2*cos(x) + 1/2)*sin(x) - 12*(a^5 - a*b^4)*cos(x))/((a^6 - a^ 
4*b^2 - (a^6 - a^4*b^2)*cos(x)^2)*sin(x))]
 

Sympy [F]

\[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\int \frac {\csc ^{4}{\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \] Input:

integrate(csc(x)**4/(a+b*sin(x)),x)
 

Output:

Integral(csc(x)**4/(a + b*sin(x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(csc(x)^4/(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 = 0.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.73 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{2} + 9 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, a^{3}} - \frac {{\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a^{4}} + \frac {22 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 44 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right ) - a^{3}}{24 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3}} \] Input:

integrate(csc(x)^4/(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^4/(sqrt(a^2 - b^2)*a^4) + 1/24*(a^2*tan(1/2*x)^3 - 3*a*b*tan(1/2* 
x)^2 + 9*a^2*tan(1/2*x) + 12*b^2*tan(1/2*x))/a^3 - 1/2*(a^2*b + 2*b^3)*log 
(abs(tan(1/2*x)))/a^4 + 1/24*(22*a^2*b*tan(1/2*x)^3 + 44*b^3*tan(1/2*x)^3 
- 9*a^3*tan(1/2*x)^2 - 12*a*b^2*tan(1/2*x)^2 + 3*a^2*b*tan(1/2*x) - a^3)/( 
a^4*tan(1/2*x)^3)
 

Mupad [B] (verification not implemented)

Time = 17.31 (sec) , antiderivative size = 586, normalized size of antiderivative = 5.23 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx =\text {Too large to display} \] Input:

int(1/(sin(x)^4*(a + b*sin(x))),x)
 

Output:

(a^5*(cos(3*x)/12 - cos(x)/4) - a*((b^4*cos(3*x))/8 - (b^4*cos(x))/8) + a^ 
4*((b*sin(2*x))/8 - (3*b*log(sin(x/2)/cos(x/2))*sin(x))/16 + (b*sin(3*x)*l 
og(sin(x/2)/cos(x/2)))/16) - a^2*((b^3*sin(2*x))/8 - (b^3*sin(3*x)*log(sin 
(x/2)/cos(x/2)))/16 + (3*b^3*log(sin(x/2)/cos(x/2))*sin(x))/16) + a^3*((b^ 
2*cos(3*x))/24 + (b^2*cos(x))/8) - (b^5*sin(3*x)*log(sin(x/2)/cos(x/2)))/8 
 + (3*b^5*log(sin(x/2)/cos(x/2))*sin(x))/8 + (b^4*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)))*sin(3*x)*(b^2 - a^2)^(1/2)*1i)/4 - (b^4*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)))*sin(x)*(b^2 - a^2)^(1/2)*3i)/4)/((3*a^6*sin(x))/ 
8 - (a^6*sin(3*x))/8 + (a^4*b^2*sin(3*x))/8 - (3*a^4*b^2*sin(x))/8)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.64 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{3} b^{4}-4 \cos \left (x \right ) \sin \left (x \right )^{2} a^{5}-2 \cos \left (x \right ) \sin \left (x \right )^{2} a^{3} b^{2}+6 \cos \left (x \right ) \sin \left (x \right )^{2} a \,b^{4}+3 \cos \left (x \right ) \sin \left (x \right ) a^{4} b -3 \cos \left (x \right ) \sin \left (x \right ) a^{2} b^{3}-2 \cos \left (x \right ) a^{5}+2 \cos \left (x \right ) a^{3} b^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{3} a^{4} b -3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{3} a^{2} b^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{3} b^{5}}{6 \sin \left (x \right )^{3} a^{4} \left (a^{2}-b^{2}\right )} \] Input:

int(csc(x)^4/(a+b*sin(x)),x)
 

Output:

(12*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)**3*b 
**4 - 4*cos(x)*sin(x)**2*a**5 - 2*cos(x)*sin(x)**2*a**3*b**2 + 6*cos(x)*si 
n(x)**2*a*b**4 + 3*cos(x)*sin(x)*a**4*b - 3*cos(x)*sin(x)*a**2*b**3 - 2*co 
s(x)*a**5 + 2*cos(x)*a**3*b**2 - 3*log(tan(x/2))*sin(x)**3*a**4*b - 3*log( 
tan(x/2))*sin(x)**3*a**2*b**3 + 6*log(tan(x/2))*sin(x)**3*b**5)/(6*sin(x)* 
*3*a**4*(a**2 - b**2))