\(\int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx\) [138]

Optimal result

Integrand size = 17, antiderivative size = 72 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {2 \left (a c^2+b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {x}{2}\right )}{\sqrt {c^2-d^2}}\right )}{c^2 \sqrt {c^2-d^2}}+\frac {b d \text {arctanh}(\cos (x))}{c^2}-\frac {b \cot (x)}{c} \] Output:

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {\csc \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right ) \left (\frac {2 \left (a c^2+b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {x}{2}\right )}{\sqrt {c^2-d^2}}\right ) \sin (x)}{\sqrt {c^2-d^2}}-b \left (c \cos (x)+d \left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right ) \sin (x)\right )\right )}{2 c^2} \] Input:

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

Output:

(Csc[x/2]*Sec[x/2]*((2*(a*c^2 + b*d^2)*ArcTan[(d + c*Tan[x/2])/Sqrt[c^2 - 
d^2]]*Sin[x])/Sqrt[c^2 - d^2] - b*(c*Cos[x] + d*(-Log[Cos[x/2]] + Log[Sin[ 
x/2]])*Sin[x])))/(2*c^2)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.18, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {3042, 4721, 3042, 3535, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4721

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {b d-a c \sin (x)}{\sin (x) (c+d \sin (x))}dx}{c}-\frac {b \cot (x)}{c}\)

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

-(((-2*(a*c^2 + b*d^2)*ArcTan[(2*d + 2*c*Tan[x/2])/(2*Sqrt[c^2 - d^2])])/( 
c*Sqrt[c^2 - d^2]) - (b*d*ArcTanh[Cos[x]])/c)/c) - (b*Cot[x])/c
 

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 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 3535

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[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*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, 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]
 

rule 4721

Int[(csc[(a_.) + (b_.)*(x_)]^2*(C_.) + (A_))*(u_), x_Symbol] :> Int[Activat 
eTrig[u]*((C + A*Sin[a + b*x]^2)/Sin[a + b*x]^2), x] /; FreeQ[{a, b, A, C}, 
 x] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {b \tan \left (\frac {x}{2}\right )}{2 c}-\frac {b}{2 c \tan \left (\frac {x}{2}\right )}-\frac {d b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{c^{2}}+\frac {\left (4 a \,c^{2}+4 b \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 c^{2} \sqrt {c^{2}-d^{2}}}\)	\(90\)
parts	\(\frac {2 a \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\sqrt {c^{2}-d^{2}}}+b \left (\frac {\tan \left (\frac {x}{2}\right )}{2 c}+\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{c^{2} \sqrt {c^{2}-d^{2}}}-\frac {1}{2 c \tan \left (\frac {x}{2}\right )}-\frac {d \ln \left (\tan \left (\frac {x}{2}\right )\right )}{c^{2}}\right )\)	\(119\)
risch	\(-\frac {2 i b}{c \left ({\mathrm e}^{2 i x}-1\right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) a}{\sqrt {-c^{2}+d^{2}}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) b \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, c^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) a}{\sqrt {-c^{2}+d^{2}}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) b \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, c^{2}}+\frac {d b \ln \left ({\mathrm e}^{i x}+1\right )}{c^{2}}-\frac {d b \ln \left ({\mathrm e}^{i x}-1\right )}{c^{2}}\)	\(297\)

Input:

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

Output:

1/2*b/c*tan(1/2*x)-1/2*b/c/tan(1/2*x)-d*b/c^2*ln(tan(1/2*x))+1/2/c^2*(4*a* 
c^2+4*b*d^2)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*x)+2*d)/(c^2-d^2)^(1/ 
2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (66) = 132\).

Time = 0.40 (sec) , antiderivative size = 332, normalized size of antiderivative = 4.61 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\left [-\frac {{\left (a c^{2} + b d^{2}\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (x\right )^{2} - 2 \, c d \sin \left (x\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (x\right ) \sin \left (x\right ) + d \cos \left (x\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (x\right )^{2} - 2 \, c d \sin \left (x\right ) - c^{2} - d^{2}}\right ) \sin \left (x\right ) - {\left (b c^{2} d - b d^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + {\left (b c^{2} d - b d^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + 2 \, {\left (b c^{3} - b c d^{2}\right )} \cos \left (x\right )}{2 \, {\left (c^{4} - c^{2} d^{2}\right )} \sin \left (x\right )}, -\frac {2 \, {\left (a c^{2} + b d^{2}\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (x\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (x\right )}\right ) \sin \left (x\right ) - {\left (b c^{2} d - b d^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + {\left (b c^{2} d - b d^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + 2 \, {\left (b c^{3} - b c d^{2}\right )} \cos \left (x\right )}{2 \, {\left (c^{4} - c^{2} d^{2}\right )} \sin \left (x\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*csc(x)^2)/(c+d*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*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.53 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=-\frac {b d \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{c^{2}} + \frac {b \tan \left (\frac {1}{2} \, x\right )}{2 \, c} + \frac {2 \, {\left (a c^{2} + b d^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, x\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} c^{2}} + \frac {2 \, b d \tan \left (\frac {1}{2} \, x\right ) - b c}{2 \, c^{2} \tan \left (\frac {1}{2} \, x\right )} \] Input:

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

Output:

-b*d*log(abs(tan(1/2*x)))/c^2 + 1/2*b*tan(1/2*x)/c + 2*(a*c^2 + b*d^2)*(pi 
*floor(1/2*x/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*x) + d)/sqrt(c^2 - d^2)) 
)/(sqrt(c^2 - d^2)*c^2) + 1/2*(2*b*d*tan(1/2*x) - b*c)/(c^2*tan(1/2*x))
 

Mupad [B] (verification not implemented)

Time = 16.63 (sec) , antiderivative size = 463, normalized size of antiderivative = 6.43 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {b\,d^3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )-b\,c^2\,d\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )+a\,c^2\,\mathrm {atan}\left (\frac {a\,c^3\,\sqrt {d^2-c^2}\,1{}\mathrm {i}+b\,d^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,4{}\mathrm {i}+b\,c\,d^2\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+a\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}-b\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,1{}\mathrm {i}}{4\,b\,d^4\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,c^4\,\mathrm {tan}\left (\frac {x}{2}\right )+a\,c^3\,d+2\,b\,c\,d^3-b\,c^3\,d+2\,a\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )-3\,b\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+b\,d^2\,\mathrm {atan}\left (\frac {a\,c^3\,\sqrt {d^2-c^2}\,1{}\mathrm {i}+b\,d^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,4{}\mathrm {i}+b\,c\,d^2\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+a\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}-b\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,1{}\mathrm {i}}{4\,b\,d^4\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,c^4\,\mathrm {tan}\left (\frac {x}{2}\right )+a\,c^3\,d+2\,b\,c\,d^3-b\,c^3\,d+2\,a\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )-3\,b\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}}{c^4-c^2\,d^2}-\frac {b\,c^3-b\,c\,d^2}{c^4\,\mathrm {tan}\left (x\right )-c^2\,d^2\,\mathrm {tan}\left (x\right )} \] Input:

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

Output:

(b*d^3*log(tan(x/2)) + a*c^2*atan((a*c^3*(d^2 - c^2)^(1/2)*1i + b*d^3*tan( 
x/2)*(d^2 - c^2)^(1/2)*4i + b*c*d^2*(d^2 - c^2)^(1/2)*2i + a*c^2*d*tan(x/2 
)*(d^2 - c^2)^(1/2)*2i - b*c^2*d*tan(x/2)*(d^2 - c^2)^(1/2)*1i)/(4*b*d^4*t 
an(x/2) - a*c^4*tan(x/2) + a*c^3*d + 2*b*c*d^3 - b*c^3*d + 2*a*c^2*d^2*tan 
(x/2) - 3*b*c^2*d^2*tan(x/2)))*(d^2 - c^2)^(1/2)*2i + b*d^2*atan((a*c^3*(d 
^2 - c^2)^(1/2)*1i + b*d^3*tan(x/2)*(d^2 - c^2)^(1/2)*4i + b*c*d^2*(d^2 - 
c^2)^(1/2)*2i + a*c^2*d*tan(x/2)*(d^2 - c^2)^(1/2)*2i - b*c^2*d*tan(x/2)*( 
d^2 - c^2)^(1/2)*1i)/(4*b*d^4*tan(x/2) - a*c^4*tan(x/2) + a*c^3*d + 2*b*c* 
d^3 - b*c^3*d + 2*a*c^2*d^2*tan(x/2) - 3*b*c^2*d^2*tan(x/2)))*(d^2 - c^2)^ 
(1/2)*2i - b*c^2*d*log(tan(x/2)))/(c^4 - c^2*d^2) - (b*c^3 - b*c*d^2)/(c^4 
*tan(x) - c^2*d^2*tan(x))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.97 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (x \right ) a \,c^{2}+2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (x \right ) b \,d^{2}-\cos \left (x \right ) b \,c^{3}+\cos \left (x \right ) b c \,d^{2}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right ) b \,c^{2} d +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right ) b \,d^{3}}{\sin \left (x \right ) c^{2} \left (c^{2}-d^{2}\right )} \] Input:

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

Output:

(2*sqrt(c**2 - d**2)*atan((tan(x/2)*c + d)/sqrt(c**2 - d**2))*sin(x)*a*c** 
2 + 2*sqrt(c**2 - d**2)*atan((tan(x/2)*c + d)/sqrt(c**2 - d**2))*sin(x)*b* 
d**2 - cos(x)*b*c**3 + cos(x)*b*c*d**2 - log(tan(x/2))*sin(x)*b*c**2*d + l 
og(tan(x/2))*sin(x)*b*d**3)/(sin(x)*c**2*(c**2 - d**2))