\(\int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1465]

Optimal result

Integrand size = 29, antiderivative size = 248 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2} d}-\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2} d}+\frac {2 b \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \] Output:

-2*b^4*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/(a^2-b^2)^(5/2)/ 
d-4*b^4*(2*a^2-b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/( 
a^2-b^2)^(5/2)/d+2*b*arctanh(cos(d*x+c))/a^3/d-cot(d*x+c)/a^2/d+1/2*cos(d* 
x+c)/(a+b)^2/d/(1-sin(d*x+c))-1/2*cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))-b^5* 
cos(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 4.09 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {4 b^4 \left (-5 a^2+2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{a^2}+\frac {4 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}-\frac {4 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 b^5 \cos (c+d x)}{a^2 (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{a^2}}{2 d} \] Input:

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

Output:

((4*b^4*(-5*a^2 + 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]) 
/(a^3*(a^2 - b^2)^(5/2)) - Cot[(c + d*x)/2]/a^2 + (4*b*Log[Cos[(c + d*x)/2 
]])/a^3 - (4*b*Log[Sin[(c + d*x)/2]])/a^3 + (2*Sin[(c + d*x)/2])/((a + b)^ 
2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Sin[(c + d*x)/2])/((a - b)^2 
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (2*b^5*Cos[c + d*x])/(a^2*(a - b 
)^2*(a + b)^2*(a + b*Sin[c + d*x])) + Tan[(c + d*x)/2]/a^2)/(2*d)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3376, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3376

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 b \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {2 b^4 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{5/2}}-\frac {b^5 \cos (c+d x)}{a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a^2 d}-\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}\)

Input:

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

Output:

(-2*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^( 
5/2)*d) - (4*b^4*(2*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - 
b^2]])/(a^3*(a^2 - b^2)^(5/2)*d) + (2*b*ArcTanh[Cos[c + d*x]])/(a^3*d) - C 
ot[c + d*x]/(a^2*d) + Cos[c + d*x]/(2*(a + b)^2*d*(1 - Sin[c + d*x])) - Co 
s[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) - (b^5*Cos[c + d*x])/(a^2*(a 
^2 - b^2)^2*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3376

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) 
 + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ 
e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr 
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( 
LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
 

Maple [A] (verified)

Time = 1.58 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{4} \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (5 a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a^{3}}-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(215\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 b^{4} \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (5 a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a^{3}}-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(215\)
risch	\(-\frac {2 i \left (-2 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a^{4} b +2 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-4 i a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+4 \,{\mathrm e}^{i \left (d x +c \right )} a^{5}+3 i a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+2 i b^{5}-4 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}-2 i b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+3 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}-i a^{2} b^{3}-a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) \left (a^{2}-b^{2}\right )^{2} d \,a^{2}}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {2 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}-\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}+\frac {2 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}\)	\(674\)

Input:

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

Output:

1/d*(1/2*tan(1/2*d*x+1/2*c)/a^2-1/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-1/2/a^2/t 
an(1/2*d*x+1/2*c)-2/a^3*b*ln(tan(1/2*d*x+1/2*c))-2*b^4/(a-b)^2/(a+b)^2/a^3 
*((b^2*tan(1/2*d*x+1/2*c)+a*b)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2 
*c)+a)+(5*a^2-2*b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2* 
b)/(a^2-b^2)^(1/2)))-1/(a-b)^2/(tan(1/2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 634 vs. \(2 (232) = 464\).

Time = 0.57 (sec) , antiderivative size = 1355, normalized size of antiderivative = 5.46 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/2*(2*a^8 - 4*a^6*b^2 + 2*a^4*b^4 - 2*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 - 
a^2*b^6)*cos(d*x + c)^2 - ((5*a^2*b^5 - 2*b^7)*cos(d*x + c)^3 - (5*a^3*b^4 
 - 2*a*b^6)*cos(d*x + c)*sin(d*x + c) - (5*a^2*b^5 - 2*b^7)*cos(d*x + c))* 
sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - 
a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b 
^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*((a^6*b^2 
- 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^3 - (a^7*b - 3*a^5*b^3 + 3*a^3 
*b^5 - a*b^7)*cos(d*x + c)*sin(d*x + c) - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 
 - b^8)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 2*((a^6*b^2 - 3*a^4*b^ 
4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^3 - (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b 
^7)*cos(d*x + c)*sin(d*x + c) - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*co 
s(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5 
+ (2*a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - 2*a*b^7)*cos(d*x + c)^2)*sin(d*x + c) 
)/((a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(d*x + c)^3 - (a^10 - 3* 
a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c)*sin(d*x + c) - (a^9*b - 3*a^ 
7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(d*x + c)), -(a^8 - 2*a^6*b^2 + a^4*b^4 
- (2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 - ((5*a^2*b^5 - 
 2*b^7)*cos(d*x + c)^3 - (5*a^3*b^4 - 2*a*b^6)*cos(d*x + c)*sin(d*x + c) - 
 (5*a^2*b^5 - 2*b^7)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) 
 + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - ((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 523 vs. \(2 (232) = 464\).

Time = 0.43 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.11 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/10*(20*(5*a^2*b^4 - 2*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + a 
rctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a 
^3*b^4)*sqrt(a^2 - b^2)) - (4*a^5*b*tan(1/2*d*x + 1/2*c)^5 - 8*a^3*b^3*tan 
(1/2*d*x + 1/2*c)^5 + 4*a*b^5*tan(1/2*d*x + 1/2*c)^5 - 25*a^6*tan(1/2*d*x 
+ 1/2*c)^4 - 2*a^4*b^2*tan(1/2*d*x + 1/2*c)^4 - 21*a^2*b^4*tan(1/2*d*x + 1 
/2*c)^4 - 12*b^6*tan(1/2*d*x + 1/2*c)^4 - 10*a^5*b*tan(1/2*d*x + 1/2*c)^3 
- 20*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 30*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 20 
*a^6*tan(1/2*d*x + 1/2*c)^2 + 52*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 + 16*a^2*b 
^4*tan(1/2*d*x + 1/2*c)^2 + 12*b^6*tan(1/2*d*x + 1/2*c)^2 + 46*a^5*b*tan(1 
/2*d*x + 1/2*c) - 12*a^3*b^3*tan(1/2*d*x + 1/2*c) + 26*a*b^5*tan(1/2*d*x + 
 1/2*c) + 5*a^6 - 10*a^4*b^2 + 5*a^2*b^4)/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a* 
tan(1/2*d*x + 1/2*c)^5 + 2*b*tan(1/2*d*x + 1/2*c)^4 - 2*b*tan(1/2*d*x + 1/ 
2*c)^2 - a*tan(1/2*d*x + 1/2*c))) + 20*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^ 
3 - 5*tan(1/2*d*x + 1/2*c)/a^2)/d
 

Mupad [B] (verification not implemented)

Time = 20.40 (sec) , antiderivative size = 2151, normalized size of antiderivative = 8.67 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(a + (2*tan(c/2 + (d*x)/2)*(5*a^4*b + 3*b^5 - 2*a^2*b^3))/(a^4 + b^4 - 2*a 
^2*b^2) + (4*tan(c/2 + (d*x)/2)^2*(b^6 - a^6 + 3*a^4*b^2))/(a*(a^2 - b^2)^ 
2) - (2*b*tan(c/2 + (d*x)/2)^3*(a^4 + 3*b^4 + 2*a^2*b^2))/(a^4 + b^4 - 2*a 
^2*b^2) - (tan(c/2 + (d*x)/2)^4*(5*a^6 + 4*b^6 + a^2*b^4 + 2*a^4*b^2))/(a* 
(a^4 + b^4 - 2*a^2*b^2)))/(d*(2*a^3*tan(c/2 + (d*x)/2)^5 - 2*a^3*tan(c/2 + 
 (d*x)/2) - 4*a^2*b*tan(c/2 + (d*x)/2)^2 + 4*a^2*b*tan(c/2 + (d*x)/2)^4)) 
+ tan(c/2 + (d*x)/2)/(2*a^2*d) - (2*b*log(tan(c/2 + (d*x)/2)))/(a^3*d) + ( 
b^4*atan(((b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d* 
x)/2)*(4*a^18*b - 16*a^4*b^15 + 104*a^6*b^13 - 272*a^8*b^11 + 372*a^10*b^9 
 - 288*a^12*b^7 + 128*a^14*b^5 - 32*a^16*b^3) - 8*a^5*b^14 + 50*a^7*b^12 - 
 124*a^9*b^10 + 156*a^11*b^8 - 104*a^13*b^6 + 34*a^15*b^4 - 4*a^17*b^2 + ( 
b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a^20*b - tan(c/2 + (d* 
x)/2)*(6*a^21 - 8*a^7*b^14 + 54*a^9*b^12 - 156*a^11*b^10 + 250*a^13*b^8 - 
240*a^15*b^6 + 138*a^17*b^4 - 44*a^19*b^2) + 2*a^8*b^13 - 12*a^10*b^11 + 3 
0*a^12*b^9 - 40*a^14*b^7 + 30*a^16*b^5 - 12*a^18*b^3))/(a^13 - a^3*b^10 + 
5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2))*1i)/(a^13 - a^3*b^10 + 
5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2) - (b^4*(5*a^2 - 2*b^2)*( 
-(a + b)^5*(a - b)^5)^(1/2)*(8*a^5*b^14 - tan(c/2 + (d*x)/2)*(4*a^18*b - 1 
6*a^4*b^15 + 104*a^6*b^13 - 272*a^8*b^11 + 372*a^10*b^9 - 288*a^12*b^7 + 1 
28*a^14*b^5 - 32*a^16*b^3) - 50*a^7*b^12 + 124*a^9*b^10 - 156*a^11*b^8 ...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 947, normalized size of antiderivative = 3.82 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 40*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* 
cos(c + d*x)*sin(c + d*x)**2*a**2*b**6 + 16*sqrt(a**2 - b**2)*atan((tan((c 
 + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*b**8 - 4 
0*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c 
 + d*x)*sin(c + d*x)*a**3*b**5 + 16*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 
2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)*a*b**7 - 8*cos(c + 
d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**6*b**3 + 24*cos(c + d*x)*log 
(tan((c + d*x)/2))*sin(c + d*x)**2*a**4*b**5 - 24*cos(c + d*x)*log(tan((c 
+ d*x)/2))*sin(c + d*x)**2*a**2*b**7 + 8*cos(c + d*x)*log(tan((c + d*x)/2) 
)*sin(c + d*x)**2*b**9 - 8*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x) 
*a**7*b**2 + 24*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**5*b**4 
- 24*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**3*b**6 + 8*cos(c + 
 d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**8 + 5*cos(c + d*x)*sin(c + d 
*x)**2*a**8*b - 11*cos(c + d*x)*sin(c + d*x)**2*a**6*b**3 + 7*cos(c + d*x) 
*sin(c + d*x)**2*a**4*b**5 - cos(c + d*x)*sin(c + d*x)**2*a**2*b**7 + 5*co 
s(c + d*x)*sin(c + d*x)*a**9 - 11*cos(c + d*x)*sin(c + d*x)*a**7*b**2 + 7* 
cos(c + d*x)*sin(c + d*x)*a**5*b**4 - cos(c + d*x)*sin(c + d*x)*a**3*b**6 
+ 8*sin(c + d*x)**3*a**7*b**2 - 12*sin(c + d*x)**3*a**5*b**4 + 12*sin(c + 
d*x)**3*a**3*b**6 - 8*sin(c + d*x)**3*a*b**8 + 8*sin(c + d*x)**2*a**8*b - 
20*sin(c + d*x)**2*a**6*b**3 + 16*sin(c + d*x)**2*a**4*b**5 - 4*sin(c +...