\(\int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1466]

Optimal result

Integrand size = 29, antiderivative size = 295 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}+\frac {2 b^5 \left (5 a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2} d}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}-\frac {\left (a^2+3 b^2\right ) \text {arctanh}(\cos (c+d x))}{a^4 d}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^6 \cos (c+d x)}{a^3 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \] Output:

2*b^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(5/2) 
/d+2*b^5*(5*a^2-3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^ 
4/(a^2-b^2)^(5/2)/d-1/2*arctanh(cos(d*x+c))/a^2/d-(a^2+3*b^2)*arctanh(cos( 
d*x+c))/a^4/d+2*b*cot(d*x+c)/a^3/d-1/2*cot(d*x+c)*csc(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 
^6*cos(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 7.51 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.21 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {6 b^5 \left (2 a^2-b^2\right ) \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2} d}+\frac {b \cot \left (\frac {1}{2} (c+d x)\right )}{a^3 d}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}-\frac {3 \left (a^2+2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {3 \left (a^2+2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^6 \cos (c+d x)}{a^3 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))}-\frac {b \tan \left (\frac {1}{2} (c+d x)\right )}{a^3 d} \] Input:

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

Output:

(6*b^5*(2*a^2 - b^2)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[ 
(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(5/2)*d) + (b*Cot[(c + d 
*x)/2])/(a^3*d) - Csc[(c + d*x)/2]^2/(8*a^2*d) - (3*(a^2 + 2*b^2)*Log[Cos[ 
(c + d*x)/2]])/(2*a^4*d) + (3*(a^2 + 2*b^2)*Log[Sin[(c + d*x)/2]])/(2*a^4* 
d) + Sec[(c + d*x)/2]^2/(8*a^2*d) + Sin[(c + d*x)/2]/((a + b)^2*d*(Cos[(c 
+ d*x)/2] - Sin[(c + d*x)/2])) - Sin[(c + d*x)/2]/((a - b)^2*d*(Cos[(c + d 
*x)/2] + Sin[(c + d*x)/2])) + (b^6*Cos[c + d*x])/(a^3*(a - b)^2*(a + b)^2* 
d*(a + b*Sin[c + d*x])) - (b*Tan[(c + d*x)/2])/(a^3*d)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 295, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3376

\(\displaystyle \int \left (-\frac {2 b \csc ^2(c+d x)}{a^3}+\frac {\csc ^3(c+d x)}{a^2}+\frac {\left (a^2+3 b^2\right ) \csc (c+d x)}{a^4}+\frac {b^5 \left (5 a^2-3 b^2\right )}{a^4 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b^5}{a^3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {1}{2 (a+b)^2 (\sin (c+d x)-1)}-\frac {1}{2 (a-b)^2 (\sin (c+d x)+1)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 b \cot (c+d x)}{a^3 d}+\frac {2 b^5 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 b^5 \left (5 a^2-3 b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}-\frac {\left (a^2+3 b^2\right ) \text {arctanh}(\cos (c+d x))}{a^4 d}+\frac {b^6 \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}\)

Input:

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

Output:

(2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^ 
(5/2)*d) + (2*b^5*(5*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 
 - b^2]])/(a^4*(a^2 - b^2)^(5/2)*d) - ArcTanh[Cos[c + d*x]]/(2*a^2*d) - (( 
a^2 + 3*b^2)*ArcTanh[Cos[c + d*x]])/(a^4*d) + (2*b*Cot[c + d*x])/(a^3*d) - 
 (Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + Cos[c + d*x]/(2*(a + b)^2*d*(1 - 
Sin[c + d*x])) + Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) + (b^6*Co 
s[c + d*x])/(a^3*(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 = 1.86 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-4 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{3}}+\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {4 b^{5} \left (\frac {\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a b}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {3 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (6 a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{4}}+\frac {b}{a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(261\)
default	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-4 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{3}}+\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {4 b^{5} \left (\frac {\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a b}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {3 \left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (6 a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{4}}+\frac {b}{a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(261\)
risch	\(\frac {8 a^{4} b^{2}-8 a^{2} b^{4}+6 b^{6}-8 i a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+3 i a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}-3 i a^{5} b \,{\mathrm e}^{7 i \left (d x +c \right )}-13 i a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}-9 i a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}+16 i a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a \,b^{5} {\mathrm e}^{7 i \left (d x +c \right )}+9 i a \,b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+11 i a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}-8 i a^{3} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+5 i a^{5} b \,{\mathrm e}^{5 i \left (d x +c \right )}-16 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 a^{4} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 a^{2} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+10 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-4 a^{6} {\mathrm e}^{4 i \left (d x +c \right )}+6 a^{6} {\mathrm e}^{6 i \left (d x +c \right )}+6 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{6} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{6} {\mathrm e}^{6 i \left (d x +c \right )}-6 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \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^{3}}-\frac {6 i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}+\frac {3 i b^{7} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{4}}+\frac {6 i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}-\frac {3 i b^{7} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\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}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{a^{4} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{a^{4} d}\)	\(877\)

Input:

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

Output:

1/d*(1/4/a^3*(1/2*tan(1/2*d*x+1/2*c)^2*a-4*b*tan(1/2*d*x+1/2*c))+1/(a-b)^2 
/(tan(1/2*d*x+1/2*c)+1)+4/a^4*b^5/(a-b)^2/(a+b)^2*((1/2*b^2*tan(1/2*d*x+1/ 
2*c)+1/2*a*b)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+3/2*(2*a^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)-1/8/a^2/tan(1/2*d*x+1/2*c)^2+1/4/a^ 
4*(6*a^2+12*b^2)*ln(tan(1/2*d*x+1/2*c))+b/a^3/tan(1/2*d*x+1/2*c))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 880 vs. \(2 (275) = 550\).

Time = 1.13 (sec) , antiderivative size = 1844, normalized size of antiderivative = 6.25 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

[-1/4*(4*a^9 - 8*a^7*b^2 + 4*a^5*b^4 - 4*(4*a^7*b^2 - 8*a^5*b^4 + 7*a^3*b^ 
6 - 3*a*b^8)*cos(d*x + c)^4 - 6*(a^9 - 5*a^7*b^2 + 7*a^5*b^4 - 5*a^3*b^6 + 
 2*a*b^8)*cos(d*x + c)^2 - 6*((2*a^3*b^5 - a*b^7)*cos(d*x + c)^3 - (2*a^3* 
b^5 - a*b^7)*cos(d*x + c) + ((2*a^2*b^6 - b^8)*cos(d*x + c)^3 - (2*a^2*b^6 
 - b^8)*cos(d*x + c))*sin(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*c 
os(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)) + 3*((a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b 
^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos 
(d*x + c) + ((a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c 
)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c))*sin( 
d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((a^9 - a^7*b^2 - 3*a^5*b^4 + 5* 
a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 
 - 2*a*b^8)*cos(d*x + c) + ((a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b 
^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos 
(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^8*b - 4*a^6 
*b^3 + 2*a^4*b^5 - (5*a^8*b - 13*a^6*b^3 + 11*a^4*b^5 - 3*a^2*b^7)*cos(d*x 
 + c)^2)*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x 
 + c)^3 - (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c) + ((a^10 
*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c)^3 - (a^10*b - 3*a^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(csc(d*x+c)^3*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 [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.43 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {48 \, {\left (2 \, a^{2} b^{5} - b^{7}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {16 \, {\left (2 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{7} - a^{5} b^{2} - a b^{6}\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}} + \frac {12 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:

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

Output:

1/8*(48*(2*a^2*b^5 - b^7)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arcta 
n((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^8 - 2*a^6*b^2 + a^4*b 
^4)*sqrt(a^2 - b^2)) + 16*(2*a^6*b*tan(1/2*d*x + 1/2*c)^3 + b^7*tan(1/2*d* 
x + 1/2*c)^3 - a^7*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*b^2*tan(1/2*d*x + 1/2*c) 
^2 + a*b^6*tan(1/2*d*x + 1/2*c)^2 - 2*a^4*b^3*tan(1/2*d*x + 1/2*c) - b^7*t 
an(1/2*d*x + 1/2*c) - a^7 - a^5*b^2 - a*b^6)/((a^8 - 2*a^6*b^2 + a^4*b^4)* 
(a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 
 1/2*c) - a)) + 12*(a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 + (a^2 
*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c))/a^4 - (18*a^2*tan(1/ 
2*d*x + 1/2*c)^2 + 36*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2 
*c) + a^2)/(a^4*tan(1/2*d*x + 1/2*c)^2))/d
 

Mupad [B] (verification not implemented)

Time = 20.94 (sec) , antiderivative size = 2302, normalized size of antiderivative = 7.80 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

tan(c/2 + (d*x)/2)^2/(8*a^2*d) + ((tan(c/2 + (d*x)/2)^4*(17*a^6 - 32*b^6 + 
 33*a^2*b^4 - 66*a^4*b^2))/(2*(a^4 + b^4 - 2*a^2*b^2)) - a^2/2 + (8*tan(c/ 
2 + (d*x)/2)^2*(a^6 + 2*b^6 - 2*a^2*b^4 + 2*a^4*b^2))/(a^4 + b^4 - 2*a^2*b 
^2) + 3*a*b*tan(c/2 + (d*x)/2) - (4*tan(c/2 + (d*x)/2)^5*(5*a^6*b + 2*b^7 
+ a^2*b^5 - 2*a^4*b^3))/(a*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^ 
3*(a^6*b + 8*b^7 + a^2*b^5 + 14*a^4*b^3))/(a*(a^2 - b^2)^2))/(d*(4*a^4*tan 
(c/2 + (d*x)/2)^2 - 4*a^4*tan(c/2 + (d*x)/2)^6 + 8*a^3*b*tan(c/2 + (d*x)/2 
)^3 - 8*a^3*b*tan(c/2 + (d*x)/2)^5)) - (b*tan(c/2 + (d*x)/2))/(a^3*d) + (l 
og(tan(c/2 + (d*x)/2))*(3*a^2 + 6*b^2))/(2*a^4*d) - (b^5*atan(((b^5*(2*a^2 
 - b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(24*a^22 - 192*a^ 
6*b^16 + 1152*a^8*b^14 - 2760*a^10*b^12 + 3312*a^12*b^10 - 1944*a^14*b^8 + 
 288*a^16*b^6 + 264*a^18*b^4 - 144*a^20*b^2) - 24*a^21*b - 96*a^7*b^15 + 5 
52*a^9*b^13 - 1248*a^11*b^11 + 1368*a^13*b^9 - 672*a^15*b^7 + 24*a^17*b^5 
+ 96*a^19*b^3 + (3*b^5*(2*a^2 - b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(16*a^23 
*b - tan(c/2 + (d*x)/2)*(48*a^24 - 64*a^10*b^14 + 432*a^12*b^12 - 1248*a^1 
4*b^10 + 2000*a^16*b^8 - 1920*a^18*b^6 + 1104*a^20*b^4 - 352*a^22*b^2) + 1 
6*a^11*b^13 - 96*a^13*b^11 + 240*a^15*b^9 - 320*a^17*b^7 + 240*a^19*b^5 - 
96*a^21*b^3))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5* 
a^12*b^2))*3i)/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5 
*a^12*b^2) - (b^5*(2*a^2 - b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(24*a^21*b...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1097, normalized size of antiderivative = 3.72 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(96*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)**3*a**2*b**6 - 48*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)**3*b**8 + 96*s 
qrt(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**3*b**5 - 48*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*b**7 + 12*cos( 
c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**8*b - 12*cos(c + d*x)*lo 
g(tan((c + d*x)/2))*sin(c + d*x)**3*a**6*b**3 - 36*cos(c + d*x)*log(tan((c 
 + d*x)/2))*sin(c + d*x)**3*a**4*b**5 + 60*cos(c + d*x)*log(tan((c + d*x)/ 
2))*sin(c + d*x)**3*a**2*b**7 - 24*cos(c + d*x)*log(tan((c + d*x)/2))*sin( 
c + d*x)**3*b**9 + 12*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a 
**9 - 12*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**7*b**2 - 36 
*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**5*b**4 + 60*cos(c + 
 d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**3*b**6 - 24*cos(c + d*x)*lo 
g(tan((c + d*x)/2))*sin(c + d*x)**2*a*b**8 - 11*cos(c + d*x)*sin(c + d*x)* 
*3*a**8*b + 25*cos(c + d*x)*sin(c + d*x)**3*a**6*b**3 - 17*cos(c + d*x)*si 
n(c + d*x)**3*a**4*b**5 + 3*cos(c + d*x)*sin(c + d*x)**3*a**2*b**7 - 11*co 
s(c + d*x)*sin(c + d*x)**2*a**9 + 25*cos(c + d*x)*sin(c + d*x)**2*a**7*b** 
2 - 17*cos(c + d*x)*sin(c + d*x)**2*a**5*b**4 + 3*cos(c + d*x)*sin(c + d*x 
)**2*a**3*b**6 - 32*sin(c + d*x)**4*a**7*b**2 + 64*sin(c + d*x)**4*a**5...