\(\int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1472]

Optimal result

Integrand size = 29, antiderivative size = 424 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {b^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{7/2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \] Output:

-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^2/(a 
^2-b^2)^(7/2)/d-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^2/(a^2-b^2)^(7/2)/d-2*b^4*(10*a^4-9*a^2*b^2+3*b^4)*arctan((b+a*ta 
n(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(7/2)/d+3*b*arctanh(cos(d 
*x+c))/a^4/d-cot(d*x+c)/a^3/d+1/2*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/2* 
cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))-1/2*b^5*cos(d*x+c)/a^2/(a^2-b^2)^2/d/( 
a+b*sin(d*x+c))^2-3/2*b^5*cos(d*x+c)/a/(a^2-b^2)^3/d/(a+b*sin(d*x+c))-2*b^ 
5*(2*a^2-b^2)*cos(d*x+c)/a^3/(a^2-b^2)^3/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 6.74 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=4 \left (-\frac {3 b^4 \left (10 a^4-7 a^2 b^2+2 b^4\right ) \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{4 a^4 \left (a^2-b^2\right )^{7/2} d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d}+\frac {3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^4 d}-\frac {3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^4 d}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{4 (a+b)^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{4 (a-b)^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {b^5 \cos (c+d x)}{8 a^2 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))^2}+\frac {-11 a^2 b^5 \cos (c+d x)+4 b^7 \cos (c+d x)}{8 a^3 (a-b)^3 (a+b)^3 d (a+b \sin (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d}\right ) \] Input:

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

Output:

4*((-3*b^4*(10*a^4 - 7*a^2*b^2 + 2*b^4)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c 
 + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(4*a^4*(a^2 - b^2)^(7/ 
2)*d) - Cot[(c + d*x)/2]/(8*a^3*d) + (3*b*Log[Cos[(c + d*x)/2]])/(4*a^4*d) 
 - (3*b*Log[Sin[(c + d*x)/2]])/(4*a^4*d) + Sin[(c + d*x)/2]/(4*(a + b)^3*d 
*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + Sin[(c + d*x)/2]/(4*(a - b)^3*d* 
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (b^5*Cos[c + d*x])/(8*a^2*(a - b) 
^2*(a + b)^2*d*(a + b*Sin[c + d*x])^2) + (-11*a^2*b^5*Cos[c + d*x] + 4*b^7 
*Cos[c + d*x])/(8*a^3*(a - b)^3*(a + b)^3*d*(a + b*Sin[c + d*x])) + Tan[(c 
 + d*x)/2]/(8*a^3*d))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 424, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3376

\(\displaystyle \int \left (-\frac {3 b \csc (c+d x)}{a^4}+\frac {\csc ^2(c+d x)}{a^3}-\frac {b^4}{a^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^3}+\frac {-10 a^4 b^4+9 a^2 b^6-3 b^8}{a^4 \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {2 b^4 \left (2 a^2-b^2\right )}{a^3 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {1}{2 (a+b)^3 (\sin (c+d x)-1)}+\frac {1}{2 (a-b)^3 (\sin (c+d x)+1)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac {b^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac {3 b^5 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{7/2}}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)}\)

Input:

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

Output:

(-4*b^4*(2*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a 
^2*(a^2 - b^2)^(7/2)*d) - (b^4*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2 
])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(7/2)*d) - (2*b^4*(10*a^4 - 9*a^2*b^ 
2 + 3*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b 
^2)^(7/2)*d) + (3*b*ArcTanh[Cos[c + d*x]])/(a^4*d) - Cot[c + d*x]/(a^3*d) 
+ Cos[c + d*x]/(2*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b 
)^3*d*(1 + Sin[c + d*x])) - (b^5*Cos[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + 
 b*Sin[c + d*x])^2) - (3*b^5*Cos[c + d*x])/(2*a*(a^2 - b^2)^3*d*(a + b*Sin 
[c + d*x])) - (2*b^5*(2*a^2 - b^2)*Cos[c + d*x])/(a^3*(a^2 - b^2)^3*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 = 3.22 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{4} \left (\frac {\left (\frac {13}{2} a^{3} b^{2}-3 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {b \left (12 a^{4}+19 a^{2} b^{2}-10 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {7 b^{2} a \left (5 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (12 a^{2}-5 b^{2}\right )}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (10 a^{4}-7 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a^{4}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}}{d}\)	\(311\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{4} \left (\frac {\left (\frac {13}{2} a^{3} b^{2}-3 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {b \left (12 a^{4}+19 a^{2} b^{2}-10 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {7 b^{2} a \left (5 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (12 a^{2}-5 b^{2}\right )}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (10 a^{4}-7 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a^{4}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}}{d}\)	\(311\)
risch	\(\text {Expression too large to display}\)	\(1106\)

Input:

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

Output:

1/d*(1/2*tan(1/2*d*x+1/2*c)/a^3-1/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-2*b^4/(a- 
b)^3/(a+b)^3/a^4*(((13/2*a^3*b^2-3*a*b^4)*tan(1/2*d*x+1/2*c)^3+1/2*b*(12*a 
^4+19*a^2*b^2-10*b^4)*tan(1/2*d*x+1/2*c)^2+7/2*b^2*a*(5*a^2-2*b^2)*tan(1/2 
*d*x+1/2*c)+1/2*a^2*b*(12*a^2-5*b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2* 
d*x+1/2*c)+a)^2+3/2*(10*a^4-7*a^2*b^2+2*b^4)/(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)^3/(tan(1/2*d*x+1/2*c) 
-1)-1/2/a^3/tan(1/2*d*x+1/2*c)-3/a^4*b*ln(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. 1028 vs. \(2 (399) = 798\).

Time = 1.40 (sec) , antiderivative size = 2140, normalized size of antiderivative = 5.05 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(4*a^11 - 12*a^9*b^2 + 12*a^7*b^4 - 4*a^5*b^6 + 2*(4*a^9*b^2 - 4*a^7 
*b^4 + 17*a^5*b^6 - 23*a^3*b^8 + 6*a*b^10)*cos(d*x + c)^4 - 2*(4*a^11 - 10 
*a^9*b^2 + 14*a^7*b^4 + 7*a^5*b^6 - 21*a^3*b^8 + 6*a*b^10)*cos(d*x + c)^2 
- 3*(2*(10*a^5*b^5 - 7*a^3*b^7 + 2*a*b^9)*cos(d*x + c)^3 - 2*(10*a^5*b^5 - 
 7*a^3*b^7 + 2*a*b^9)*cos(d*x + c) + ((10*a^4*b^6 - 7*a^2*b^8 + 2*b^10)*co 
s(d*x + c)^3 - (10*a^6*b^4 + 3*a^4*b^6 - 5*a^2*b^8 + 2*b^10)*cos(d*x + c)) 
*sin(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)) 
 - 6*(2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c 
)^3 - 2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c 
) + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*cos(d*x + c)^3 - 
 (a^10*b - 3*a^8*b^3 + 2*a^6*b^5 + 2*a^4*b^7 - 3*a^2*b^9 + b^11)*cos(d*x + 
 c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 6*(2*(a^9*b^2 - 4*a^7*b^4 
 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c)^3 - 2*(a^9*b^2 - 4*a^7*b^4 
 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c) + ((a^8*b^3 - 4*a^6*b^5 + 
6*a^4*b^7 - 4*a^2*b^9 + b^11)*cos(d*x + c)^3 - (a^10*b - 3*a^8*b^3 + 2*a^6 
*b^5 + 2*a^4*b^7 - 3*a^2*b^9 + b^11)*cos(d*x + c))*sin(d*x + c))*log(-1/2* 
cos(d*x + c) + 1/2) - 2*(2*a^10*b - 6*a^8*b^3 + 6*a^6*b^5 - 2*a^4*b^7 + (8 
*a^10*b - 14*a^8*b^3 + 28*a^6*b^5 - 31*a^4*b^7 + 9*a^2*b^9)*cos(d*x + c...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,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 = 633, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(6*(10*a^4*b^4 - 7*a^2*b^6 + 2*b^8)*(pi*floor(1/2*(d*x + c)/pi + 1/2) 
*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^10 - 3 
*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*sqrt(a^2 - b^2)) - (2*a^6*b*tan(1/2*d*x + 
1/2*c)^3 - 6*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b^5*tan(1/2*d*x + 1/2* 
c)^3 - 2*b^7*tan(1/2*d*x + 1/2*c)^3 - 5*a^7*tan(1/2*d*x + 1/2*c)^2 - 9*a^5 
*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^3*b^4*tan(1/2*d*x + 1/2*c)^2 + a*b^6*tan 
(1/2*d*x + 1/2*c)^2 + 10*a^6*b*tan(1/2*d*x + 1/2*c) + 10*a^4*b^3*tan(1/2*d 
*x + 1/2*c) - 6*a^2*b^5*tan(1/2*d*x + 1/2*c) + 2*b^7*tan(1/2*d*x + 1/2*c) 
+ a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)/((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^ 
4*b^6)*(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))) + 2*(13*a^3*b^6*ta 
n(1/2*d*x + 1/2*c)^3 - 6*a*b^8*tan(1/2*d*x + 1/2*c)^3 + 12*a^4*b^5*tan(1/2 
*d*x + 1/2*c)^2 + 19*a^2*b^7*tan(1/2*d*x + 1/2*c)^2 - 10*b^9*tan(1/2*d*x + 
 1/2*c)^2 + 35*a^3*b^6*tan(1/2*d*x + 1/2*c) - 14*a*b^8*tan(1/2*d*x + 1/2*c 
) + 12*a^4*b^5 - 5*a^2*b^7)/((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*(a*t 
an(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) + 6*b*log(abs(tan 
(1/2*d*x + 1/2*c)))/a^4 - tan(1/2*d*x + 1/2*c)/a^3)/d
 

Mupad [B] (verification not implemented)

Time = 22.84 (sec) , antiderivative size = 3122, normalized size of antiderivative = 7.36 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

tan(c/2 + (d*x)/2)/(2*a^3*d) - ((tan(c/2 + (d*x)/2)^6*(5*a^8 - 12*b^8 + 25 
*a^2*b^6 + 3*a^4*b^4 + 9*a^6*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - a 
^2 + (tan(c/2 + (d*x)/2)^4*(9*a^8 - 20*b^8 + 55*a^2*b^6 + 23*a^4*b^4 - 7*a 
^6*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (tan(c/2 + (d*x)/2)^2*(81*a 
^2*b^6 - 32*b^8 - 3*a^8 + 7*a^4*b^4 + 37*a^6*b^2))/(a^6 - b^6 + 3*a^2*b^4 
- 3*a^4*b^2) + (2*tan(c/2 + (d*x)/2)*(7*a*b^7 - 8*a^7*b - 18*a^3*b^5 + 4*a 
^5*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (4*tan(c/2 + (d*x)/2)^3*(2* 
a^8*b - 5*b^9 + 12*a^2*b^7 + 4*a^4*b^5 + 2*a^6*b^3))/(a*(a^6 - b^6 + 3*a^2 
*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(4*a^8*b - 10*b^9 + 17*a^2*b^ 
7 + 18*a^4*b^5 + 16*a^6*b^3))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/(d* 
(2*a^5*tan(c/2 + (d*x)/2)^7 - tan(c/2 + (d*x)/2)^3*(2*a^5 + 8*a^3*b^2) + t 
an(c/2 + (d*x)/2)^5*(2*a^5 + 8*a^3*b^2) - 2*a^5*tan(c/2 + (d*x)/2) - 8*a^4 
*b*tan(c/2 + (d*x)/2)^2 + 8*a^4*b*tan(c/2 + (d*x)/2)^6)) - (3*b*log(tan(c/ 
2 + (d*x)/2)))/(a^4*d) - (b^4*atan(((b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10* 
a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6*a^26*b + 24*a^6*b^21 - 228 
*a^8*b^19 + 978*a^10*b^17 - 2454*a^12*b^15 + 3936*a^14*b^13 - 4170*a^16*b^ 
11 + 2928*a^18*b^9 - 1338*a^20*b^7 + 384*a^22*b^5 - 66*a^24*b^3) + 12*a^7* 
b^20 - 111*a^9*b^18 + 462*a^11*b^16 - 1119*a^13*b^14 + 1716*a^15*b^12 - 17 
07*a^17*b^10 + 1086*a^19*b^8 - 417*a^21*b^6 + 84*a^23*b^4 - 6*a^25*b^2 - ( 
3*b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/...
 

Reduce [B] (verification not implemented)

Time = 191.67 (sec) , antiderivative size = 1923, normalized size of antiderivative = 4.54 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 120*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**4*b**7 + 84*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* 
*9 - 24*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**11 - 240*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**5*b**6 
+ 168*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c 
os(c + d*x)*sin(c + d*x)**2*a**3*b**8 - 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**10 - 
 120*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*co 
s(c + d*x)*sin(c + d*x)*a**6*b**5 + 84*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**4*b**7 - 24*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)*a**2*b**9 - 12*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c 
+ d*x)**3*a**8*b**4 + 48*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)** 
3*a**6*b**6 - 72*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**4*b 
**8 + 48*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**2*b**10 - 1 
2*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*b**12 - 24*cos(c + d* 
x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**9*b**3 + 96*cos(c + d*x)*log(t 
an((c + d*x)/2))*sin(c + d*x)**2*a**7*b**5 - 144*cos(c + d*x)*log(tan((...