3.14.57 \(\int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) [1357]

3.14.57.1 Optimal result
3.14.57.2 Mathematica [B] (verified)
3.14.57.3 Rubi [A] (verified)
3.14.57.4 Maple [A] (verified)
3.14.57.5 Fricas [A] (verification not implemented)
3.14.57.6 Sympy [F(-1)]
3.14.57.7 Maxima [F(-2)]
3.14.57.8 Giac [A] (verification not implemented)
3.14.57.9 Mupad [B] (verification not implemented)

3.14.57.1 Optimal result

Integrand size = 29, antiderivative size = 220 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^6 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a d}+\frac {b \left (-a^2+2 b^2\right ) \sec (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \sec ^3(c+d x) (-a+b \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}+\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right )^2 d}+\frac {\tan ^3(c+d x)}{3 a d} \]

output
2*b^6*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+b*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a/d+b*(-a^2+2*b^2)*sec(d*x+c)/(a 
^2-b^2)^2/d+1/3*b*sec(d*x+c)^3*(-a+b*sin(d*x+c))/a/(a^2-b^2)/d+1/3*(6*a^4- 
10*a^2*b^2+b^4)*tan(d*x+c)/a/(a^2-b^2)^2/d+1/3*tan(d*x+c)^3/a/d
 
3.14.57.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(450\) vs. \(2(220)=440\).

Time = 6.81 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.05 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^6 \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{2 a d}+\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}-\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}+\frac {1}{12 (a+b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 (a+b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 (a-b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {1}{12 (a-b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {10 a \sin \left (\frac {1}{2} (c+d x)\right )-13 b \sin \left (\frac {1}{2} (c+d x)\right )}{6 (a-b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {10 a \sin \left (\frac {1}{2} (c+d x)\right )+13 b \sin \left (\frac {1}{2} (c+d x)\right )}{6 (a+b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{2 a d} \]

input
Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]
 
output
(2*b^6*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2])) 
/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(5/2)*d) - Cot[(c + d*x)/2]/(2*a*d) + 
(b*Log[Cos[(c + d*x)/2]])/(a^2*d) - (b*Log[Sin[(c + d*x)/2]])/(a^2*d) + 1/ 
(12*(a + b)*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + Sin[(c + d*x)/2]/ 
(6*(a + b)*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + Sin[(c + d*x)/2]/( 
6*(a - b)*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/(12*(a - b)*d*(Co 
s[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (10*a*Sin[(c + d*x)/2] - 13*b*Sin[ 
(c + d*x)/2])/(6*(a - b)^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (10* 
a*Sin[(c + d*x)/2] + 13*b*Sin[(c + d*x)/2])/(6*(a + b)^2*d*(Cos[(c + d*x)/ 
2] - Sin[(c + d*x)/2])) + Tan[(c + d*x)/2]/(2*a*d)
 
3.14.57.3 Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3377, 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.

\(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^4 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

input
Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]
 
output
(2*b^6*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^ 
(5/2)*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a*d) - (b*Sec 
[c + d*x])/(a^2*d) - (b*Sec[c + d*x]^3)/(3*a^2*d) - (b^2*Sec[c + d*x]^3*(b 
 - a*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*d) + (b^2*Sec[c + d*x]*(3*b^3 + a*( 
2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*d) + (2*Tan[c + d*x])/( 
a*d) + Tan[c + d*x]^3/(3*a*d)
 

3.14.57.3.1 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 3377
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 
3.14.57.4 Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.15

method result size
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {4 a -5 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 b^{6} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {4 a +5 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) \(253\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {4 a -5 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 b^{6} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {4 a +5 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) \(253\)
risch \(-\frac {2 \left (9 i b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+3 a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}-6 b^{3} a \,{\mathrm e}^{7 i \left (d x +c \right )}+16 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-6 i a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+7 a^{3} {\mathrm e}^{5 i \left (d x +c \right )} b -10 b^{3} a \,{\mathrm e}^{5 i \left (d x +c \right )}-28 i a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 i b^{4}+3 i b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-7 b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+10 b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}-14 i a^{2} b^{2}+9 i b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+8 i a^{4}-3 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )} b^{3} a \right )}{3 d \left (a^{2}-b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}-\frac {b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}\) \(493\)

input
int(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d*(1/2*tan(1/2*d*x+1/2*c)/a-1/2*(4*a-5*b)/(a-b)^2/(tan(1/2*d*x+1/2*c)+1) 
-1/3/(a-b)/(tan(1/2*d*x+1/2*c)+1)^3+1/2/(a-b)/(tan(1/2*d*x+1/2*c)+1)^2+2/a 
^2/(a+b)^2/(a-b)^2*b^6/(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))-1/2*(4*a+5*b)/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-1/3/(a+ 
b)/(tan(1/2*d*x+1/2*c)-1)^3-1/2/(a+b)/(tan(1/2*d*x+1/2*c)-1)^2-1/2/a/tan(1 
/2*d*x+1/2*c)-1/a^2*b*ln(tan(1/2*d*x+1/2*c)))
 
3.14.57.5 Fricas [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 831, normalized size of antiderivative = 3.78 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} b^{6} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, a^{7} + 4 \, a^{5} b^{2} - 2 \, a^{3} b^{4} - 3 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 2 \, {\left (8 \, a^{7} - 22 \, a^{5} b^{2} + 17 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{7} - 11 \, a^{5} b^{2} + 7 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + 3 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )}, -\frac {6 \, \sqrt {a^{2} - b^{2}} b^{6} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{7} + 4 \, a^{5} b^{2} - 2 \, a^{3} b^{4} - 3 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 2 \, {\left (8 \, a^{7} - 22 \, a^{5} b^{2} + 17 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{7} - 11 \, a^{5} b^{2} + 7 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + 3 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )}\right ] \]

input
integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas" 
)
 
output
[-1/6*(3*sqrt(-a^2 + b^2)*b^6*cos(d*x + c)^3*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))*sin(d*x + c) - 2*a^7 + 4*a^5*b^2 - 2*a^3*b^4 - 3*(a^6*b - 3* 
a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^3*log(1/2*cos(d*x + c) + 1/2)*sin( 
d*x + c) + 3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^3*log(-1/2 
*cos(d*x + c) + 1/2)*sin(d*x + c) + 2*(8*a^7 - 22*a^5*b^2 + 17*a^3*b^4 - 3 
*a*b^6)*cos(d*x + c)^4 - 2*(4*a^7 - 11*a^5*b^2 + 7*a^3*b^4)*cos(d*x + c)^2 
 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5 + 3*(a^6*b - 3*a^4*b^3 + 2*a^2*b^5)*cos( 
d*x + c)^2)*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d 
*x + c)^3*sin(d*x + c)), -1/6*(6*sqrt(a^2 - b^2)*b^6*arctan(-(a*sin(d*x + 
c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c)^3*sin(d*x + c) - 2*a^ 
7 + 4*a^5*b^2 - 2*a^3*b^4 - 3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d* 
x + c)^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(a^6*b - 3*a^4*b^3 + 
 3*a^2*b^5 - b^7)*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) 
 + 2*(8*a^7 - 22*a^5*b^2 + 17*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^4 - 2*(4*a^7 
 - 11*a^5*b^2 + 7*a^3*b^4)*cos(d*x + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5 
 + 3*(a^6*b - 3*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^8 - 
 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^3*sin(d*x + c))]
 
3.14.57.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

input
integrate(csc(d*x+c)**2*sec(d*x+c)**4/(a+b*sin(d*x+c)),x)
 
output
Timed out
 
3.14.57.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

input
integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),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
 
3.14.57.8 Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.62 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{6}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {3 \, {\left (2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {4 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 14 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2} b + 7 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

input
integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")
 
output
1/6*(12*(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)))*b^6/((a^6 - 2*a^4*b^2 + a^2*b^4)*sqrt(a^2 - 
 b^2)) - 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*tan(1/2*d*x + 1/2*c)/a 
 + 3*(2*b*tan(1/2*d*x + 1/2*c) - a)/(a^2*tan(1/2*d*x + 1/2*c)) - 4*(6*a^3* 
tan(1/2*d*x + 1/2*c)^5 - 9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*b*tan(1/2* 
d*x + 1/2*c)^4 + 9*b^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^3*tan(1/2*d*x + 1/2*c) 
^3 + 14*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 12 
*b^3*tan(1/2*d*x + 1/2*c)^2 + 6*a^3*tan(1/2*d*x + 1/2*c) - 9*a*b^2*tan(1/2 
*d*x + 1/2*c) - 4*a^2*b + 7*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1 
/2*c)^2 - 1)^3))/d
 
3.14.57.9 Mupad [B] (verification not implemented)

Time = 17.55 (sec) , antiderivative size = 2317, normalized size of antiderivative = 10.53 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

input
int(1/(cos(c + d*x)^4*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
 
output
(a*((3*b^10)/8 + (b^10*cos(2*c + 2*d*x))/2 + (b^10*cos(4*c + 4*d*x))/8) - 
a^10*((7*b*sin(c + d*x))/12 + (b*sin(2*c + 2*d*x))/3 + (b*sin(3*c + 3*d*x) 
)/4 + (b*sin(4*c + 4*d*x))/6 + (b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2 
))*sin(2*c + 2*d*x))/4 + (b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin 
(4*c + 4*d*x))/8) - a^6*((17*b^5*sin(c + d*x))/4 + (11*b^5*sin(2*c + 2*d*x 
))/4 + (9*b^5*sin(3*c + 3*d*x))/4 + (11*b^5*sin(4*c + 4*d*x))/8 + (5*b^5*l 
og(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x))/2 + (5*b^5*log 
(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(4*c + 4*d*x))/4) - a^2*((5*b^9 
*sin(c + d*x))/6 + (7*b^9*sin(2*c + 2*d*x))/12 + (b^9*sin(3*c + 3*d*x))/2 
+ (7*b^9*sin(4*c + 4*d*x))/24 + (5*b^9*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d 
*x)/2))*sin(2*c + 2*d*x))/4 + (5*b^9*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x 
)/2))*sin(4*c + 4*d*x))/8) + a^8*((31*b^3*sin(c + d*x))/12 + (19*b^3*sin(2 
*c + 2*d*x))/12 + (5*b^3*sin(3*c + 3*d*x))/4 + (19*b^3*sin(4*c + 4*d*x))/2 
4 + (5*b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x))/4 
+ (5*b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(4*c + 4*d*x))/8) + 
 a^4*((37*b^7*sin(c + d*x))/12 + (25*b^7*sin(2*c + 2*d*x))/12 + (7*b^7*sin 
(3*c + 3*d*x))/4 + (25*b^7*sin(4*c + 4*d*x))/24 + (5*b^7*log(sin(c/2 + (d* 
x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x))/2 + (5*b^7*log(sin(c/2 + (d*x) 
/2)/cos(c/2 + (d*x)/2))*sin(4*c + 4*d*x))/4) - a^7*((9*b^4)/8 + 6*b^4*cos( 
2*c + 2*d*x) + (23*b^4*cos(4*c + 4*d*x))/8) + a^9*(b^2/4 + (19*b^2*cos(...