Integrand size = 29, antiderivative size = 181 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2} d}-\frac {\left (3 a^2+2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 a^3 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {\left (3 a^2-b^2\right ) \sec (c+d x)}{2 a \left (a^2-b^2\right ) d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d} \] Output:
2*b^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2-b^2)^(3/2) /d-1/2*(3*a^2+2*b^2)*arctanh(cos(d*x+c))/a^3/d+b*cot(d*x+c)/a^2/d+1/2*(3*a ^2-b^2)*sec(d*x+c)/a/(a^2-b^2)/d-1/2*csc(d*x+c)^2*sec(d*x+c)/a/d-b*tan(d*x +c)/(a^2-b^2)/d
Time = 5.03 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2}}+\frac {4 b \cot \left (\frac {1}{2} (c+d x)\right )}{a^2}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}-\frac {4 \left (3 a^2+2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {4 \left (3 a^2+2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{a}+\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 b \tan \left (\frac {1}{2} (c+d x)\right )}{a^2}}{8 d} \] Input:
Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
((16*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2 )^(3/2)) + (4*b*Cot[(c + d*x)/2])/a^2 - Csc[(c + d*x)/2]^2/a - (4*(3*a^2 + 2*b^2)*Log[Cos[(c + d*x)/2]])/a^3 + (4*(3*a^2 + 2*b^2)*Log[Sin[(c + d*x)/ 2]])/a^3 + Sec[(c + d*x)/2]^2/a + (8*Sin[(c + d*x)/2])/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (8*Sin[(c + d*x)/2])/((a - b)*(Cos[(c + d*x )/2] + Sin[(c + d*x)/2])) - (4*b*Tan[(c + d*x)/2])/a^2)/(8*d)
Time = 0.60 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.17, 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.
| \(\displaystyle \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^3 \cos (c+d x)^2 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (-\frac {b^3 \sec ^2(c+d x)}{a^3 (a+b \sin (c+d x))}+\frac {b^2 \csc (c+d x) \sec ^2(c+d x)}{a^3}-\frac {b \csc ^2(c+d x) \sec ^2(c+d x)}{a^2}+\frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b^2 \sec (c+d x)}{a^3 d}-\frac {b \tan (c+d x)}{a^2 d}+\frac {b \cot (c+d x)}{a^2 d}+\frac {2 b^5 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}+\frac {b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 d \left (a^2-b^2\right )}-\frac {3 \text {arctanh}(\cos (c+d x))}{2 a d}+\frac {3 \sec (c+d x)}{2 a d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d}\) |
Input:
Int[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
(2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^ (3/2)*d) - (3*ArcTanh[Cos[c + d*x]])/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]]) /(a^3*d) + (b*Cot[c + d*x])/(a^2*d) + (3*Sec[c + d*x])/(2*a*d) + (b^2*Sec[ c + d*x])/(a^3*d) - (Csc[c + d*x]^2*Sec[c + d*x])/(2*a*d) + (b^3*Sec[c + d *x]*(b - a*Sin[c + d*x]))/(a^3*(a^2 - b^2)*d) - (b*Tan[c + d*x])/(a^2*d)
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])
Time = 0.88 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{\left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}+\frac {1}{\left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (6 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(199\) |
| default | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{\left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}+\frac {1}{\left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (6 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(199\) |
| risch | \(\frac {i \left (3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-2 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +4 a^{2} b -2 b^{3}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d \,a^{3}}-\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d \,a^{3}}\) | \(437\) |
Input:
int(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/4/a^2*(1/2*tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c))-1/(a+b)/( tan(1/2*d*x+1/2*c)-1)+2/a^3/(a-b)/(a+b)*b^5/(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)/(tan(1/2*d*x+1/2*c)+1)- 1/8/a/tan(1/2*d*x+1/2*c)^2+1/4/a^3*(6*a^2+4*b^2)*ln(tan(1/2*d*x+1/2*c))+1/ 2/a^2*b/tan(1/2*d*x+1/2*c))
Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (170) = 340\).
Time = 0.42 (sec) , antiderivative size = 878, normalized size of antiderivative = 4.85 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
[-1/4*(4*a^6 - 4*a^4*b^2 - 2*(3*a^6 - 4*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 - 2*(b^5*cos(d*x + c)^3 - b^5*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)) + ((3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*co s(d*x + c)^3 - (3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c))*log(1/2 *cos(d*x + c) + 1/2) - ((3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c) ^3 - (3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^5*b - a^3*b^3 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c)^3 - (a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c)), -1/4*(4*a^6 - 4*a^4*b^2 - 2*(3*a^6 - 4*a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + 4*(b^5*cos(d*x + c)^3 - b^5*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))) + ((3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c)^3 - (3*a ^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2 ) - ((3*a^6 - 4*a^4*b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c)^3 - (3*a^6 - 4*a^4 *b^2 - a^2*b^4 + 2*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^ 5*b - a^3*b^3 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(d*x + c)^2)*sin(d*x + c) )/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c)^3 - (a^7 - 2*a^5*b^2 + a^3*b ^4)*d*cos(d*x + c))]
\[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**3*sec(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)**3*sec(c + d*x)**2/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^2/(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
Time = 0.24 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.35 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{5}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {16 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {4 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/8*(16*(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^5/((a^5 - a^3*b^2)*sqrt(a^2 - b^2)) + 16* (b*tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1)) + (a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 + 4*(3*a^2 + 2*b ^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (18*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c) + a^2)/(a^3*tan (1/2*d*x + 1/2*c)^2))/d
Time = 22.09 (sec) , antiderivative size = 1570, normalized size of antiderivative = 8.67 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^2*sin(c + d*x)^3*(a + b*sin(c + d*x))),x)
Output:
(a^5*((3*b^3*sin(c + d*x))/4 - (5*b^3*sin(3*c + 3*d*x))/4) - a^7*((b*sin(c + d*x))/2 - (b*sin(3*c + 3*d*x))/2) + a^2*(b^6/4 + (b^6*cos(2*c + 2*d*x)) /4 + (3*b^6*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - ( 3*b^6*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/8) - a* ((b^7*sin(c + d*x))/4 + (b^7*sin(3*c + 3*d*x))/4) - a^6*((b^2*cos(c + d*x) )/2 + b^2/4 - (7*b^2*cos(2*c + 2*d*x))/4 - (b^2*cos(3*c + 3*d*x))/2 + (7*b ^2*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (7*b^2*log (sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/8) - a^4*(b^4/4 - (b^4*cos(c + d*x))/4 + (5*b^4*cos(2*c + 2*d*x))/4 + (b^4*cos(3*c + 3*d*x ))/4 - (3*b^4*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (3*b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/8) + a^8*(cos(c + d*x)/4 - (3*cos(2*c + 2*d*x))/4 - cos(3*c + 3*d*x)/4 + (3*cos (c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/8 + 1/4) - (b^8*cos(c + d* x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (b^8*log(sin(c/2 + (d*x )/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + a^3*b^5*sin(3*c + 3*d*x) + (b^5*atan((3*a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1 /2) + 8*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + a^3*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 7* a^4*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + ...
Time = 0.20 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.38 \[ \int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{5}+12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{6}-16 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{4} b^{2}-4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2} b^{4}+8 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{6}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{6}+10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{4} b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{4}-16 \sin \left (d x +c \right )^{3} a^{5} b +24 \sin \left (d x +c \right )^{3} a^{3} b^{3}-8 \sin \left (d x +c \right )^{3} a \,b^{5}+12 \sin \left (d x +c \right )^{2} a^{6}-16 \sin \left (d x +c \right )^{2} a^{4} b^{2}+4 \sin \left (d x +c \right )^{2} a^{2} b^{4}+8 \sin \left (d x +c \right ) a^{5} b -16 \sin \left (d x +c \right ) a^{3} b^{3}+8 \sin \left (d x +c \right ) a \,b^{5}-4 a^{6}+8 a^{4} b^{2}-4 a^{2} b^{4}}{8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
(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**5 + 12*cos(c + d*x)*log(tan((c + d*x)/2))*sin (c + d*x)**2*a**6 - 16*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2* a**4*b**2 - 4*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b**4 + 8*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**6 - 9*cos(c + d *x)*sin(c + d*x)**2*a**6 + 10*cos(c + d*x)*sin(c + d*x)**2*a**4*b**2 - cos (c + d*x)*sin(c + d*x)**2*a**2*b**4 - 16*sin(c + d*x)**3*a**5*b + 24*sin(c + d*x)**3*a**3*b**3 - 8*sin(c + d*x)**3*a*b**5 + 12*sin(c + d*x)**2*a**6 - 16*sin(c + d*x)**2*a**4*b**2 + 4*sin(c + d*x)**2*a**2*b**4 + 8*sin(c + d *x)*a**5*b - 16*sin(c + d*x)*a**3*b**3 + 8*sin(c + d*x)*a*b**5 - 4*a**6 + 8*a**4*b**2 - 4*a**2*b**4)/(8*cos(c + d*x)*sin(c + d*x)**2*a**3*d*(a**4 - 2*a**2*b**2 + b**4))