Integrand size = 29, antiderivative size = 128 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d} \] Output:
-2*b^4*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(3/2 )/d+b*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a/d-b*sec(d*x+c)/(a^2-b^2)/d+a* tan(d*x+c)/(a^2-b^2)/d
Time = 1.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.60 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {4 b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{a}+\frac {2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}-\frac {2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{a}}{2 d} \] Input:
Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
((-4*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2 )^(3/2)) - Cot[(c + d*x)/2]/a + (2*b*Log[Cos[(c + d*x)/2]])/a^2 - (2*b*Log [Sin[(c + d*x)/2]])/a^2 + (2*Sin[(c + d*x)/2])/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Sin[(c + d*x)/2])/((a - b)*(Cos[(c + d*x)/2] + S in[(c + d*x)/2])) + Tan[(c + d*x)/2]/a)/(2*d)
Time = 0.53 (sec) , antiderivative size = 150, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^2 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {b^2 \sec ^2(c+d x)}{a^2 (a+b \sin (c+d x))}-\frac {b \csc (c+d x) \sec ^2(c+d x)}{a^2}+\frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b^4 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{a^2 d}+\frac {\tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d}\) |
Input:
Int[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
(-2*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2) ^(3/2)*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a*d) - (b*Se c[c + d*x])/(a^2*d) - (b^2*Sec[c + d*x]*(b - a*Sin[c + d*x]))/(a^2*(a^2 - b^2)*d) + Tan[c + d*x]/(a*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 = 0.69 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{\left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {1}{\left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(155\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{\left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {1}{\left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(155\) |
| risch | \(-\frac {2 \left (-i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+a b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 i a^{2}-i b^{2}-a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}\) | \(308\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*tan(1/2*d*x+1/2*c)/a-1/(a-b)/(tan(1/2*d*x+1/2*c)+1)-2/a^2/(a-b)/( a+b)*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))-1/2/a/tan(1/2*d*x+1/2*c)-1/a^2*b*ln(tan(1/2*d*x+1/2*c))-1/(a+b)/(t an(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (123) = 246\).
Time = 0.22 (sec) , antiderivative size = 582, normalized size of antiderivative = 4.55 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right ) \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^{5} - 2 \, a^{3} b^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right )}, \frac {2 \, \sqrt {a^{2} - b^{2}} b^{4} \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 ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ] \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
[1/2*(sqrt(-a^2 + b^2)*b^4*cos(d*x + c)*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^5 - 2*a^3*b^2 + (a^4*b - 2*a^2*b^3 + b^5)*cos( d*x + c)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - (a^4*b - 2*a^2*b^3 + b ^5)*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*(2*a^5 - 3* a^3*b^2 + a*b^4)*cos(d*x + c)^2 - 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*d*cos(d*x + c)*sin(d*x + c)), 1/2*(2*sqrt(a^2 - b^2 )*b^4*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c)*sin(d*x + c) + 2*a^5 - 2*a^3*b^2 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - (a^4*b - 2*a^2*b^3 + b^5) *cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*(2*a^5 - 3*a^3 *b^2 + a*b^4)*cos(d*x + c)^2 - 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^6 - 2 *a^4*b^2 + a^2*b^4)*d*cos(d*x + c)*sin(d*x + c))]
\[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)**2*sec(c + d*x)**2/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(csc(d*x+c)^2*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
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (123) = 246\).
Time = 0.23 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.02 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{3} - 3 \, a b^{2}}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}}{6 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(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^4/((a^4 - a^2*b^2)*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 - (2*a^2*b *tan(1/2*d*x + 1/2*c)^3 - 2*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*a^3*tan(1/2*d* x + 1/2*c)^2 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 10*a^2*b*tan(1/2*d*x + 1/2 *c) + 2*b^3*tan(1/2*d*x + 1/2*c) + 3*a^3 - 3*a*b^2)/((a^4 - a^2*b^2)*(tan( 1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))))/d
Time = 21.32 (sec) , antiderivative size = 778, normalized size of antiderivative = 6.08 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^2*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
Output:
-((a*b^6*1i - a^7*cos(2*c + 2*d*x)*2i - a^3*b^4*2i + a^5*b^2*1i + a*b^6*co s(2*c + 2*d*x)*1i - a^6*b*sin(2*c + 2*d*x)*1i - a^2*b^5*sin(c + d*x)*2i + a^4*b^3*sin(c + d*x)*4i - a^3*b^4*cos(2*c + 2*d*x)*4i + a^5*b^2*cos(2*c + 2*d*x)*5i + b^7*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x )*1i - a^2*b^5*sin(2*c + 2*d*x)*1i + a^4*b^3*sin(2*c + 2*d*x)*2i - a^6*b*s in(c + d*x)*2i - a^6*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x)*1i + 2*b^4*sin(2*c + 2*d*x)*atan((a^4*sin(c/2 + (d*x)/2)*(b^6 - a ^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*4i - a^2*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*3i + a*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3 *a^2*b^4 + 3*a^4*b^2)^(1/2)*2i - a^3*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a ^2*b^4 + 3*a^4*b^2)^(1/2)*1i)/(a^7*cos(c/2 + (d*x)/2) - 4*b^7*sin(c/2 + (d *x)/2) - 2*a*b^6*cos(c/2 + (d*x)/2) + 2*a^6*b*sin(c/2 + (d*x)/2) + 4*a^3*b ^4*cos(c/2 + (d*x)/2) - 3*a^5*b^2*cos(c/2 + (d*x)/2) + 9*a^2*b^5*sin(c/2 + (d*x)/2) - 7*a^4*b^3*sin(c/2 + (d*x)/2)))*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4* b^2)^(1/2) - a^2*b^5*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x)*3i + a^4*b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2 *d*x)*3i)*1i)/(a^2*d*sin(2*c + 2*d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2))
Time = 0.20 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.37 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{4} b +2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} b^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b^{5}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b -\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{3}+2 \sin \left (d x +c \right )^{2} a^{5}-3 \sin \left (d x +c \right )^{2} a^{3} b^{2}+\sin \left (d x +c \right )^{2} a \,b^{4}-\sin \left (d x +c \right ) a^{4} b +\sin \left (d x +c \right ) a^{2} b^{3}-a^{5}+2 a^{3} b^{2}-a \,b^{4}}{\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
( - 2*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)*b**4 - cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**4*b + 2*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**2*b**3 - cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*b**5 + cos(c + d*x)*sin (c + d*x)*a**4*b - cos(c + d*x)*sin(c + d*x)*a**2*b**3 + 2*sin(c + d*x)**2 *a**5 - 3*sin(c + d*x)**2*a**3*b**2 + sin(c + d*x)**2*a*b**4 - sin(c + d*x )*a**4*b + sin(c + d*x)*a**2*b**3 - a**5 + 2*a**3*b**2 - a*b**4)/(cos(c + d*x)*sin(c + d*x)*a**2*d*(a**4 - 2*a**2*b**2 + b**4))