Integrand size = 29, antiderivative size = 93 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\cos (c+d x))}{a d}-\frac {\cot (c+d x)}{a d}-\frac {\sec (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}+\frac {2 \tan (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{3 a d} \] Output:
arctanh(cos(d*x+c))/a/d-cot(d*x+c)/a/d-sec(d*x+c)/a/d-1/3*sec(d*x+c)^3/a/d +2*tan(d*x+c)/a/d+1/3*tan(d*x+c)^3/a/d
Leaf count is larger than twice the leaf count of optimal. \(245\) vs. \(2(93)=186\).
Time = 1.51 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.63 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^3(c+d x) \left (2+10 \cos (2 (c+d x))+8 \cos (3 (c+d x))+3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (-8-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 \sin (c+d x)-16 \sin (2 (c+d x))-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+8 \sin (3 (c+d x))\right )}{3 a d \left (\csc \left (\frac {1}{2} (c+d x)\right )-\sec \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right ) (1+\sin (c+d x))} \] Input:
Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-1/3*(Csc[c + d*x]^3*(2 + 10*Cos[2*(c + d*x)] + 8*Cos[3*(c + d*x)] + 3*Cos [3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*Cos[3*(c + d*x)]*Log[Sin[(c + d*x) /2]] + Cos[c + d*x]*(-8 - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2] ]) + 4*Sin[c + d*x] - 16*Sin[2*(c + d*x)] - 6*Log[Cos[(c + d*x)/2]]*Sin[2* (c + d*x)] + 6*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 8*Sin[3*(c + d*x)] ))/(a*d*(Csc[(c + d*x)/2] - Sec[(c + d*x)/2])*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2])*(1 + Sin[c + d*x]))
Time = 0.47 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3318, 3042, 3100, 244, 2009, 3102, 25, 254, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^2 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle \frac {\int \csc ^2(c+d x) \sec ^4(c+d x)dx}{a}-\frac {\int \csc (c+d x) \sec ^4(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc (c+d x)^2 \sec (c+d x)^4dx}{a}-\frac {\int \csc (c+d x) \sec (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3100 |
| \(\displaystyle \frac {\int \cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right )^2d\tan (c+d x)}{a d}-\frac {\int \csc (c+d x) \sec (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {\int \left (\cot ^2(c+d x)+\tan ^2(c+d x)+2\right )d\tan (c+d x)}{a d}-\frac {\int \csc (c+d x) \sec (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} \tan ^3(c+d x)+2 \tan (c+d x)-\cot (c+d x)}{a d}-\frac {\int \csc (c+d x) \sec (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\frac {1}{3} \tan ^3(c+d x)+2 \tan (c+d x)-\cot (c+d x)}{a d}-\frac {\int -\frac {\sec ^4(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{a d}+\frac {\frac {1}{3} \tan ^3(c+d x)+2 \tan (c+d x)-\cot (c+d x)}{a d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {\int \left (-\sec ^2(c+d x)+\frac {1}{1-\sec ^2(c+d x)}-1\right )d\sec (c+d x)}{a d}+\frac {\frac {1}{3} \tan ^3(c+d x)+2 \tan (c+d x)-\cot (c+d x)}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} \tan ^3(c+d x)+2 \tan (c+d x)-\cot (c+d x)}{a d}-\frac {-\text {arctanh}(\sec (c+d x))+\frac {1}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\) |
Input:
Int[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-((-ArcTanh[Sec[c + d*x]] + Sec[c + d*x] + Sec[c + d*x]^3/3)/(a*d)) + (-Co t[c + d*x] + 2*Tan[c + d*x] + Tan[c + d*x]^3/3)/(a*d)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[1/f Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] , x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Time = 0.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12
| method | result | size |
| derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {7}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\) | \(104\) |
| default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {7}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\) | \(104\) |
| parallelrisch | \(\frac {-6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-36 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+28}{6 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(134\) |
| risch | \(-\frac {2 \left (-2 \,{\mathrm e}^{3 i \left (d x +c \right )}+6 i {\mathrm e}^{4 i \left (d x +c \right )}-13 \,{\mathrm e}^{i \left (d x +c \right )}-8 i+2 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}\) | \(150\) |
| norman | \(\frac {\frac {1}{2 a d}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 d a}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}-\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d a}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\) | \(160\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/2/d/a*(tan(1/2*d*x+1/2*c)-4/3/(tan(1/2*d*x+1/2*c)+1)^3+2/(tan(1/2*d*x+1/ 2*c)+1)^2-7/(tan(1/2*d*x+1/2*c)+1)-1/(tan(1/2*d*x+1/2*c)-1)-1/tan(1/2*d*x+ 1/2*c)-2*ln(tan(1/2*d*x+1/2*c)))
Time = 0.08 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.74 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {10 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 4}{6 \, {\left (a d \cos \left (d x + c\right )^{3} - a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a d \cos \left (d x + c\right )\right )}} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/6*(10*cos(d*x + c)^2 + 3*(cos(d*x + c)^3 - cos(d*x + c)*sin(d*x + c) - c os(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(cos(d*x + c)^3 - cos(d*x + c )*sin(d*x + c) - cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(8*cos(d*x + c)^2 - 1)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^3 - a*d*cos(d*x + c)*sin( d*x + c) - a*d*cos(d*x + c))
\[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)**2*sec(c + d*x)**2/(sin(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (89) = 178\).
Time = 0.04 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.31 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {22 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 3}{\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/6*((22*sin(d*x + c)/(cos(d*x + c) + 1) + 8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 30*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*sin(d*x + c)^4/(cos( d*x + c) + 1)^4 + 3)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a*sin(d*x + c) ^2/(cos(d*x + c) + 1)^2 - 2*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin( d*x + c)^5/(cos(d*x + c) + 1)^5) + 6*log(sin(d*x + c)/(cos(d*x + c) + 1))/ a - 3*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d
Time = 0.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.43 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a} + \frac {21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/6*(6*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*tan(1/2*d*x + 1/2*c)/a - 3*(t an(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 1)/((tan(1/2*d*x + 1/2*c) ^2 - tan(1/2*d*x + 1/2*c))*a) + (21*tan(1/2*d*x + 1/2*c)^2 + 36*tan(1/2*d* x + 1/2*c) + 19)/(a*(tan(1/2*d*x + 1/2*c) + 1)^3))/d
Time = 31.47 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {22\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1}{d\,\left (-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d} \] Input:
int(1/(cos(c + d*x)^2*sin(c + d*x)^2*(a + a*sin(c + d*x))),x)
Output:
tan(c/2 + (d*x)/2)/(2*a*d) - ((22*tan(c/2 + (d*x)/2))/3 + (8*tan(c/2 + (d* x)/2)^2)/3 - 10*tan(c/2 + (d*x)/2)^3 - 9*tan(c/2 + (d*x)/2)^4 + 1)/(d*(2*a *tan(c/2 + (d*x)/2) + 4*a*tan(c/2 + (d*x)/2)^2 - 4*a*tan(c/2 + (d*x)/2)^4 - 2*a*tan(c/2 + (d*x)/2)^5)) - log(tan(c/2 + (d*x)/2))/(a*d)
Time = 0.17 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.55 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+11 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+11 \cos \left (d x +c \right ) \sin \left (d x +c \right )+32 \sin \left (d x +c \right )^{3}+20 \sin \left (d x +c \right )^{2}-28 \sin \left (d x +c \right )-12}{12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
( - 12*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 12*cos(c + d*x )*log(tan((c + d*x)/2))*sin(c + d*x) + 11*cos(c + d*x)*sin(c + d*x)**2 + 1 1*cos(c + d*x)*sin(c + d*x) + 32*sin(c + d*x)**3 + 20*sin(c + d*x)**2 - 28 *sin(c + d*x) - 12)/(12*cos(c + d*x)*sin(c + d*x)*a*d*(sin(c + d*x) + 1))