Integrand size = 29, antiderivative size = 130 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d} \] Output:
2*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a^2/d-2*sec(d*x+c)/a^2/d-2/3*sec(d* x+c)^3/a^2/d-2/5*sec(d*x+c)^5/a^2/d+4*tan(d*x+c)/a^2/d+5/3*tan(d*x+c)^3/a^ 2/d+2/5*tan(d*x+c)^5/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(130)=260\).
Time = 1.68 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.22 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^3(c+d x) \left (40+48 \cos (2 (c+d x))+112 \cos (3 (c+d x))-28 \cos (4 (c+d x))+60 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \cos (c+d x) \left (28+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-60 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+58 \sin (c+d x)-168 \sin (2 (c+d x))-90 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+90 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+82 \sin (3 (c+d x))+28 \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))\right )}{15 a^2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\sin (c+d x))^2} \] Input:
Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/15*(Csc[c + d*x]^3*(40 + 48*Cos[2*(c + d*x)] + 112*Cos[3*(c + d*x)] - 2 8*Cos[4*(c + d*x)] + 60*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 4*Cos[c + d*x]*(28 + 15*Log[Cos[(c + d*x)/2]] - 15*Log[Sin[(c + d*x)/2]]) - 60*Cos[ 3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 58*Sin[c + d*x] - 168*Sin[2*(c + d*x) ] - 90*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] + 90*Log[Sin[(c + d*x)/2]]*S in[2*(c + d*x)] + 82*Sin[3*(c + d*x)] + 28*Sin[4*(c + d*x)] + 15*Log[Cos[( c + d*x)/2]]*Sin[4*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)]) )/(a^2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2)*(1 + Sin[c + d*x])^2)
Time = 0.59 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3352, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^2 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \csc ^2(c+d x) \sec ^6(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2}{\cos (c+d x)^6 \sin (c+d x)^2}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (a^2 \sec ^6(c+d x)+a^2 \csc ^2(c+d x) \sec ^6(c+d x)-2 a^2 \csc (c+d x) \sec ^6(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \tan ^5(c+d x)}{5 d}+\frac {5 a^2 \tan ^3(c+d x)}{3 d}+\frac {4 a^2 \tan (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}-\frac {2 a^2 \sec ^5(c+d x)}{5 d}-\frac {2 a^2 \sec ^3(c+d x)}{3 d}-\frac {2 a^2 \sec (c+d x)}{d}}{a^4}\) |
Input:
Int[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
((2*a^2*ArcTanh[Cos[c + d*x]])/d - (a^2*Cot[c + d*x])/d - (2*a^2*Sec[c + d *x])/d - (2*a^2*Sec[c + d*x]^3)/(3*d) - (2*a^2*Sec[c + d*x]^5)/(5*d) + (4* a^2*Tan[c + d*x])/d + (5*a^2*Tan[c + d*x]^3)/(3*d) + (2*a^2*Tan[c + d*x]^5 )/(5*d))/a^4
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Time = 0.85 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03
| method | result | size |
| derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {26}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{2 d \,a^{2}}\) | \(134\) |
| default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {26}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{2 d \,a^{2}}\) | \(134\) |
| parallelrisch | \(\frac {-60 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-420 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-960 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-550 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+560 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+15 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+796 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+304}{30 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(160\) |
| risch | \(-\frac {4 \left (-85 \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i {\mathrm e}^{6 i \left (d x +c \right )}-40 i {\mathrm e}^{4 i \left (d x +c \right )}+97 \,{\mathrm e}^{i \left (d x +c \right )}-108 i {\mathrm e}^{2 i \left (d x +c \right )}+28 i-27 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d \,a^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) | \(174\) |
| norman | \(\frac {\frac {1}{2 a d}-\frac {55 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 d a}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d a}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d a}+\frac {199 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{30 d a}+\frac {188 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d a}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(201\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/2/d/a^2*(tan(1/2*d*x+1/2*c)-1/tan(1/2*d*x+1/2*c)-4*ln(tan(1/2*d*x+1/2*c) )-8/5/(tan(1/2*d*x+1/2*c)+1)^5+4/(tan(1/2*d*x+1/2*c)+1)^4-26/3/(tan(1/2*d* x+1/2*c)+1)^3+9/(tan(1/2*d*x+1/2*c)+1)^2-31/2/(tan(1/2*d*x+1/2*c)+1)-1/2/( tan(1/2*d*x+1/2*c)-1))
Time = 0.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {56 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (41 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 9}{15 \, {\left (2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) + {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/15*(56*cos(d*x + c)^4 - 80*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^3 + (cos (d*x + c)^3 - 2*cos(d*x + c))*sin(d*x + c) - 2*cos(d*x + c))*log(1/2*cos(d *x + c) + 1/2) + 15*(2*cos(d*x + c)^3 + (cos(d*x + c)^3 - 2*cos(d*x + c))* sin(d*x + c) - 2*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(41*cos(d* x + c)^2 - 3)*sin(d*x + c) + 9)/(2*a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c) + (a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c))*sin(d*x + c))
\[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Integral(csc(c + d*x)**2*sec(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (122) = 244\).
Time = 0.04 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.38 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {244 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {571 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {320 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {475 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {660 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {255 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 15}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {15 \, \sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/30*((244*sin(d*x + c)/(cos(d*x + c) + 1) + 571*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 320*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 475*sin(d*x + c)^4 /(cos(d*x + c) + 1)^4 - 660*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 255*sin( d*x + c)^6/(cos(d*x + c) + 1)^6 + 15)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*a^2*sin(d*x + c)^3/(cos(d *x + c) + 1)^3 - 5*a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) + 60*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 15*sin(d*x + c)/(a^2*(cos(d* x + c) + 1)))/d
Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.24 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {15 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{2}} + \frac {465 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1590 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 383}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/60*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 30*tan(1/2*d*x + 1/2*c)/a^ 2 - 15*(4*tan(1/2*d*x + 1/2*c)^2 - 7*tan(1/2*d*x + 1/2*c) + 2)/((tan(1/2*d *x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))*a^2) + (465*tan(1/2*d*x + 1/2*c)^4 + 1590*tan(1/2*d*x + 1/2*c)^3 + 2240*tan(1/2*d*x + 1/2*c)^2 + 1450*tan(1/2* d*x + 1/2*c) + 383)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 33.21 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.66 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {-17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {95\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {571\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {244\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+1}{d\,\left (-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \] Input:
int(1/(cos(c + d*x)^2*sin(c + d*x)^2*(a + a*sin(c + d*x))^2),x)
Output:
tan(c/2 + (d*x)/2)/(2*a^2*d) - (2*log(tan(c/2 + (d*x)/2)))/(a^2*d) - ((244 *tan(c/2 + (d*x)/2))/15 + (571*tan(c/2 + (d*x)/2)^2)/15 + (64*tan(c/2 + (d *x)/2)^3)/3 - (95*tan(c/2 + (d*x)/2)^4)/3 - 44*tan(c/2 + (d*x)/2)^5 - 17*t an(c/2 + (d*x)/2)^6 + 1)/(d*(8*a^2*tan(c/2 + (d*x)/2)^2 + 10*a^2*tan(c/2 + (d*x)/2)^3 - 10*a^2*tan(c/2 + (d*x)/2)^5 - 8*a^2*tan(c/2 + (d*x)/2)^6 - 2 *a^2*tan(c/2 + (d*x)/2)^7 + 2*a^2*tan(c/2 + (d*x)/2)))
Time = 0.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.58 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-47 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-94 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-47 \cos \left (d x +c \right ) \sin \left (d x +c \right )+112 \sin \left (d x +c \right )^{4}+164 \sin \left (d x +c \right )^{3}-64 \sin \left (d x +c \right )^{2}-152 \sin \left (d x +c \right )-30}{30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - 60*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3 - 120*cos(c + d* x)*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 60*cos(c + d*x)*log(tan((c + d* x)/2))*sin(c + d*x) - 47*cos(c + d*x)*sin(c + d*x)**3 - 94*cos(c + d*x)*si n(c + d*x)**2 - 47*cos(c + d*x)*sin(c + d*x) + 112*sin(c + d*x)**4 + 164*s in(c + d*x)**3 - 64*sin(c + d*x)**2 - 152*sin(c + d*x) - 30)/(30*cos(c + d *x)*sin(c + d*x)*a**2*d*(sin(c + d*x)**2 + 2*sin(c + d*x) + 1))