Integrand size = 27, antiderivative size = 108 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {6 \log (\sin (c+d x))}{a^3 d}-\frac {6 \log (1+\sin (c+d x))}{a^3 d}+\frac {1}{2 a d (a+a \sin (c+d x))^2}+\frac {3}{d \left (a^3+a^3 \sin (c+d x)\right )} \] Output:
3*csc(d*x+c)/a^3/d-1/2*csc(d*x+c)^2/a^3/d+6*ln(sin(d*x+c))/a^3/d-6*ln(1+si n(d*x+c))/a^3/d+1/2/a/d/(a+a*sin(d*x+c))^2+3/d/(a^3+a^3*sin(d*x+c))
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6 \csc (c+d x)-\csc ^2(c+d x)+12 \log (\sin (c+d x))-12 \log (1+\sin (c+d x))+\frac {1}{(1+\sin (c+d x))^2}+\frac {6}{1+\sin (c+d x)}}{2 a^3 d} \] Input:
Integrate[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(6*Csc[c + d*x] - Csc[c + d*x]^2 + 12*Log[Sin[c + d*x]] - 12*Log[1 + Sin[c + d*x]] + (1 + Sin[c + d*x])^(-2) + 6/(1 + Sin[c + d*x]))/(2*a^3*d)
Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 54, 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.
| \(\displaystyle \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^3 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \frac {\csc ^3(c+d x)}{(\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^2 \int \frac {\csc ^3(c+d x)}{a^3 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^2 \int \left (\frac {\csc ^3(c+d x)}{a^6}-\frac {3 \csc ^2(c+d x)}{a^6}+\frac {6 \csc (c+d x)}{a^6}-\frac {6}{a^5 (\sin (c+d x) a+a)}-\frac {3}{a^4 (\sin (c+d x) a+a)^2}-\frac {1}{a^3 (\sin (c+d x) a+a)^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \left (-\frac {\csc ^2(c+d x)}{2 a^5}+\frac {3 \csc (c+d x)}{a^5}+\frac {6 \log (a \sin (c+d x))}{a^5}-\frac {6 \log (a \sin (c+d x)+a)}{a^5}+\frac {3}{a^4 (a \sin (c+d x)+a)}+\frac {1}{2 a^3 (a \sin (c+d x)+a)^2}\right )}{d}\) |
Input:
Int[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(a^2*((3*Csc[c + d*x])/a^5 - Csc[c + d*x]^2/(2*a^5) + (6*Log[a*Sin[c + d*x ]])/a^5 - (6*Log[a + a*Sin[c + d*x]])/a^5 + 1/(2*a^3*(a + a*Sin[c + d*x])^ 2) + 3/(a^4*(a + a*Sin[c + d*x]))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 2.54 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.58
| method | result | size |
| derivativedivides | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-3 \csc \left (d x +c \right )-\frac {1}{2 \left (1+\csc \left (d x +c \right )\right )^{2}}+6 \ln \left (1+\csc \left (d x +c \right )\right )+\frac {4}{1+\csc \left (d x +c \right )}}{d \,a^{3}}\) | \(63\) |
| default | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-3 \csc \left (d x +c \right )-\frac {1}{2 \left (1+\csc \left (d x +c \right )\right )^{2}}+6 \ln \left (1+\csc \left (d x +c \right )\right )+\frac {4}{1+\csc \left (d x +c \right )}}{d \,a^{3}}\) | \(63\) |
| risch | \(\frac {4 i \left (9 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-16 i {\mathrm e}^{4 i \left (d x +c \right )}-13 \,{\mathrm e}^{5 i \left (d x +c \right )}+9 i {\mathrm e}^{2 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} d \,a^{3}}-\frac {12 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\) | \(160\) |
Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-1/d/a^3*(1/2*csc(d*x+c)^2-3*csc(d*x+c)-1/2/(1+csc(d*x+c))^2+6*ln(1+csc(d* x+c))+4/(1+csc(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.81 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {18 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 17}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d - 2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
-1/2*(18*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 3*cos(d*x + c)^2 - 2*(cos(d *x + c)^2 - 1)*sin(d*x + c) + 2)*log(1/2*sin(d*x + c)) + 12*(cos(d*x + c)^ 4 - 3*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 1)*sin(d*x + c) + 2)*log(sin(d* x + c) + 1) + 4*(3*cos(d*x + c)^2 - 4)*sin(d*x + c) - 17)/(a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*x + c)^2 + 2*a^3*d - 2*(a^3*d*cos(d*x + c)^2 - a^3*d )*sin(d*x + c))
\[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)**2/(a+a*sin(d*x+c))**3,x)
Output:
Integral(cot(c + d*x)*csc(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3
Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 18 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2}} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{2 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/2*((12*sin(d*x + c)^3 + 18*sin(d*x + c)^2 + 4*sin(d*x + c) - 1)/(a^3*sin (d*x + c)^4 + 2*a^3*sin(d*x + c)^3 + a^3*sin(d*x + c)^2) - 12*log(sin(d*x + c) + 1)/a^3 + 12*log(sin(d*x + c))/a^3)/d
Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} + \frac {12 \, \sin \left (d x + c\right )^{3} + 18 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1}{2 \, {\left (\sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )\right )}^{2} a^{3} d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-6*log(abs(sin(d*x + c) + 1))/(a^3*d) + 6*log(abs(sin(d*x + c)))/(a^3*d) + 1/2*(12*sin(d*x + c)^3 + 18*sin(d*x + c)^2 + 4*sin(d*x + c) - 1)/((sin(d* x + c)^2 + sin(d*x + c))^2*a^3*d)
Time = 18.12 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.10 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {-26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {65\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+24\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {12\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \] Input:
int(cot(c + d*x)/(sin(c + d*x)^2*(a + a*sin(c + d*x))^3),x)
Output:
(6*log(tan(c/2 + (d*x)/2)))/(a^3*d) - tan(c/2 + (d*x)/2)^2/(8*a^3*d) + (4* tan(c/2 + (d*x)/2) + 21*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^3 - (6 5*tan(c/2 + (d*x)/2)^4)/2 - 26*tan(c/2 + (d*x)/2)^5 - 1/2)/(d*(4*a^3*tan(c /2 + (d*x)/2)^2 + 16*a^3*tan(c/2 + (d*x)/2)^3 + 24*a^3*tan(c/2 + (d*x)/2)^ 4 + 16*a^3*tan(c/2 + (d*x)/2)^5 + 4*a^3*tan(c/2 + (d*x)/2)^6)) - (12*log(t an(c/2 + (d*x)/2) + 1))/(a^3*d) + (3*tan(c/2 + (d*x)/2))/(2*a^3*d)
Time = 0.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.06 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-2 \csc \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}-\csc \left (d x +c \right )^{2} \sin \left (d x +c \right )-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-3 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}+15 \sin \left (d x +c \right )+6}{2 \sin \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
Output:
( - 2*csc(c + d*x)**2*sin(c + d*x)**2 - csc(c + d*x)**2*sin(c + d*x) - 24* log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 48*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**2 - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 12*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 24*log(tan((c + d*x)/2))*sin(c + d*x)**2 + 1 2*log(tan((c + d*x)/2))*sin(c + d*x) - 3*sin(c + d*x)**3 + 6*sin(c + d*x)* *2 + 15*sin(c + d*x) + 6)/(2*sin(c + d*x)*a**3*d*(sin(c + d*x)**2 + 2*sin( c + d*x) + 1))