Integrand size = 25, antiderivative size = 111 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\csc (c+d x)}{a^4 d}-\frac {4 \log (\sin (c+d x))}{a^4 d}+\frac {4 \log (1+\sin (c+d x))}{a^4 d}-\frac {1}{3 a d (a+a \sin (c+d x))^3}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{d \left (a^4+a^4 \sin (c+d x)\right )} \]
-csc(d*x+c)/a^4/d-4*ln(sin(d*x+c))/a^4/d+4*ln(1+sin(d*x+c))/a^4/d-1/3/a/d/ (a+a*sin(d*x+c))^3-1/d/(a^2+a^2*sin(d*x+c))^2-3/d/(a^4+a^4*sin(d*x+c))
Time = 0.65 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \csc (c+d x)+12 \log (\sin (c+d x))-12 \log (1+\sin (c+d x))+\frac {1}{(1+\sin (c+d x))^3}+\frac {3}{(1+\sin (c+d x))^2}+\frac {9}{1+\sin (c+d x)}}{3 a^4 d} \]
-1/3*(3*Csc[c + d*x] + 12*Log[Sin[c + d*x]] - 12*Log[1 + Sin[c + d*x]] + ( 1 + Sin[c + d*x])^(-3) + 3/(1 + Sin[c + d*x])^2 + 9/(1 + Sin[c + d*x]))/(a ^4*d)
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
| \(\displaystyle \int \frac {\cot (c+d x) \csc (c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^2 (a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {\csc ^2(c+d x)}{a^2 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a \int \left (\frac {\csc ^2(c+d x)}{a^6}-\frac {4 \csc (c+d x)}{a^6}+\frac {4}{a^5 (\sin (c+d x) a+a)}+\frac {3}{a^4 (\sin (c+d x) a+a)^2}+\frac {2}{a^3 (\sin (c+d x) a+a)^3}+\frac {1}{a^2 (\sin (c+d x) a+a)^4}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (-\frac {\csc (c+d x)}{a^5}-\frac {4 \log (a \sin (c+d x))}{a^5}+\frac {4 \log (a \sin (c+d x)+a)}{a^5}-\frac {3}{a^4 (a \sin (c+d x)+a)}-\frac {1}{a^3 (a \sin (c+d x)+a)^2}-\frac {1}{3 a^2 (a \sin (c+d x)+a)^3}\right )}{d}\) |
(a*(-(Csc[c + d*x]/a^5) - (4*Log[a*Sin[c + d*x]])/a^5 + (4*Log[a + a*Sin[c + d*x]])/a^5 - 1/(3*a^2*(a + a*Sin[c + d*x])^3) - 1/(a^3*(a + a*Sin[c + d *x])^2) - 3/(a^4*(a + a*Sin[c + d*x]))))/d
3.3.55.3.1 Defintions of rubi rules used
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 = 0.53 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.57
| method | result | size |
| derivativedivides | \(-\frac {\csc \left (d x +c \right )+\frac {2}{\left (\csc \left (d x +c \right )+1\right )^{2}}-\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}-\frac {6}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\) | \(63\) |
| default | \(-\frac {\csc \left (d x +c \right )+\frac {2}{\left (\csc \left (d x +c \right )+1\right )^{2}}-\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}-\frac {6}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\) | \(63\) |
| risch | \(-\frac {8 i \left (15 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-36 i {\mathrm e}^{4 i \left (d x +c \right )}-31 \,{\mathrm e}^{5 i \left (d x +c \right )}+15 i {\mathrm e}^{2 i \left (d x +c \right )}+31 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} d \,a^{4}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\) | \(160\) |
| parallelrisch | \(\frac {\left (144 \cos \left (2 d x +2 c \right )-360 \sin \left (d x +c \right )+24 \sin \left (3 d x +3 c \right )-240\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-72 \cos \left (2 d x +2 c \right )+180 \sin \left (d x +c \right )-12 \sin \left (3 d x +3 c \right )+120\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (166 \cos \left (d x +c \right )+44 \cos \left (2 d x +2 c \right )-22 \cos \left (3 d x +3 c \right )-188\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+108 \cos \left (2 d x +2 c \right )-108}{3 d \,a^{4} \left (-10+6 \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right )}\) | \(204\) |
| norman | \(\frac {-\frac {1}{2 a d}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {51 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {51 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {209 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {209 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {1159 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {1159 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}\) | \(210\) |
-1/d/a^4*(csc(d*x+c)+2/(csc(d*x+c)+1)^2-1/3/(csc(d*x+c)+1)^3-6/(csc(d*x+c) +1)-4*ln(csc(d*x+c)+1))
Time = 0.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.81 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {30 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 17\right )} \sin \left (d x + c\right ) - 33}{3 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 5 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d - {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]
1/3*(30*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 5*cos(d*x + c)^2 - (3*cos(d* x + c)^2 - 4)*sin(d*x + c) + 4)*log(1/2*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 5*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 4)*sin(d*x + c) + 4)*log(sin(d*x + c) + 1) + 2*(6*cos(d*x + c)^2 - 17)*sin(d*x + c) - 33)/(a^4*d*cos(d*x + c)^4 - 5*a^4*d*cos(d*x + c)^2 + 4*a^4*d - (3*a^4*d*cos(d*x + c)^2 - 4*a^4 *d)*sin(d*x + c))
\[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
Integral(cos(c + d*x)*csc(c + d*x)**2/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a**4
Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2} + 22 \, \sin \left (d x + c\right ) + 3}{a^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + a^{4} \sin \left (d x + c\right )} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{3 \, d} \]
-1/3*((12*sin(d*x + c)^3 + 30*sin(d*x + c)^2 + 22*sin(d*x + c) + 3)/(a^4*s in(d*x + c)^4 + 3*a^4*sin(d*x + c)^3 + 3*a^4*sin(d*x + c)^2 + a^4*sin(d*x + c)) - 12*log(sin(d*x + c) + 1)/a^4 + 12*log(sin(d*x + c))/a^4)/d
Time = 0.33 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {12 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2} + 22 \, \sin \left (d x + c\right ) + 3}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3} \sin \left (d x + c\right )}}{3 \, d} \]
1/3*(12*log(abs(sin(d*x + c) + 1))/a^4 - 12*log(abs(sin(d*x + c)))/a^4 - ( 12*sin(d*x + c)^3 + 30*sin(d*x + c)^2 + 22*sin(d*x + c) + 3)/(a^4*(sin(d*x + c) + 1)^3*sin(d*x + c)))/d
Time = 10.13 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.26 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+74\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {307\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+60\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1}{d\,\left (2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+40\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+30\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+12\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^4\,d} \]
(9*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) + 60*tan(c/2 + (d*x)/2)^3 + (307*tan(c/2 + (d*x)/2)^4)/3 + 74*tan(c/2 + (d*x)/2)^5 + 23*tan(c/2 + (d* x)/2)^6 - 1)/(d*(12*a^4*tan(c/2 + (d*x)/2)^2 + 30*a^4*tan(c/2 + (d*x)/2)^3 + 40*a^4*tan(c/2 + (d*x)/2)^4 + 30*a^4*tan(c/2 + (d*x)/2)^5 + 12*a^4*tan( c/2 + (d*x)/2)^6 + 2*a^4*tan(c/2 + (d*x)/2)^7 + 2*a^4*tan(c/2 + (d*x)/2))) - (4*log(tan(c/2 + (d*x)/2)))/(a^4*d) + (8*log(tan(c/2 + (d*x)/2) + 1))/( a^4*d) - tan(c/2 + (d*x)/2)/(2*a^4*d)