Integrand size = 27, antiderivative size = 131 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {4 \csc (c+d x)}{a^4 d}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {10 \log (\sin (c+d x))}{a^4 d}-\frac {10 \log (1+\sin (c+d x))}{a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {3}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {6}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
4*csc(d*x+c)/a^4/d-1/2*csc(d*x+c)^2/a^4/d+10*ln(sin(d*x+c))/a^4/d-10*ln(1+ sin(d*x+c))/a^4/d+1/3/a/d/(a+a*sin(d*x+c))^3+3/2/d/(a^2+a^2*sin(d*x+c))^2+ 6/d/(a^4+a^4*sin(d*x+c))
Time = 2.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {24 \csc (c+d x)-3 \csc ^2(c+d x)+60 \log (\sin (c+d x))-60 \log (1+\sin (c+d x))+\frac {2}{(1+\sin (c+d x))^3}+\frac {9}{(1+\sin (c+d x))^2}+\frac {36}{1+\sin (c+d x)}}{6 a^4 d} \] Input:
Integrate[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]
Output:
(24*Csc[c + d*x] - 3*Csc[c + d*x]^2 + 60*Log[Sin[c + d*x]] - 60*Log[1 + Si n[c + d*x]] + 2/(1 + Sin[c + d*x])^3 + 9/(1 + Sin[c + d*x])^2 + 36/(1 + Si n[c + d*x]))/(6*a^4*d)
Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.91, 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.
| \(\displaystyle \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^3 (a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3312 |
| \(\displaystyle \frac {\int \frac {\csc ^3(c+d x)}{(\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^2 \int \frac {\csc ^3(c+d x)}{a^3 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^2 \int \left (\frac {\csc ^3(c+d x)}{a^7}-\frac {4 \csc ^2(c+d x)}{a^7}+\frac {10 \csc (c+d x)}{a^7}-\frac {10}{a^6 (\sin (c+d x) a+a)}-\frac {6}{a^5 (\sin (c+d x) a+a)^2}-\frac {3}{a^4 (\sin (c+d x) a+a)^3}-\frac {1}{a^3 (\sin (c+d x) a+a)^4}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \left (-\frac {\csc ^2(c+d x)}{2 a^6}+\frac {4 \csc (c+d x)}{a^6}+\frac {10 \log (a \sin (c+d x))}{a^6}-\frac {10 \log (a \sin (c+d x)+a)}{a^6}+\frac {6}{a^5 (a \sin (c+d x)+a)}+\frac {3}{2 a^4 (a \sin (c+d x)+a)^2}+\frac {1}{3 a^3 (a \sin (c+d x)+a)^3}\right )}{d}\) |
Input:
Int[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]
Output:
(a^2*((4*Csc[c + d*x])/a^6 - Csc[c + d*x]^2/(2*a^6) + (10*Log[a*Sin[c + d* x]])/a^6 - (10*Log[a + a*Sin[c + d*x]])/a^6 + 1/(3*a^3*(a + a*Sin[c + d*x] )^3) + 3/(2*a^4*(a + a*Sin[c + d*x])^2) + 6/(a^5*(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 = 6.89 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.57
| method | result | size |
| derivativedivides | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-4 \csc \left (d x +c \right )+\frac {10}{1+\csc \left (d x +c \right )}-\frac {5}{2 \left (1+\csc \left (d x +c \right )\right )^{2}}+\frac {1}{3 \left (1+\csc \left (d x +c \right )\right )^{3}}+10 \ln \left (1+\csc \left (d x +c \right )\right )}{d \,a^{4}}\) | \(75\) |
| default | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-4 \csc \left (d x +c \right )+\frac {10}{1+\csc \left (d x +c \right )}-\frac {5}{2 \left (1+\csc \left (d x +c \right )\right )^{2}}+\frac {1}{3 \left (1+\csc \left (d x +c \right )\right )^{3}}+10 \ln \left (1+\csc \left (d x +c \right )\right )}{d \,a^{4}}\) | \(75\) |
| risch | \(\frac {4 i \left (75 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}-255 i {\mathrm e}^{6 i \left (d x +c \right )}-170 \,{\mathrm e}^{7 i \left (d x +c \right )}+255 i {\mathrm e}^{4 i \left (d x +c \right )}+298 \,{\mathrm e}^{5 i \left (d x +c \right )}-75 i {\mathrm e}^{2 i \left (d x +c \right )}-170 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} d \,a^{4}}-\frac {20 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}+\frac {10 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}\) | \(183\) |
Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-1/d/a^4*(1/2*csc(d*x+c)^2-4*csc(d*x+c)+10/(1+csc(d*x+c))-5/2/(1+csc(d*x+c ))^2+1/3/(1+csc(d*x+c))^3+10*ln(1+csc(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.85 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {60 \, \cos \left (d x + c\right )^{4} - 230 \, \cos \left (d x + c\right )^{2} + 60 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 60 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (10 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) + 167}{6 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{4} - 7 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{4} - 5 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
1/6*(60*cos(d*x + c)^4 - 230*cos(d*x + c)^2 + 60*(3*cos(d*x + c)^4 - 7*cos (d*x + c)^2 + (cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 4)*sin(d*x + c) + 4)*lo g(1/2*sin(d*x + c)) - 60*(3*cos(d*x + c)^4 - 7*cos(d*x + c)^2 + (cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 4)*sin(d*x + c) + 4)*log(sin(d*x + c) + 1) - 15 *(10*cos(d*x + c)^2 - 11)*sin(d*x + c) + 167)/(3*a^4*d*cos(d*x + c)^4 - 7* a^4*d*cos(d*x + c)^2 + 4*a^4*d + (a^4*d*cos(d*x + c)^4 - 5*a^4*d*cos(d*x + c)^2 + 4*a^4*d)*sin(d*x + c))
\[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)**2/(a+a*sin(d*x+c))**4,x)
Output:
Integral(cot(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 = 0.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {60 \, \sin \left (d x + c\right )^{4} + 150 \, \sin \left (d x + c\right )^{3} + 110 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 3}{a^{4} \sin \left (d x + c\right )^{5} + 3 \, a^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} \sin \left (d x + c\right )^{3} + a^{4} \sin \left (d x + c\right )^{2}} - \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{6 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
1/6*((60*sin(d*x + c)^4 + 150*sin(d*x + c)^3 + 110*sin(d*x + c)^2 + 15*sin (d*x + c) - 3)/(a^4*sin(d*x + c)^5 + 3*a^4*sin(d*x + c)^4 + 3*a^4*sin(d*x + c)^3 + a^4*sin(d*x + c)^2) - 60*log(sin(d*x + c) + 1)/a^4 + 60*log(sin(d *x + c))/a^4)/d
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {10 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} d} + \frac {10 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4} d} + \frac {60 \, \sin \left (d x + c\right )^{4} + 150 \, \sin \left (d x + c\right )^{3} + 110 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 3}{6 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3} \sin \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-10*log(abs(sin(d*x + c) + 1))/(a^4*d) + 10*log(abs(sin(d*x + c)))/(a^4*d) + 1/6*(60*sin(d*x + c)^4 + 150*sin(d*x + c)^3 + 110*sin(d*x + c)^2 + 15*s in(d*x + c) - 3)/(a^4*d*(sin(d*x + c) + 1)^3*sin(d*x + c)^2)
Time = 18.08 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.18 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {10\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^4\,d}-\frac {72\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {465\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {881\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {255\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {81\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{2}}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+60\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+80\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+60\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {20\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \] Input:
int(cot(c + d*x)/(sin(c + d*x)^2*(a + a*sin(c + d*x))^4),x)
Output:
(10*log(tan(c/2 + (d*x)/2)))/(a^4*d) - tan(c/2 + (d*x)/2)^2/(8*a^4*d) - (( 255*tan(c/2 + (d*x)/2)^4)/2 - (81*tan(c/2 + (d*x)/2)^2)/2 - 30*tan(c/2 + ( d*x)/2)^3 - 5*tan(c/2 + (d*x)/2) + (881*tan(c/2 + (d*x)/2)^5)/3 + (465*tan (c/2 + (d*x)/2)^6)/2 + 72*tan(c/2 + (d*x)/2)^7 + 1/2)/(d*(4*a^4*tan(c/2 + (d*x)/2)^2 + 24*a^4*tan(c/2 + (d*x)/2)^3 + 60*a^4*tan(c/2 + (d*x)/2)^4 + 8 0*a^4*tan(c/2 + (d*x)/2)^5 + 60*a^4*tan(c/2 + (d*x)/2)^6 + 24*a^4*tan(c/2 + (d*x)/2)^7 + 4*a^4*tan(c/2 + (d*x)/2)^8)) - (20*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) + (2*tan(c/2 + (d*x)/2))/(a^4*d)
Time = 0.18 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.17 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-9 \csc \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}-3 \csc \left (d x +c \right )^{2} \sin \left (d x +c \right )-120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-360 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-58 \sin \left (d x +c \right )^{4}-114 \sin \left (d x +c \right )^{3}-24 \sin \left (d x +c \right )^{2}+52 \sin \left (d x +c \right )+24}{6 \sin \left (d x +c \right ) a^{4} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x)
Output:
( - 9*csc(c + d*x)**2*sin(c + d*x)**2 - 3*csc(c + d*x)**2*sin(c + d*x) - 1 20*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 360*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 360*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 120*l og(tan((c + d*x)/2) + 1)*sin(c + d*x) + 60*log(tan((c + d*x)/2))*sin(c + d *x)**4 + 180*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 180*log(tan((c + d*x) /2))*sin(c + d*x)**2 + 60*log(tan((c + d*x)/2))*sin(c + d*x) - 58*sin(c + d*x)**4 - 114*sin(c + d*x)**3 - 24*sin(c + d*x)**2 + 52*sin(c + d*x) + 24) /(6*sin(c + d*x)*a**4*d*(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d *x) + 1))