Integrand size = 21, antiderivative size = 106 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {4 \csc (c+d x)}{a^4 d}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {9 \log (\sin (c+d x))}{a^4 d}-\frac {9 \log (1+\sin (c+d x))}{a^4 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {5}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
4*csc(d*x+c)/a^4/d-1/2*csc(d*x+c)^2/a^4/d+9*ln(sin(d*x+c))/a^4/d-9*ln(1+si n(d*x+c))/a^4/d+1/d/(a^2+a^2*sin(d*x+c))^2+5/d/(a^4+a^4*sin(d*x+c))
Time = 0.84 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.69 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {8 \csc (c+d x)-\csc ^2(c+d x)+18 \log (\sin (c+d x))-18 \log (1+\sin (c+d x))+\frac {2}{(1+\sin (c+d x))^2}+\frac {10}{1+\sin (c+d x)}}{2 a^4 d} \] Input:
Integrate[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]
Output:
(8*Csc[c + d*x] - Csc[c + d*x]^2 + 18*Log[Sin[c + d*x]] - 18*Log[1 + Sin[c + d*x]] + 2/(1 + Sin[c + d*x])^2 + 10/(1 + Sin[c + d*x]))/(2*a^4*d)
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 86, 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.
| \(\displaystyle \int \frac {\cot ^3(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^3 (a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))}{a^3 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^3(c+d x)}{a^5}-\frac {4 \csc ^2(c+d x)}{a^5}+\frac {9 \csc (c+d x)}{a^5}-\frac {9}{a^4 (\sin (c+d x) a+a)}-\frac {5}{a^3 (\sin (c+d x) a+a)^2}-\frac {2}{a^2 (\sin (c+d x) a+a)^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\csc ^2(c+d x)}{2 a^4}+\frac {4 \csc (c+d x)}{a^4}+\frac {9 \log (a \sin (c+d x))}{a^4}-\frac {9 \log (a \sin (c+d x)+a)}{a^4}+\frac {5}{a^3 (a \sin (c+d x)+a)}+\frac {1}{a^2 (a \sin (c+d x)+a)^2}}{d}\) |
Input:
Int[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]
Output:
((4*Csc[c + d*x])/a^4 - Csc[c + d*x]^2/(2*a^4) + (9*Log[a*Sin[c + d*x]])/a ^4 - (9*Log[a + a*Sin[c + d*x]])/a^4 + 1/(a^2*(a + a*Sin[c + d*x])^2) + 5/ (a^3*(a + a*Sin[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 16.51 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{1+\sin \left (d x +c \right )}-9 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {4}{\sin \left (d x +c \right )}+9 \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{4}}\) | \(71\) |
| default | \(\frac {\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{1+\sin \left (d x +c \right )}-9 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {4}{\sin \left (d x +c \right )}+9 \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{4}}\) | \(71\) |
| risch | \(\frac {2 i \left (27 i {\mathrm e}^{6 i \left (d x +c \right )}+9 \,{\mathrm e}^{7 i \left (d x +c \right )}-50 i {\mathrm e}^{4 i \left (d x +c \right )}-39 \,{\mathrm e}^{5 i \left (d x +c \right )}+27 i {\mathrm e}^{2 i \left (d x +c \right )}+39 \,{\mathrm e}^{3 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} a^{4} d}-\frac {18 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}+\frac {9 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\) | \(160\) |
Input:
int(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d/a^4*(1/(1+sin(d*x+c))^2+5/(1+sin(d*x+c))-9*ln(1+sin(d*x+c))-1/2/sin(d* x+c)^2+4/sin(d*x+c)+9*ln(sin(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.85 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {27 \, \cos \left (d x + c\right )^{2} - 18 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 18 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 26}{2 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 3 \, a^{4} d \cos \left (d x + c\right )^{2} + 2 \, a^{4} d - 2 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
-1/2*(27*cos(d*x + c)^2 - 18*(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)) + 18*(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) + 6*(3*cos(d*x + c)^2 - 4)*sin(d*x + c) - 26)/(a^4*d*cos(d*x + c)^4 - 3*a^4*d*cos(d*x + c)^2 + 2*a^4*d - 2*(a^4*d*cos(d*x + c)^2 - a^4*d )*sin(d*x + c))
\[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(cot(d*x+c)**3/(a+a*sin(d*x+c))**4,x)
Output:
Integral(cot(c + d*x)**3/(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 = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {18 \, \sin \left (d x + c\right )^{3} + 27 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 1}{a^{4} \sin \left (d x + c\right )^{4} + 2 \, a^{4} \sin \left (d x + c\right )^{3} + a^{4} \sin \left (d x + c\right )^{2}} - \frac {18 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {18 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
1/2*((18*sin(d*x + c)^3 + 27*sin(d*x + c)^2 + 6*sin(d*x + c) - 1)/(a^4*sin (d*x + c)^4 + 2*a^4*sin(d*x + c)^3 + a^4*sin(d*x + c)^2) - 18*log(sin(d*x + c) + 1)/a^4 + 18*log(sin(d*x + c))/a^4)/d
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {9 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} d} + \frac {9 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4} d} + \frac {18 \, \sin \left (d x + c\right )^{3} + 27 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 1}{2 \, {\left (\sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )\right )}^{2} a^{4} d} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-9*log(abs(sin(d*x + c) + 1))/(a^4*d) + 9*log(abs(sin(d*x + c)))/(a^4*d) + 1/2*(18*sin(d*x + c)^3 + 27*sin(d*x + c)^2 + 6*sin(d*x + c) - 1)/((sin(d* x + c)^2 + sin(d*x + c))^2*a^4*d)
Time = 17.79 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.15 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^4\,d}-\frac {48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {129\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{2}}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {18\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \] Input:
int(cot(c + d*x)^3/(a + a*sin(c + d*x))^4,x)
Output:
(9*log(tan(c/2 + (d*x)/2)))/(a^4*d) - tan(c/2 + (d*x)/2)^2/(8*a^4*d) - (10 *tan(c/2 + (d*x)/2)^3 - 29*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) + ( 129*tan(c/2 + (d*x)/2)^4)/2 + 48*tan(c/2 + (d*x)/2)^5 + 1/2)/(d*(4*a^4*tan (c/2 + (d*x)/2)^2 + 16*a^4*tan(c/2 + (d*x)/2)^3 + 24*a^4*tan(c/2 + (d*x)/2 )^4 + 16*a^4*tan(c/2 + (d*x)/2)^5 + 4*a^4*tan(c/2 + (d*x)/2)^6)) - (18*log (tan(c/2 + (d*x)/2) + 1))/(a^4*d) + (2*tan(c/2 + (d*x)/2))/(a^4*d)
Time = 0.20 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.91 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-144 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-288 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-144 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+144 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-45 \sin \left (d x +c \right )^{4}-18 \sin \left (d x +c \right )^{3}+63 \sin \left (d x +c \right )^{2}+24 \sin \left (d x +c \right )-4}{8 \sin \left (d x +c \right )^{2} a^{4} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x)
Output:
( - 144*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 288*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**3 - 144*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 72*log(tan((c + d*x)/2))*sin(c + d*x)**4 + 144*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 72*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 45*sin(c + d*x)**4 - 18*sin(c + d*x)**3 + 63*sin(c + d*x)**2 + 24*sin(c + d*x) - 4)/(8*sin(c + d*x)**2*a**4*d*(sin(c + d*x)**2 + 2*sin(c + d*x) + 1))