Integrand size = 21, antiderivative size = 86 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {5 \log (\sin (c+d x))}{a^3 d}-\frac {5 \log (1+\sin (c+d x))}{a^3 d}+\frac {2}{d \left (a^3+a^3 \sin (c+d x)\right )} \] Output:
3*csc(d*x+c)/a^3/d-1/2*csc(d*x+c)^2/a^3/d+5*ln(sin(d*x+c))/a^3/d-5*ln(1+si n(d*x+c))/a^3/d+2/d/(a^3+a^3*sin(d*x+c))
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6 \csc (c+d x)-\csc ^2(c+d x)+10 \log (\sin (c+d x))-10 \log (1+\sin (c+d x))+\frac {4}{1+\sin (c+d x)}}{2 a^3 d} \] Input:
Integrate[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^3,x]
Output:
(6*Csc[c + d*x] - Csc[c + d*x]^2 + 10*Log[Sin[c + d*x]] - 10*Log[1 + Sin[c + d*x]] + 4/(1 + Sin[c + d*x]))/(2*a^3*d)
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91, 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.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^3 (a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))}{a^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {\csc ^3(c+d x)}{a^4}-\frac {3 \csc ^2(c+d x)}{a^4}+\frac {5 \csc (c+d x)}{a^4}-\frac {5}{a^3 (\sin (c+d x) a+a)}-\frac {2}{a^2 (\sin (c+d x) a+a)^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\csc ^2(c+d x)}{2 a^3}+\frac {3 \csc (c+d x)}{a^3}+\frac {5 \log (a \sin (c+d x))}{a^3}-\frac {5 \log (a \sin (c+d x)+a)}{a^3}+\frac {2}{a^2 (a \sin (c+d x)+a)}}{d}\) |
Input:
Int[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^3,x]
Output:
((3*Csc[c + d*x])/a^3 - Csc[c + d*x]^2/(2*a^3) + (5*Log[a*Sin[c + d*x]])/a ^3 - (5*Log[a + a*Sin[c + d*x]])/a^3 + 2/(a^2*(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 = 6.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {3}{\sin \left (d x +c \right )}+5 \ln \left (\sin \left (d x +c \right )\right )+\frac {2}{1+\sin \left (d x +c \right )}-5 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(61\) |
default | \(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {3}{\sin \left (d x +c \right )}+5 \ln \left (\sin \left (d x +c \right )\right )+\frac {2}{1+\sin \left (d x +c \right )}-5 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(61\) |
risch | \(\frac {2 i \left (5 i {\mathrm e}^{4 i \left (d x +c \right )}+5 \,{\mathrm e}^{5 i \left (d x +c \right )}-5 i {\mathrm e}^{2 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} a^{3} d}-\frac {10 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {5 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}\) | \(137\) |
Input:
int(cot(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(-1/2/sin(d*x+c)^2+3/sin(d*x+c)+5*ln(sin(d*x+c))+2/(1+sin(d*x+c))- 5*ln(1+sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.71 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {10 \, \cos \left (d x + c\right )^{2} + 10 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 10 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, \sin \left (d x + c\right ) - 9}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
1/2*(10*cos(d*x + c)^2 + 10*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) - 10*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(sin(d*x + c) + 1) - 5*sin(d*x + c) - 9)/(a^3*d*co s(d*x + c)^2 - a^3*d + (a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))
\[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cot(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
Output:
Integral(cot(c + d*x)**3/(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 = 80, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {10 \, \sin \left (d x + c\right )^{2} + 5 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2}} - \frac {10 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {10 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
1/2*((10*sin(d*x + c)^2 + 5*sin(d*x + c) - 1)/(a^3*sin(d*x + c)^3 + a^3*si n(d*x + c)^2) - 10*log(sin(d*x + c) + 1)/a^3 + 10*log(sin(d*x + c))/a^3)/d
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} + \frac {10 \, \sin \left (d x + c\right )^{2} + 5 \, \sin \left (d x + c\right ) - 1}{2 \, a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-5*log(abs(sin(d*x + c) + 1))/(a^3*d) + 5*log(abs(sin(d*x + c)))/(a^3*d) + 1/2*(10*sin(d*x + c)^2 + 5*sin(d*x + c) - 1)/(a^3*d*(sin(d*x + c) + 1)*si n(d*x + c)^2)
Time = 18.03 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.97 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {10\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}+\frac {-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \] Input:
int(cot(c + d*x)^3/(a + a*sin(c + d*x))^3,x)
Output:
(5*log(tan(c/2 + (d*x)/2)))/(a^3*d) - tan(c/2 + (d*x)/2)^2/(8*a^3*d) - (10 *log(tan(c/2 + (d*x)/2) + 1))/(a^3*d) + (5*tan(c/2 + (d*x)/2) + (23*tan(c/ 2 + (d*x)/2)^2)/2 - 10*tan(c/2 + (d*x)/2)^3 - 1/2)/(d*(4*a^3*tan(c/2 + (d* x)/2)^2 + 8*a^3*tan(c/2 + (d*x)/2)^3 + 4*a^3*tan(c/2 + (d*x)/2)^4)) + (3*t an(c/2 + (d*x)/2))/(2*a^3*d)
Time = 0.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.63 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-40 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-40 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+20 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+20 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-5 \sin \left (d x +c \right )^{3}+15 \sin \left (d x +c \right )^{2}+10 \sin \left (d x +c \right )-2}{4 \sin \left (d x +c \right )^{2} a^{3} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
Output:
( - 40*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 40*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 20*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 20*log( tan((c + d*x)/2))*sin(c + d*x)**2 - 5*sin(c + d*x)**3 + 15*sin(c + d*x)**2 + 10*sin(c + d*x) - 2)/(4*sin(c + d*x)**2*a**3*d*(sin(c + d*x) + 1))