Integrand size = 27, antiderivative size = 117 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \csc (c+d x)}{a^3 d}+\frac {2 \csc ^2(c+d x)}{a^3 d}-\frac {4 \csc ^3(c+d x)}{3 a^3 d}+\frac {3 \csc ^4(c+d x)}{4 a^3 d}-\frac {\csc ^5(c+d x)}{5 a^3 d}-\frac {4 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (1+\sin (c+d x))}{a^3 d} \] Output:
-4*csc(d*x+c)/a^3/d+2*csc(d*x+c)^2/a^3/d-4/3*csc(d*x+c)^3/a^3/d+3/4*csc(d* x+c)^4/a^3/d-1/5*csc(d*x+c)^5/a^3/d-4*ln(sin(d*x+c))/a^3/d+4*ln(1+sin(d*x+ c))/a^3/d
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {240 \csc (c+d x)-120 \csc ^2(c+d x)+80 \csc ^3(c+d x)-45 \csc ^4(c+d x)+12 \csc ^5(c+d x)+240 \log (\sin (c+d x))-240 \log (1+\sin (c+d x))}{60 a^3 d} \] Input:
Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]
Output:
-1/60*(240*Csc[c + d*x] - 120*Csc[c + d*x]^2 + 80*Csc[c + d*x]^3 - 45*Csc[ c + d*x]^4 + 12*Csc[c + d*x]^5 + 240*Log[Sin[c + d*x]] - 240*Log[1 + Sin[c + d*x]])/(a^3*d)
Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^6 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2}{a^6 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (\frac {\csc ^6(c+d x)}{a^5}-\frac {3 \csc ^5(c+d x)}{a^5}+\frac {4 \csc ^4(c+d x)}{a^5}-\frac {4 \csc ^3(c+d x)}{a^5}+\frac {4 \csc ^2(c+d x)}{a^5}-\frac {4 \csc (c+d x)}{a^5}+\frac {4}{a^4 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (-\frac {\csc ^5(c+d x)}{5 a^4}+\frac {3 \csc ^4(c+d x)}{4 a^4}-\frac {4 \csc ^3(c+d x)}{3 a^4}+\frac {2 \csc ^2(c+d x)}{a^4}-\frac {4 \csc (c+d x)}{a^4}-\frac {4 \log (a \sin (c+d x))}{a^4}+\frac {4 \log (a \sin (c+d x)+a)}{a^4}\right )}{d}\) |
Input:
Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]
Output:
(a*((-4*Csc[c + d*x])/a^4 + (2*Csc[c + d*x]^2)/a^4 - (4*Csc[c + d*x]^3)/(3 *a^4) + (3*Csc[c + d*x]^4)/(4*a^4) - Csc[c + d*x]^5/(5*a^4) - (4*Log[a*Sin [c + d*x]])/a^4 + (4*Log[a + a*Sin[c + d*x]])/a^4))/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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 6.97 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.58
| method | result | size |
| derivativedivides | \(\frac {-\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {3 \csc \left (d x +c \right )^{4}}{4}-\frac {4 \csc \left (d x +c \right )^{3}}{3}+2 \csc \left (d x +c \right )^{2}-4 \csc \left (d x +c \right )+4 \ln \left (1+\csc \left (d x +c \right )\right )}{d \,a^{3}}\) | \(68\) |
| default | \(\frac {-\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {3 \csc \left (d x +c \right )^{4}}{4}-\frac {4 \csc \left (d x +c \right )^{3}}{3}+2 \csc \left (d x +c \right )^{2}-4 \csc \left (d x +c \right )+4 \ln \left (1+\csc \left (d x +c \right )\right )}{d \,a^{3}}\) | \(68\) |
| risch | \(-\frac {4 i \left (30 \,{\mathrm e}^{9 i \left (d x +c \right )}-160 \,{\mathrm e}^{7 i \left (d x +c \right )}-30 i {\mathrm e}^{8 i \left (d x +c \right )}+284 \,{\mathrm e}^{5 i \left (d x +c \right )}+135 i {\mathrm e}^{6 i \left (d x +c \right )}-160 \,{\mathrm e}^{3 i \left (d x +c \right )}-135 i {\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{i \left (d x +c \right )}+30 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\) | \(169\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(-1/5*csc(d*x+c)^5+3/4*csc(d*x+c)^4-4/3*csc(d*x+c)^3+2*csc(d*x+c)^ 2-4*csc(d*x+c)+4*ln(1+csc(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.38 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {240 \, \cos \left (d x + c\right )^{4} + 240 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 240 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 560 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) + 332}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
-1/60*(240*cos(d*x + c)^4 + 240*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*lo g(1/2*sin(d*x + c))*sin(d*x + c) - 240*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(sin(d*x + c) + 1)*sin(d*x + c) - 560*cos(d*x + c)^2 + 15*(8*cos(d *x + c)^2 - 11)*sin(d*x + c) + 332)/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d *x + c)^2 + a^3*d)*sin(d*x + c))
\[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)/(a+a*sin(d*x+c))**3,x)
Output:
Integral(cot(c + d*x)**5*csc(c + d*x)/(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 = 85, normalized size of antiderivative = 0.73 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {240 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {240 \, \sin \left (d x + c\right )^{4} - 120 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} - 45 \, \sin \left (d x + c\right ) + 12}{a^{3} \sin \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/60*(240*log(sin(d*x + c) + 1)/a^3 - 240*log(sin(d*x + c))/a^3 - (240*sin (d*x + c)^4 - 120*sin(d*x + c)^3 + 80*sin(d*x + c)^2 - 45*sin(d*x + c) + 1 2)/(a^3*sin(d*x + c)^5))/d
Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} - \frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} - \frac {240 \, \sin \left (d x + c\right )^{4} - 120 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} - 45 \, \sin \left (d x + c\right ) + 12}{60 \, a^{3} d \sin \left (d x + c\right )^{5}} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
4*log(abs(sin(d*x + c) + 1))/(a^3*d) - 4*log(abs(sin(d*x + c)))/(a^3*d) - 1/60*(240*sin(d*x + c)^4 - 120*sin(d*x + c)^3 + 80*sin(d*x + c)^2 - 45*sin (d*x + c) + 12)/(a^3*d*sin(d*x + c)^5)
Time = 34.52 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.74 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a^3\,d}-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^3\,d}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^3\,d}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {41\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a^3\,d} \] Input:
int(cot(c + d*x)^5/(sin(c + d*x)*(a + a*sin(c + d*x))^3),x)
Output:
(11*tan(c/2 + (d*x)/2)^2)/(16*a^3*d) - (19*tan(c/2 + (d*x)/2)^3)/(96*a^3*d ) + (3*tan(c/2 + (d*x)/2)^4)/(64*a^3*d) - tan(c/2 + (d*x)/2)^5/(160*a^3*d) - (4*log(tan(c/2 + (d*x)/2)))/(a^3*d) + (8*log(tan(c/2 + (d*x)/2) + 1))/( a^3*d) - (41*tan(c/2 + (d*x)/2))/(16*a^3*d) - (cot(c/2 + (d*x)/2)^5*((19*t an(c/2 + (d*x)/2)^2)/3 - (3*tan(c/2 + (d*x)/2))/2 - 22*tan(c/2 + (d*x)/2)^ 3 + 82*tan(c/2 + (d*x)/2)^4 + 1/5))/(32*a^3*d)
Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3840 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5}-1920 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}-615 \sin \left (d x +c \right )^{5}-1920 \sin \left (d x +c \right )^{4}+960 \sin \left (d x +c \right )^{3}-640 \sin \left (d x +c \right )^{2}+360 \sin \left (d x +c \right )-96}{480 \sin \left (d x +c \right )^{5} a^{3} d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x)
Output:
(3840*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 1920*log(tan((c + d*x)/2 ))*sin(c + d*x)**5 - 615*sin(c + d*x)**5 - 1920*sin(c + d*x)**4 + 960*sin( c + d*x)**3 - 640*sin(c + d*x)**2 + 360*sin(c + d*x) - 96)/(480*sin(c + d* x)**5*a**3*d)