Integrand size = 27, antiderivative size = 135 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {16 \csc (c+d x)}{a^4 d}+\frac {6 \csc ^2(c+d x)}{a^4 d}-\frac {8 \csc ^3(c+d x)}{3 a^4 d}+\frac {\csc ^4(c+d x)}{a^4 d}-\frac {\csc ^5(c+d x)}{5 a^4 d}-\frac {20 \log (\sin (c+d x))}{a^4 d}+\frac {20 \log (1+\sin (c+d x))}{a^4 d}-\frac {4}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-16*csc(d*x+c)/a^4/d+6*csc(d*x+c)^2/a^4/d-8/3*csc(d*x+c)^3/a^4/d+csc(d*x+c )^4/a^4/d-1/5*csc(d*x+c)^5/a^4/d-20*ln(sin(d*x+c))/a^4/d+20*ln(1+sin(d*x+c ))/a^4/d-4/d/(a^4+a^4*sin(d*x+c))
Time = 0.31 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.67 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {240 \csc (c+d x)-90 \csc ^2(c+d x)+40 \csc ^3(c+d x)-15 \csc ^4(c+d x)+3 \csc ^5(c+d x)+300 \log (\sin (c+d x))-300 \log (1+\sin (c+d x))+\frac {60}{1+\sin (c+d x)}}{15 a^4 d} \] Input:
Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^4,x]
Output:
-1/15*(240*Csc[c + d*x] - 90*Csc[c + d*x]^2 + 40*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 + 3*Csc[c + d*x]^5 + 300*Log[Sin[c + d*x]] - 300*Log[1 + Sin[c + d*x]] + 60/(1 + Sin[c + d*x]))/(a^4*d)
Time = 0.35 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^6 (a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2}{a^6 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (\frac {\csc ^6(c+d x)}{a^6}-\frac {4 \csc ^5(c+d x)}{a^6}+\frac {8 \csc ^4(c+d x)}{a^6}-\frac {12 \csc ^3(c+d x)}{a^6}+\frac {16 \csc ^2(c+d x)}{a^6}-\frac {20 \csc (c+d x)}{a^6}+\frac {20}{a^5 (\sin (c+d x) a+a)}+\frac {4}{a^4 (\sin (c+d x) a+a)^2}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (-\frac {\csc ^5(c+d x)}{5 a^5}+\frac {\csc ^4(c+d x)}{a^5}-\frac {8 \csc ^3(c+d x)}{3 a^5}+\frac {6 \csc ^2(c+d x)}{a^5}-\frac {16 \csc (c+d x)}{a^5}-\frac {20 \log (a \sin (c+d x))}{a^5}+\frac {20 \log (a \sin (c+d x)+a)}{a^5}-\frac {4}{a^4 (a \sin (c+d x)+a)}\right )}{d}\) |
Input:
Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^4,x]
Output:
(a*((-16*Csc[c + d*x])/a^5 + (6*Csc[c + d*x]^2)/a^5 - (8*Csc[c + d*x]^3)/( 3*a^5) + Csc[c + d*x]^4/a^5 - Csc[c + d*x]^5/(5*a^5) - (20*Log[a*Sin[c + d *x]])/a^5 + (20*Log[a + a*Sin[c + d*x]])/a^5 - 4/(a^4*(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_)*((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 = 19.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.58
| method | result | size |
| derivativedivides | \(\frac {-\frac {\csc \left (d x +c \right )^{5}}{5}+\csc \left (d x +c \right )^{4}-\frac {8 \csc \left (d x +c \right )^{3}}{3}+6 \csc \left (d x +c \right )^{2}-16 \csc \left (d x +c \right )+\frac {4}{1+\csc \left (d x +c \right )}+20 \ln \left (1+\csc \left (d x +c \right )\right )}{d \,a^{4}}\) | \(78\) |
| default | \(\frac {-\frac {\csc \left (d x +c \right )^{5}}{5}+\csc \left (d x +c \right )^{4}-\frac {8 \csc \left (d x +c \right )^{3}}{3}+6 \csc \left (d x +c \right )^{2}-16 \csc \left (d x +c \right )+\frac {4}{1+\csc \left (d x +c \right )}+20 \ln \left (1+\csc \left (d x +c \right )\right )}{d \,a^{4}}\) | \(78\) |
| risch | \(-\frac {8 i \left (75 i {\mathrm e}^{10 i \left (d x +c \right )}+75 \,{\mathrm e}^{11 i \left (d x +c \right )}-350 i {\mathrm e}^{8 i \left (d x +c \right )}-325 \,{\mathrm e}^{9 i \left (d x +c \right )}+574 i {\mathrm e}^{6 i \left (d x +c \right )}+552 \,{\mathrm e}^{7 i \left (d x +c \right )}-350 i {\mathrm e}^{4 i \left (d x +c \right )}-552 \,{\mathrm e}^{5 i \left (d x +c \right )}+75 i {\mathrm e}^{2 i \left (d x +c \right )}+325 \,{\mathrm e}^{3 i \left (d x +c \right )}-75 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{4}}-\frac {20 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}+\frac {40 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\) | \(206\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d/a^4*(-1/5*csc(d*x+c)^5+csc(d*x+c)^4-8/3*csc(d*x+c)^3+6*csc(d*x+c)^2-16 *csc(d*x+c)+4/(1+csc(d*x+c))+20*ln(1+csc(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (131) = 262\).
Time = 0.09 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.10 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {150 \, \cos \left (d x + c\right )^{4} - 325 \, \cos \left (d x + c\right )^{2} - 300 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 300 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (150 \, \cos \left (d x + c\right )^{4} - 275 \, \cos \left (d x + c\right )^{2} + 119\right )} \sin \left (d x + c\right ) + 178}{15 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d - {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
1/15*(150*cos(d*x + c)^4 - 325*cos(d*x + c)^2 - 300*(cos(d*x + c)^6 - 3*co s(d*x + c)^4 + 3*cos(d*x + c)^2 - (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)* sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) + 300*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d* x + c) - 1)*log(sin(d*x + c) + 1) + 2*(150*cos(d*x + c)^4 - 275*cos(d*x + c)^2 + 119)*sin(d*x + c) + 178)/(a^4*d*cos(d*x + c)^6 - 3*a^4*d*cos(d*x + c)^4 + 3*a^4*d*cos(d*x + c)^2 - a^4*d - (a^4*d*cos(d*x + c)^4 - 2*a^4*d*co s(d*x + c)^2 + a^4*d)*sin(d*x + c))
\[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)/(a+a*sin(d*x+c))**4,x)
Output:
Integral(cot(c + d*x)**5*csc(c + d*x)/(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 = 110, normalized size of antiderivative = 0.81 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {300 \, \sin \left (d x + c\right )^{5} + 150 \, \sin \left (d x + c\right )^{4} - 50 \, \sin \left (d x + c\right )^{3} + 25 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 3}{a^{4} \sin \left (d x + c\right )^{6} + a^{4} \sin \left (d x + c\right )^{5}} - \frac {300 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {300 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{15 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
-1/15*((300*sin(d*x + c)^5 + 150*sin(d*x + c)^4 - 50*sin(d*x + c)^3 + 25*s in(d*x + c)^2 - 12*sin(d*x + c) + 3)/(a^4*sin(d*x + c)^6 + a^4*sin(d*x + c )^5) - 300*log(sin(d*x + c) + 1)/a^4 + 300*log(sin(d*x + c))/a^4)/d
Time = 0.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {20 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} d} - \frac {20 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4} d} - \frac {300 \, \sin \left (d x + c\right )^{5} + 150 \, \sin \left (d x + c\right )^{4} - 50 \, \sin \left (d x + c\right )^{3} + 25 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 3}{15 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{5}} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
20*log(abs(sin(d*x + c) + 1))/(a^4*d) - 20*log(abs(sin(d*x + c)))/(a^4*d) - 1/15*(300*sin(d*x + c)^5 + 150*sin(d*x + c)^4 - 50*sin(d*x + c)^3 + 25*s in(d*x + c)^2 - 12*sin(d*x + c) + 3)/(a^4*d*(sin(d*x + c) + 1)*sin(d*x + c )^5)
Time = 34.17 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.97 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^4\,d}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^4\,d}-\frac {20\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+524\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {569\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {104\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {118\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {1}{5}}{d\,\left (32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+64\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {40\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {145\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^4\,d} \] Input:
int(cot(c + d*x)^5/(sin(c + d*x)*(a + a*sin(c + d*x))^4),x)
Output:
(7*tan(c/2 + (d*x)/2)^2)/(4*a^4*d) - (35*tan(c/2 + (d*x)/2)^3)/(96*a^4*d) + tan(c/2 + (d*x)/2)^4/(16*a^4*d) - tan(c/2 + (d*x)/2)^5/(160*a^4*d) - (20 *log(tan(c/2 + (d*x)/2)))/(a^4*d) - ((118*tan(c/2 + (d*x)/2)^2)/15 - (8*ta n(c/2 + (d*x)/2))/5 - (104*tan(c/2 + (d*x)/2)^3)/3 + (569*tan(c/2 + (d*x)/ 2)^4)/3 + 524*tan(c/2 + (d*x)/2)^5 + 34*tan(c/2 + (d*x)/2)^6 + 1/5)/(d*(32 *a^4*tan(c/2 + (d*x)/2)^5 + 64*a^4*tan(c/2 + (d*x)/2)^6 + 32*a^4*tan(c/2 + (d*x)/2)^7)) + (40*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) - (145*tan(c/2 + (d*x)/2))/(16*a^4*d)
Time = 0.16 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.26 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {9600 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{6}+9600 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5}-4800 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{6}-4800 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+1365 \sin \left (d x +c \right )^{6}-3435 \sin \left (d x +c \right )^{5}-2400 \sin \left (d x +c \right )^{4}+800 \sin \left (d x +c \right )^{3}-400 \sin \left (d x +c \right )^{2}+192 \sin \left (d x +c \right )-48}{240 \sin \left (d x +c \right )^{5} a^{4} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^4,x)
Output:
(9600*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 9600*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**5 - 4800*log(tan((c + d*x)/2))*sin(c + d*x)**6 - 4800 *log(tan((c + d*x)/2))*sin(c + d*x)**5 + 1365*sin(c + d*x)**6 - 3435*sin(c + d*x)**5 - 2400*sin(c + d*x)**4 + 800*sin(c + d*x)**3 - 400*sin(c + d*x) **2 + 192*sin(c + d*x) - 48)/(240*sin(c + d*x)**5*a**4*d*(sin(c + d*x) + 1 ))