Integrand size = 27, antiderivative size = 133 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {5 a^3 \csc (c+d x)}{d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^5(c+d x)}{5 d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \]
5*a^3*csc(d*x+c)/d+5/2*a^3*csc(d*x+c)^2/d-1/3*a^3*csc(d*x+c)^3/d-3/4*a^3*c sc(d*x+c)^4/d-1/5*a^3*csc(d*x+c)^5/d+a^3*ln(sin(d*x+c))/d+3*a^3*sin(d*x+c) /d+1/2*a^3*sin(d*x+c)^2/d
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (300 \csc (c+d x)+150 \csc ^2(c+d x)-20 \csc ^3(c+d x)-45 \csc ^4(c+d x)-12 \csc ^5(c+d x)+60 \log (\sin (c+d x))+180 \sin (c+d x)+30 \sin ^2(c+d x)\right )}{60 d} \]
(a^3*(300*Csc[c + d*x] + 150*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 - 12*Csc[c + d*x]^5 + 60*Log[Sin[c + d*x]] + 180*Sin[c + d*x] + 30*Sin[c + d*x]^2))/(60*d)
Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87, 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.
| \(\displaystyle \int \cot ^5(c+d x) \csc (c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^3}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5}{a^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (a \csc ^6(c+d x)+3 a \csc ^5(c+d x)+a \csc ^4(c+d x)-5 a \csc ^3(c+d x)-5 a \csc ^2(c+d x)+a \csc (c+d x)+3 a+a \sin (c+d x)\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {1}{2} a^2 \sin ^2(c+d x)+3 a^2 \sin (c+d x)-\frac {1}{5} a^2 \csc ^5(c+d x)-\frac {3}{4} a^2 \csc ^4(c+d x)-\frac {1}{3} a^2 \csc ^3(c+d x)+\frac {5}{2} a^2 \csc ^2(c+d x)+5 a^2 \csc (c+d x)+a^2 \log (a \sin (c+d x))\right )}{d}\) |
(a*(5*a^2*Csc[c + d*x] + (5*a^2*Csc[c + d*x]^2)/2 - (a^2*Csc[c + d*x]^3)/3 - (3*a^2*Csc[c + d*x]^4)/4 - (a^2*Csc[c + d*x]^5)/5 + a^2*Log[a*Sin[c + d *x]] + 3*a^2*Sin[c + d*x] + (a^2*Sin[c + d*x]^2)/2))/d
3.6.28.3.1 Defintions of rubi rules used
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 = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64
| method | result | size |
| derivativedivides | \(-\frac {a^{3} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {5 \left (\csc ^{2}\left (d x +c \right )\right )}{2}-5 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\) | \(85\) |
| default | \(-\frac {a^{3} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {5 \left (\csc ^{2}\left (d x +c \right )\right )}{2}-5 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\) | \(85\) |
| parallelrisch | \(\frac {\left (\left (-\sin \left (5 d x +5 c \right )+5 \sin \left (3 d x +3 c \right )-10 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\sin \left (5 d x +5 c \right )-5 \sin \left (3 d x +3 c \right )+10 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {59 \sin \left (5 d x +5 c \right )}{64}-\frac {\sin \left (7 d x +7 c \right )}{8}-\frac {359 \cos \left (2 d x +2 c \right )}{6}+19 \cos \left (4 d x +4 c \right )-\frac {3 \cos \left (6 d x +6 c \right )}{2}+\frac {141 \sin \left (d x +c \right )}{32}-\frac {233 \sin \left (3 d x +3 c \right )}{64}+\frac {587}{15}\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{512 d}\) | \(191\) |
| risch | \(-i a^{3} x -\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3} c}{d}+\frac {2 i a^{3} \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}-280 \,{\mathrm e}^{7 i \left (d x +c \right )}+75 i {\mathrm e}^{8 i \left (d x +c \right )}+362 \,{\mathrm e}^{5 i \left (d x +c \right )}-135 i {\mathrm e}^{6 i \left (d x +c \right )}-280 \,{\mathrm e}^{3 i \left (d x +c \right )}+135 i {\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}-75 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(234\) |
| norman | \(\frac {-\frac {a^{3}}{160 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}-\frac {11 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d}+\frac {19 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {83 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {601 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {1235 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d}+\frac {601 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {83 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {19 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {11 a^{3} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d}-\frac {3 a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {a^{3} \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {3 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {3 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(341\) |
-a^3/d*(1/5*csc(d*x+c)^5+3/4*csc(d*x+c)^4+1/3*csc(d*x+c)^3-5/2*csc(d*x+c)^ 2-5*csc(d*x+c)+ln(csc(d*x+c))-3/csc(d*x+c)-1/2/csc(d*x+c)^2)
Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.35 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {180 \, a^{3} \cos \left (d x + c\right )^{6} - 840 \, a^{3} \cos \left (d x + c\right )^{4} + 1120 \, a^{3} \cos \left (d x + c\right )^{2} - 448 \, a^{3} - 60 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{6} - 5 \, a^{3} \cos \left (d x + c\right )^{4} + 14 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]
-1/60*(180*a^3*cos(d*x + c)^6 - 840*a^3*cos(d*x + c)^4 + 1120*a^3*cos(d*x + c)^2 - 448*a^3 - 60*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*lo g(1/2*sin(d*x + c))*sin(d*x + c) + 15*(2*a^3*cos(d*x + c)^6 - 5*a^3*cos(d* x + c)^4 + 14*a^3*cos(d*x + c)^2 - 8*a^3)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))
Timed out. \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 180 \, a^{3} \sin \left (d x + c\right ) + \frac {300 \, a^{3} \sin \left (d x + c\right )^{4} + 150 \, a^{3} \sin \left (d x + c\right )^{3} - 20 \, a^{3} \sin \left (d x + c\right )^{2} - 45 \, a^{3} \sin \left (d x + c\right ) - 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]
1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c)) + 180*a^3*sin(d*x + c) + (300*a^3*sin(d*x + c)^4 + 150*a^3*sin(d*x + c)^3 - 20*a^3*sin(d*x + c)^2 - 45*a^3*sin(d*x + c) - 12*a^3)/sin(d*x + c)^5)/d
Time = 0.42 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a^{3} \sin \left (d x + c\right ) - \frac {137 \, a^{3} \sin \left (d x + c\right )^{5} - 300 \, a^{3} \sin \left (d x + c\right )^{4} - 150 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 45 \, a^{3} \sin \left (d x + c\right ) + 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]
1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(abs(sin(d*x + c))) + 180*a^3*sin( d*x + c) - (137*a^3*sin(d*x + c)^5 - 300*a^3*sin(d*x + c)^4 - 150*a^3*sin( d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 45*a^3*sin(d*x + c) + 12*a^3)/sin(d*x + c)^5)/d
Time = 9.85 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.34 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {266\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+78\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {1013\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {53\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {1037\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {41\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {a^3}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {37\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
(7*a^3*tan(c/2 + (d*x)/2)^2)/(16*d) - (7*a^3*tan(c/2 + (d*x)/2)^3)/(96*d) - (3*a^3*tan(c/2 + (d*x)/2)^4)/(64*d) - (a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (a^3*log(tan(c/2 + (d*x)/2)))/d + (11*a^3*tan(c/2 + (d*x)/2)^3 - (41*a^ 3*tan(c/2 + (d*x)/2)^2)/15 + (1037*a^3*tan(c/2 + (d*x)/2)^4)/15 + (53*a^3* tan(c/2 + (d*x)/2)^5)/2 + (1013*a^3*tan(c/2 + (d*x)/2)^6)/3 + 78*a^3*tan(c /2 + (d*x)/2)^7 + 266*a^3*tan(c/2 + (d*x)/2)^8 - a^3/5 - (3*a^3*tan(c/2 + (d*x)/2))/2)/(d*(32*tan(c/2 + (d*x)/2)^5 + 64*tan(c/2 + (d*x)/2)^7 + 32*ta n(c/2 + (d*x)/2)^9)) + (37*a^3*tan(c/2 + (d*x)/2))/(16*d) - (a^3*log(tan(c /2 + (d*x)/2)^2 + 1))/d