Integrand size = 27, antiderivative size = 146 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {10 a^4 \csc (c+d x)}{d}+\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {4 a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {a^4 \sin ^3(c+d x)}{3 d} \] Output:
10*a^4*csc(d*x+c)/d+2*a^4*csc(d*x+c)^2/d-4/3*a^4*csc(d*x+c)^3/d-a^4*csc(d* x+c)^4/d-1/5*a^4*csc(d*x+c)^5/d-4*a^4*ln(sin(d*x+c))/d+4*a^4*sin(d*x+c)/d+ 2*a^4*sin(d*x+c)^2/d+1/3*a^4*sin(d*x+c)^3/d
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.66 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \left (150 \csc (c+d x)+30 \csc ^2(c+d x)-20 \csc ^3(c+d x)-15 \csc ^4(c+d x)-3 \csc ^5(c+d x)-60 \log (\sin (c+d x))+60 \sin (c+d x)+30 \sin ^2(c+d x)+5 \sin ^3(c+d x)\right )}{15 d} \] Input:
Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]
Output:
(a^4*(150*Csc[c + d*x] + 30*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 - 3*Csc[c + d*x]^5 - 60*Log[Sin[c + d*x]] + 60*Sin[c + d*x] + 30* Sin[c + d*x]^2 + 5*Sin[c + d*x]^3))/(15*d)
Time = 0.35 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86, 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.
| \(\displaystyle \int \cot ^5(c+d x) \csc (c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^4}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^6d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^6}{a^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (a^2 \csc ^6(c+d x)+4 a^2 \csc ^5(c+d x)+4 a^2 \csc ^4(c+d x)-4 a^2 \csc ^3(c+d x)-10 a^2 \csc ^2(c+d x)-4 a^2 \csc (c+d x)+4 a^2+a^2 \sin ^2(c+d x)+4 a^2 \sin (c+d x)\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {1}{3} a^3 \sin ^3(c+d x)+2 a^3 \sin ^2(c+d x)+4 a^3 \sin (c+d x)-\frac {1}{5} a^3 \csc ^5(c+d x)-a^3 \csc ^4(c+d x)-\frac {4}{3} a^3 \csc ^3(c+d x)+2 a^3 \csc ^2(c+d x)+10 a^3 \csc (c+d x)-4 a^3 \log (a \sin (c+d x))\right )}{d}\) |
Input:
Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]
Output:
(a*(10*a^3*Csc[c + d*x] + 2*a^3*Csc[c + d*x]^2 - (4*a^3*Csc[c + d*x]^3)/3 - a^3*Csc[c + d*x]^4 - (a^3*Csc[c + d*x]^5)/5 - 4*a^3*Log[a*Sin[c + d*x]] + 4*a^3*Sin[c + d*x] + 2*a^3*Sin[c + d*x]^2 + (a^3*Sin[c + d*x]^3)/3))/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 = 2.63 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(-\frac {a^{4} \left (\frac {\csc \left (d x +c \right )^{5}}{5}+\csc \left (d x +c \right )^{4}+\frac {4 \csc \left (d x +c \right )^{3}}{3}-2 \csc \left (d x +c \right )^{2}-10 \csc \left (d x +c \right )-\frac {4}{\csc \left (d x +c \right )}-4 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{3 \csc \left (d x +c \right )^{3}}-\frac {2}{\csc \left (d x +c \right )^{2}}\right )}{d}\) | \(95\) |
| default | \(-\frac {a^{4} \left (\frac {\csc \left (d x +c \right )^{5}}{5}+\csc \left (d x +c \right )^{4}+\frac {4 \csc \left (d x +c \right )^{3}}{3}-2 \csc \left (d x +c \right )^{2}-10 \csc \left (d x +c \right )-\frac {4}{\csc \left (d x +c \right )}-4 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{3 \csc \left (d x +c \right )^{3}}-\frac {2}{\csc \left (d x +c \right )^{2}}\right )}{d}\) | \(95\) |
| risch | \(4 i a^{4} x +\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {17 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {17 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {8 i a^{4} c}{d}+\frac {4 i a^{4} \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}-260 \,{\mathrm e}^{7 i \left (d x +c \right )}+30 i {\mathrm e}^{8 i \left (d x +c \right )}+346 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{6 i \left (d x +c \right )}-260 \,{\mathrm e}^{3 i \left (d x +c \right )}+30 i {\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {4 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(271\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-a^4/d*(1/5*csc(d*x+c)^5+csc(d*x+c)^4+4/3*csc(d*x+c)^3-2*csc(d*x+c)^2-10*c sc(d*x+c)-4/csc(d*x+c)-4*ln(csc(d*x+c))-1/3/csc(d*x+c)^3-2/csc(d*x+c)^2)
Time = 0.10 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.32 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \cos \left (d x + c\right )^{8} - 80 \, a^{4} \cos \left (d x + c\right )^{6} + 360 \, a^{4} \cos \left (d x + c\right )^{4} - 480 \, a^{4} \cos \left (d x + c\right )^{2} + 192 \, a^{4} - 60 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{6} - 5 \, a^{4} \cos \left (d x + c\right )^{4} + 6 \, a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="fricas" )
Output:
1/15*(5*a^4*cos(d*x + c)^8 - 80*a^4*cos(d*x + c)^6 + 360*a^4*cos(d*x + c)^ 4 - 480*a^4*cos(d*x + c)^2 + 192*a^4 - 60*(a^4*cos(d*x + c)^4 - 2*a^4*cos( d*x + c)^2 + a^4)*log(1/2*sin(d*x + c))*sin(d*x + c) - 15*(2*a^4*cos(d*x + c)^6 - 5*a^4*cos(d*x + c)^4 + 6*a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c)) /((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))
Timed out. \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)*(a+a*sin(d*x+c))**4,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.82 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{4} \sin \left (d x + c\right ) + \frac {150 \, a^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{4} \sin \left (d x + c\right )^{3} - 20 \, a^{4} \sin \left (d x + c\right )^{2} - 15 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{5}}}{15 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="maxima" )
Output:
1/15*(5*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^2 - 60*a^4*log(sin(d*x + c)) + 60*a^4*sin(d*x + c) + (150*a^4*sin(d*x + c)^4 + 30*a^4*sin(d*x + c)^ 3 - 20*a^4*sin(d*x + c)^2 - 15*a^4*sin(d*x + c) - 3*a^4)/sin(d*x + c)^5)/d
Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.83 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{4} \sin \left (d x + c\right ) + \frac {150 \, a^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{4} \sin \left (d x + c\right )^{3} - 20 \, a^{4} \sin \left (d x + c\right )^{2} - 15 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{5}}}{15 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
1/15*(5*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^2 - 60*a^4*log(abs(sin(d* x + c))) + 60*a^4*sin(d*x + c) + (150*a^4*sin(d*x + c)^4 + 30*a^4*sin(d*x + c)^3 - 20*a^4*sin(d*x + c)^2 - 15*a^4*sin(d*x + c) - 3*a^4)/sin(d*x + c) ^5)/d
Time = 35.36 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.45 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {19\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {4\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {398\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+264\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+1017\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+278\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {3314\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+18\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {612\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {104\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^4}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {71\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {4\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int((cot(c + d*x)^5*(a + a*sin(c + d*x))^4)/sin(c + d*x),x)
Output:
(a^4*tan(c/2 + (d*x)/2)^2)/(4*d) - (19*a^4*tan(c/2 + (d*x)/2)^3)/(96*d) - (a^4*tan(c/2 + (d*x)/2)^4)/(16*d) - (a^4*tan(c/2 + (d*x)/2)^5)/(160*d) - ( 4*a^4*log(tan(c/2 + (d*x)/2)))/d + (2*a^4*tan(c/2 + (d*x)/2)^3 - (104*a^4* tan(c/2 + (d*x)/2)^2)/15 + (612*a^4*tan(c/2 + (d*x)/2)^4)/5 + 18*a^4*tan(c /2 + (d*x)/2)^5 + (3314*a^4*tan(c/2 + (d*x)/2)^6)/5 + 278*a^4*tan(c/2 + (d *x)/2)^7 + 1017*a^4*tan(c/2 + (d*x)/2)^8 + 264*a^4*tan(c/2 + (d*x)/2)^9 + 398*a^4*tan(c/2 + (d*x)/2)^10 - a^4/5 - 2*a^4*tan(c/2 + (d*x)/2))/(d*(32*t an(c/2 + (d*x)/2)^5 + 96*tan(c/2 + (d*x)/2)^7 + 96*tan(c/2 + (d*x)/2)^9 + 32*tan(c/2 + (d*x)/2)^11)) + (71*a^4*tan(c/2 + (d*x)/2))/(16*d) + (4*a^4*l og(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.96 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \left (960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{5}-960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+80 \sin \left (d x +c \right )^{8}+480 \sin \left (d x +c \right )^{7}+960 \sin \left (d x +c \right )^{6}-285 \sin \left (d x +c \right )^{5}+2400 \sin \left (d x +c \right )^{4}+480 \sin \left (d x +c \right )^{3}-320 \sin \left (d x +c \right )^{2}-240 \sin \left (d x +c \right )-48\right )}{240 \sin \left (d x +c \right )^{5} d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^4,x)
Output:
(a**4*(960*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**5 - 960*log(tan((c + d*x)/2))*sin(c + d*x)**5 + 80*sin(c + d*x)**8 + 480*sin(c + d*x)**7 + 960 *sin(c + d*x)**6 - 285*sin(c + d*x)**5 + 2400*sin(c + d*x)**4 + 480*sin(c + d*x)**3 - 320*sin(c + d*x)**2 - 240*sin(c + d*x) - 48))/(240*sin(c + d*x )**5*d)