Integrand size = 27, antiderivative size = 112 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \csc ^2(c+d x)}{d}+\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^5(c+d x)}{5 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x)}{d} \] Output:
a^2*csc(d*x+c)/d+2*a^2*csc(d*x+c)^2/d+1/3*a^2*csc(d*x+c)^3/d-1/2*a^2*csc(d *x+c)^4/d-1/5*a^2*csc(d*x+c)^5/d+2*a^2*ln(sin(d*x+c))/d+a^2*sin(d*x+c)/d
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.68 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (30 \csc (c+d x)+60 \csc ^2(c+d x)+10 \csc ^3(c+d x)-15 \csc ^4(c+d x)-6 \csc ^5(c+d x)+60 \log (\sin (c+d x))+30 \sin (c+d x)\right )}{30 d} \] Input:
Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(30*Csc[c + d*x] + 60*Csc[c + d*x]^2 + 10*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 - 6*Csc[c + d*x]^5 + 60*Log[Sin[c + d*x]] + 30*Sin[c + d*x]))/(30* d)
Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.75, 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.
| \(\displaystyle \int \cot ^5(c+d x) \csc (c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^2}{\sin (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}{a^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (\csc ^6(c+d x)+2 \csc ^5(c+d x)-\csc ^4(c+d x)-4 \csc ^3(c+d x)-\csc ^2(c+d x)+2 \csc (c+d x)+1\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (a \sin (c+d x)-\frac {1}{5} a \csc ^5(c+d x)-\frac {1}{2} a \csc ^4(c+d x)+\frac {1}{3} a \csc ^3(c+d x)+2 a \csc ^2(c+d x)+a \csc (c+d x)+2 a \log (a \sin (c+d x))\right )}{d}\) |
Input:
Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(a*(a*Csc[c + d*x] + 2*a*Csc[c + d*x]^2 + (a*Csc[c + d*x]^3)/3 - (a*Csc[c + d*x]^4)/2 - (a*Csc[c + d*x]^5)/5 + 2*a*Log[a*Sin[c + d*x]] + 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 = 0.66 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69
| method | result | size |
| derivativedivides | \(-\frac {a^{2} \left (\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {\csc \left (d x +c \right )^{4}}{2}-\frac {\csc \left (d x +c \right )^{3}}{3}-2 \csc \left (d x +c \right )^{2}-\csc \left (d x +c \right )+2 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\) | \(77\) |
| default | \(-\frac {a^{2} \left (\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {\csc \left (d x +c \right )^{4}}{2}-\frac {\csc \left (d x +c \right )^{3}}{3}-2 \csc \left (d x +c \right )^{2}-\csc \left (d x +c \right )+2 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\) | \(77\) |
| risch | \(-2 i a^{2} x -\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {4 i a^{2} c}{d}+\frac {2 i a^{2} \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}-80 \,{\mathrm e}^{7 i \left (d x +c \right )}+60 i {\mathrm e}^{8 i \left (d x +c \right )}+82 \,{\mathrm e}^{5 i \left (d x +c \right )}-120 i {\mathrm e}^{6 i \left (d x +c \right )}-80 \,{\mathrm e}^{3 i \left (d x +c \right )}+120 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}-60 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(201\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-a^2/d*(1/5*csc(d*x+c)^5+1/2*csc(d*x+c)^4-1/3*csc(d*x+c)^3-2*csc(d*x+c)^2- csc(d*x+c)+2*ln(csc(d*x+c))-1/csc(d*x+c))
Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.37 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {30 \, a^{2} \cos \left (d x + c\right )^{6} - 120 \, a^{2} \cos \left (d x + c\right )^{4} + 160 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 64 \, a^{2} + 15 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-1/30*(30*a^2*cos(d*x + c)^6 - 120*a^2*cos(d*x + c)^4 + 160*a^2*cos(d*x + c)^2 - 60*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*sin(d* x + c))*sin(d*x + c) - 64*a^2 + 15*(4*a^2*cos(d*x + c)^2 - 3*a^2)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))
\[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cot(c + d*x)**5*csc(c + d*x), x) + Integral(2*sin(c + d*x)* cot(c + d*x)**5*csc(c + d*x), x) + Integral(sin(c + d*x)**2*cot(c + d*x)** 5*csc(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 30 \, a^{2} \sin \left (d x + c\right ) + \frac {30 \, a^{2} \sin \left (d x + c\right )^{4} + 60 \, a^{2} \sin \left (d x + c\right )^{3} + 10 \, a^{2} \sin \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/30*(60*a^2*log(sin(d*x + c)) + 30*a^2*sin(d*x + c) + (30*a^2*sin(d*x + c )^4 + 60*a^2*sin(d*x + c)^3 + 10*a^2*sin(d*x + c)^2 - 15*a^2*sin(d*x + c) - 6*a^2)/sin(d*x + c)^5)/d
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 30 \, a^{2} \sin \left (d x + c\right ) + \frac {30 \, a^{2} \sin \left (d x + c\right )^{4} + 60 \, a^{2} \sin \left (d x + c\right )^{3} + 10 \, a^{2} \sin \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/30*(60*a^2*log(abs(sin(d*x + c))) + 30*a^2*sin(d*x + c) + (30*a^2*sin(d* x + c)^4 + 60*a^2*sin(d*x + c)^3 + 10*a^2*sin(d*x + c)^2 - 15*a^2*sin(d*x + c) - 6*a^2)/sin(d*x + c)^5)/d
Time = 32.25 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.38 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {82\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {55\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^2}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {2\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int((cot(c + d*x)^5*(a + a*sin(c + d*x))^2)/sin(c + d*x),x)
Output:
((2*a^2*tan(c/2 + (d*x)/2)^2)/15 + 11*a^2*tan(c/2 + (d*x)/2)^3 + (55*a^2*t an(c/2 + (d*x)/2)^4)/3 + 12*a^2*tan(c/2 + (d*x)/2)^5 + 82*a^2*tan(c/2 + (d *x)/2)^6 - a^2/5 - a^2*tan(c/2 + (d*x)/2))/(d*(32*tan(c/2 + (d*x)/2)^5 + 3 2*tan(c/2 + (d*x)/2)^7)) + (3*a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^2*tan(c /2 + (d*x)/2)^3)/(96*d) - (a^2*tan(c/2 + (d*x)/2)^4)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^5)/(160*d) + (2*a^2*log(tan(c/2 + (d*x)/2)))/d + (9*a^2*tan(c/ 2 + (d*x)/2))/(16*d) - (2*a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{5}+480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+240 \sin \left (d x +c \right )^{6}-285 \sin \left (d x +c \right )^{5}+240 \sin \left (d x +c \right )^{4}+480 \sin \left (d x +c \right )^{3}+80 \sin \left (d x +c \right )^{2}-120 \sin \left (d x +c \right )-48\right )}{240 \sin \left (d x +c \right )^{5} d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 480*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**5 + 480*log(tan(( c + d*x)/2))*sin(c + d*x)**5 + 240*sin(c + d*x)**6 - 285*sin(c + d*x)**5 + 240*sin(c + d*x)**4 + 480*sin(c + d*x)**3 + 80*sin(c + d*x)**2 - 120*sin( c + d*x) - 48))/(240*sin(c + d*x)**5*d)