Integrand size = 29, antiderivative size = 114 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x)}{d}-\frac {2 a^2 \sin ^2(c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^4(c+d x)}{2 d}+\frac {a^2 \sin ^5(c+d x)}{5 d} \] Output:
-a^2*csc(d*x+c)/d+2*a^2*ln(sin(d*x+c))/d-a^2*sin(d*x+c)/d-2*a^2*sin(d*x+c) ^2/d-1/3*a^2*sin(d*x+c)^3/d+1/2*a^2*sin(d*x+c)^4/d+1/5*a^2*sin(d*x+c)^5/d
Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x)}{d}-\frac {2 a^2 \sin ^2(c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^4(c+d x)}{2 d}+\frac {a^2 \sin ^5(c+d x)}{5 d} \] Input:
Integrate[Cos[c + d*x]^3*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
-((a^2*Csc[c + d*x])/d) + (2*a^2*Log[Sin[c + d*x]])/d - (a^2*Sin[c + d*x]) /d - (2*a^2*Sin[c + d*x]^2)/d - (a^2*Sin[c + d*x]^3)/(3*d) + (a^2*Sin[c + d*x]^4)/(2*d) + (a^2*Sin[c + d*x]^5)/(5*d)
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int \cos ^3(c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^2}{\sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \csc ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}{a^2}d(a \sin (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\sin ^4(c+d x) a^4+2 \sin ^3(c+d x) a^4+\csc ^2(c+d x) a^4-\sin ^2(c+d x) a^4+2 \csc (c+d x) a^4-4 \sin (c+d x) a^4-a^4\right )d(a \sin (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{5} a^5 \sin ^5(c+d x)+\frac {1}{2} a^5 \sin ^4(c+d x)-\frac {1}{3} a^5 \sin ^3(c+d x)-2 a^5 \sin ^2(c+d x)-a^5 \sin (c+d x)-a^5 \csc (c+d x)+2 a^5 \log (a \sin (c+d x))}{a^3 d}\) |
Input:
Int[Cos[c + d*x]^3*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
(-(a^5*Csc[c + d*x]) + 2*a^5*Log[a*Sin[c + d*x]] - a^5*Sin[c + d*x] - 2*a^ 5*Sin[c + d*x]^2 - (a^5*Sin[c + d*x]^3)/3 + (a^5*Sin[c + d*x]^4)/2 + (a^5* Sin[c + d*x]^5)/5)/(a^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 = 4.92 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a^{2} \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(121\) |
| default | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a^{2} \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(121\) |
| risch | \(-2 i a^{2} x +\frac {3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {9 i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {9 i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {3 a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {4 i a^{2} c}{d}-\frac {2 i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} \cos \left (4 d x +4 c \right )}{16 d}+\frac {a^{2} \sin \left (3 d x +3 c \right )}{48 d}\) | \(191\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/5*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+2*a^2*(1/4*cos (d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c)))+a^2*(-1/sin(d*x+c)*cos(d*x+c)^6 -(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {48 \, a^{2} \cos \left (d x + c\right )^{6} - 64 \, a^{2} \cos \left (d x + c\right )^{4} - 256 \, a^{2} \cos \left (d x + c\right )^{2} - 480 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 512 \, a^{2} - 15 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/240*(48*a^2*cos(d*x + c)^6 - 64*a^2*cos(d*x + c)^4 - 256*a^2*cos(d*x + c)^2 - 480*a^2*log(1/2*sin(d*x + c))*sin(d*x + c) + 512*a^2 - 15*(8*a^2*co s(d*x + c)^4 + 16*a^2*cos(d*x + c)^2 - 11*a^2)*sin(d*x + c))/(d*sin(d*x + c))
\[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{3}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cos(c + d*x)**3*cot(c + d*x)**2, x) + Integral(2*sin(c + d* x)*cos(c + d*x)**3*cot(c + d*x)**2, x) + Integral(sin(c + d*x)**2*cos(c + d*x)**3*cot(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{3} - 60 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 30 \, a^{2} \sin \left (d x + c\right ) - \frac {30 \, a^{2}}{\sin \left (d x + c\right )}}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/30*(6*a^2*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^3 - 60*a^2*sin(d*x + c)^2 + 60*a^2*log(sin(d*x + c)) - 30*a^2*sin(d*x + c) - 30*a^2/sin(d*x + c))/d
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{3} - 60 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 30 \, a^{2} \sin \left (d x + c\right ) - \frac {30 \, a^{2}}{\sin \left (d x + c\right )}}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/30*(6*a^2*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^3 - 60*a^2*sin(d*x + c)^2 + 60*a^2*log(abs(sin(d*x + c))) - 30*a^2*sin(d*x + c) - 30*a^2/sin(d*x + c))/d
Time = 18.62 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.92 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {16\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {8\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}-\frac {16\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {8\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}-\frac {2\,a^2\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {2\,a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {176\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {328\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{15\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {96\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {32\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {5\,a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \] Input:
int(cos(c + d*x)^3*cot(c + d*x)^2*(a + a*sin(c + d*x))^2,x)
Output:
(16*a^2*cos(c/2 + (d*x)/2)^4)/d - (8*a^2*cos(c/2 + (d*x)/2)^2)/d - (16*a^2 *cos(c/2 + (d*x)/2)^6)/d + (8*a^2*cos(c/2 + (d*x)/2)^8)/d - (2*a^2*log(1/c os(c/2 + (d*x)/2)^2))/d + (2*a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) ))/d - (2*a^2*cos(c/2 + (d*x)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) + (176*a^2*co s(c/2 + (d*x)/2)^5)/(15*d*sin(c/2 + (d*x)/2)) - (328*a^2*cos(c/2 + (d*x)/2 )^7)/(15*d*sin(c/2 + (d*x)/2)) + (96*a^2*cos(c/2 + (d*x)/2)^9)/(5*d*sin(c/ 2 + (d*x)/2)) - (32*a^2*cos(c/2 + (d*x)/2)^11)/(5*d*sin(c/2 + (d*x)/2)) - (5*a^2*cos(c/2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a^2*sin(c/2 + (d*x) /2))/(2*d*cos(c/2 + (d*x)/2))
Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+6 \sin \left (d x +c \right )^{6}+15 \sin \left (d x +c \right )^{5}-10 \sin \left (d x +c \right )^{4}-60 \sin \left (d x +c \right )^{3}-30 \sin \left (d x +c \right )^{2}-30\right )}{30 \sin \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 60*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 60*log(tan((c + d *x)/2))*sin(c + d*x) + 6*sin(c + d*x)**6 + 15*sin(c + d*x)**5 - 10*sin(c + d*x)**4 - 60*sin(c + d*x)**3 - 30*sin(c + d*x)**2 - 30))/(30*sin(c + d*x) *d)