Integrand size = 29, antiderivative size = 116 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {a^2 \log (\sin (c+d x))}{d}-\frac {4 a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^4(c+d x)}{4 d} \] Output:
-2*a^2*csc(d*x+c)/d-1/2*a^2*csc(d*x+c)^2/d-a^2*ln(sin(d*x+c))/d-4*a^2*sin( d*x+c)/d-1/2*a^2*sin(d*x+c)^2/d+2/3*a^2*sin(d*x+c)^3/d+1/4*a^2*sin(d*x+c)^ 4/d
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \left (24 \csc (c+d x)+6 \csc ^2(c+d x)+12 \log (\sin (c+d x))+48 \sin (c+d x)+6 \sin ^2(c+d x)-8 \sin ^3(c+d x)-3 \sin ^4(c+d x)\right )}{12 d} \] Input:
Integrate[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]
Output:
-1/12*(a^2*(24*Csc[c + d*x] + 6*Csc[c + d*x]^2 + 12*Log[Sin[c + d*x]] + 48 *Sin[c + d*x] + 6*Sin[c + d*x]^2 - 8*Sin[c + d*x]^3 - 3*Sin[c + d*x]^4))/d
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90, 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.
| \(\displaystyle \int \cos ^2(c+d x) \cot ^3(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^2}{\sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}{a^3}d(a \sin (c+d x))}{a^2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\csc ^3(c+d x) a^3+\sin ^3(c+d x) a^3+2 \csc ^2(c+d x) a^3+2 \sin ^2(c+d x) a^3-\csc (c+d x) a^3-\sin (c+d x) a^3-4 a^3\right )d(a \sin (c+d x))}{a^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{4} a^4 \sin ^4(c+d x)+\frac {2}{3} a^4 \sin ^3(c+d x)-\frac {1}{2} a^4 \sin ^2(c+d x)-4 a^4 \sin (c+d x)-\frac {1}{2} a^4 \csc ^2(c+d x)-2 a^4 \csc (c+d x)-a^4 \log (a \sin (c+d x))}{a^2 d}\) |
Input:
Int[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]
Output:
(-2*a^4*Csc[c + d*x] - (a^4*Csc[c + d*x]^2)/2 - a^4*Log[a*Sin[c + d*x]] - 4*a^4*Sin[c + d*x] - (a^4*Sin[c + d*x]^2)/2 + (2*a^4*Sin[c + d*x]^3)/3 + ( a^4*Sin[c + d*x]^4)/4)/(a^2*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 = 3.60 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.22
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(142\) |
| default | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(142\) |
| risch | \(i a^{2} x +\frac {i a^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{12 d}+\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {7 i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{4 d}-\frac {7 i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{4 d}+\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {i a^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{12 d}+\frac {2 i a^{2} c}{d}-\frac {2 i a^{2} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {a^{2} \cos \left (4 d x +4 c \right )}{32 d}\) | \(219\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c)))+2*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))+a^2*(- 1/2/sin(d*x+c)^2*cos(d*x+c)^6-1/2*cos(d*x+c)^4-cos(d*x+c)^2-2*ln(sin(d*x+c ))))
Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.13 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {24 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \, a^{2} \cos \left (d x + c\right )^{4} - 9 \, a^{2} \cos \left (d x + c\right )^{2} + 57 \, a^{2} - 96 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 64 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/96*(24*a^2*cos(d*x + c)^6 - 24*a^2*cos(d*x + c)^4 - 9*a^2*cos(d*x + c)^2 + 57*a^2 - 96*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c)) - 64*(a^2* cos(d*x + c)^4 + 4*a^2*cos(d*x + c)^2 - 8*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2 - d)
\[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**3*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cos(c + d*x)**2*cot(c + d*x)**3, x) + Integral(2*sin(c + d* x)*cos(c + d*x)**2*cot(c + d*x)**3, x) + Integral(sin(c + d*x)**2*cos(c + d*x)**2*cot(c + d*x)**3, x))
Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \sin \left (d x + c\right )^{4} + 8 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 48 \, a^{2} \sin \left (d x + c\right ) - \frac {6 \, {\left (4 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/12*(3*a^2*sin(d*x + c)^4 + 8*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) - 48*a^2*sin(d*x + c) - 6*(4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \sin \left (d x + c\right )^{4} + 8 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 48 \, a^{2} \sin \left (d x + c\right ) - \frac {6 \, {\left (4 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{2}}}{12 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/12*(3*a^2*sin(d*x + c)^4 + 8*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) - 48*a^2*sin(d*x + c) - 6*(4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d
Time = 18.15 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.56 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {272\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {296\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+48\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d} \] Input:
int(cos(c + d*x)^2*cot(c + d*x)^3*(a + a*sin(c + d*x))^2,x)
Output:
(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a^2*log(tan(c/2 + (d*x)/2)))/d - (2*a^2*tan(c/2 + (d*x)/2)^2 + 48*a^2*tan(c/2 + (d*x)/2)^3 + 11*a^2*tan(c/2 + (d*x)/2)^4 + (296*a^2*tan(c/2 + (d*x)/2)^5)/3 + 2*a^2*tan(c/2 + (d*x)/2 )^6 + (272*a^2*tan(c/2 + (d*x)/2)^7)/3 + (17*a^2*tan(c/2 + (d*x)/2)^8)/2 + 36*a^2*tan(c/2 + (d*x)/2)^9 + a^2/2 + 4*a^2*tan(c/2 + (d*x)/2))/(d*(4*tan (c/2 + (d*x)/2)^2 + 16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6 + 16 *tan(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) - (a^2*tan(c/2 + (d*x)/2 ))/d - (a^2*tan(c/2 + (d*x)/2)^2)/(8*d)
Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )^{6}+8 \sin \left (d x +c \right )^{5}-6 \sin \left (d x +c \right )^{4}-48 \sin \left (d x +c \right )^{3}+9 \sin \left (d x +c \right )^{2}-24 \sin \left (d x +c \right )-6\right )}{12 \sin \left (d x +c \right )^{2} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 12*log(tan((c + d *x)/2))*sin(c + d*x)**2 + 3*sin(c + d*x)**6 + 8*sin(c + d*x)**5 - 6*sin(c + d*x)**4 - 48*sin(c + d*x)**3 + 9*sin(c + d*x)**2 - 24*sin(c + d*x) - 6)) /(12*sin(c + d*x)**2*d)