Integrand size = 27, antiderivative size = 86 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {2 a \log (\sin (c+d x))}{d}-\frac {2 a \sin (c+d x)}{d}+\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{3 d} \] Output:
-a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-2*a*ln(sin(d*x+c))/d-2*a*sin(d*x+c)/d +1/2*a*sin(d*x+c)^2/d+1/3*a*sin(d*x+c)^3/d
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {2 a \sin (c+d x)}{d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \left (\csc ^2(c+d x)+4 \log (\sin (c+d x))-\sin ^2(c+d x)\right )}{2 d} \] Input:
Integrate[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]
Output:
-((a*Csc[c + d*x])/d) - (2*a*Sin[c + d*x])/d + (a*Sin[c + d*x]^3)/(3*d) - (a*(Csc[c + d*x]^2 + 4*Log[Sin[c + d*x]] - Sin[c + d*x]^2))/(2*d)
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \cos ^2(c+d x) \cot ^3(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)}{\sin (c+d x)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}{a^3}d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (a^2 \csc ^3(c+d x)+a^2 \csc ^2(c+d x)-2 a^2 \csc (c+d x)-2 a^2+a^2 \sin ^2(c+d x)+a^2 \sin (c+d x)\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} a^3 \sin ^3(c+d x)+\frac {1}{2} a^3 \sin ^2(c+d x)-2 a^3 \sin (c+d x)-\frac {1}{2} a^3 \csc ^2(c+d x)-a^3 \csc (c+d x)-2 a^3 \log (a \sin (c+d x))}{a^2 d}\) |
Input:
Int[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]
Output:
(-(a^3*Csc[c + d*x]) - (a^3*Csc[c + d*x]^2)/2 - 2*a^3*Log[a*Sin[c + d*x]] - 2*a^3*Sin[c + d*x] + (a^3*Sin[c + d*x]^2)/2 + (a^3*Sin[c + d*x]^3)/3)/(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 = 1.79 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(105\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(105\) |
risch | \(2 i a x +\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {4 i a c}{d}-\frac {2 i a \left (i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(180\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-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*(-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 = 102, normalized size of antiderivative = 1.19 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{4} - 9 \, a \cos \left (d x + c\right )^{2} + 24 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right ) - 3 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/12*(6*a*cos(d*x + c)^4 - 9*a*cos(d*x + c)^2 + 24*(a*cos(d*x + c)^2 - a) *log(1/2*sin(d*x + c)) + 4*(a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 - 8*a)*s in(d*x + c) - 3*a)/(d*cos(d*x + c)^2 - d)
\[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**3*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)**2*cot(c + d*x)**3, x) + Integral(sin(c + d*x)*co s(c + d*x)**2*cot(c + d*x)**3, x))
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left (\sin \left (d x + c\right )\right ) - 12 \, a \sin \left (d x + c\right ) - \frac {3 \, {\left (2 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(sin(d*x + c)) - 12 *a*sin(d*x + c) - 3*(2*a*sin(d*x + c) + a)/sin(d*x + c)^2)/d
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 12 \, a \sin \left (d x + c\right ) - \frac {3 \, {\left (2 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(abs(sin(d*x + c))) - 12*a*sin(d*x + c) - 3*(2*a*sin(d*x + c) + a)/sin(d*x + c)^2)/d
Time = 18.20 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.66 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {82\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {2\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:
int(cos(c + d*x)^2*cot(c + d*x)^3*(a + a*sin(c + d*x)),x)
Output:
(2*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a/2 + 2*a*tan(c/2 + (d*x)/2) + (3 *a*tan(c/2 + (d*x)/2)^2)/2 + 22*a*tan(c/2 + (d*x)/2)^3 - (13*a*tan(c/2 + ( d*x)/2)^4)/2 + (82*a*tan(c/2 + (d*x)/2)^5)/3 - (15*a*tan(c/2 + (d*x)/2)^6) /2 + 18*a*tan(c/2 + (d*x)/2)^7)/(d*(4*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)^6 + 4*tan(c/2 + (d*x)/2)^8)) - (a*tan(c /2 + (d*x)/2))/(2*d) - (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (2*a*log(tan(c/2 + (d*x)/2)))/d
Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.26 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+8 \sin \left (d x +c \right )^{5}+12 \sin \left (d x +c \right )^{4}-48 \sin \left (d x +c \right )^{3}+15 \sin \left (d x +c \right )^{2}-24 \sin \left (d x +c \right )-12\right )}{24 \sin \left (d x +c \right )^{2} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^3*(a+a*sin(d*x+c)),x)
Output:
(a*(48*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 48*log(tan((c + d*x) /2))*sin(c + d*x)**2 + 8*sin(c + d*x)**5 + 12*sin(c + d*x)**4 - 48*sin(c + d*x)**3 + 15*sin(c + d*x)**2 - 24*sin(c + d*x) - 12))/(24*sin(c + d*x)**2 *d)