Integrand size = 27, antiderivative size = 86 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {2 a \log (\sin (c+d x))}{d}-\frac {2 b \sin (c+d x)}{d}+\frac {a \sin ^2(c+d x)}{2 d}+\frac {b \sin ^3(c+d x)}{3 d} \] Output:
-b*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-2*a*ln(sin(d*x+c))/d-2*b*sin(d*x+c)/d +1/2*a*sin(d*x+c)^2/d+1/3*b*sin(d*x+c)^3/d
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \csc (c+d x)}{d}-\frac {2 b \sin (c+d x)}{d}+\frac {b \sin ^3(c+d x)}{3 d}-\frac {a \left (\csc ^2(c+d x)+4 \log (\sin (c+d x))-\sin ^2(c+d x)\right )}{2 d} \] Input:
Integrate[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + b*Sin[c + d*x]),x]
Output:
-((b*Csc[c + d*x])/d) - (2*b*Sin[c + d*x])/d + (b*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 = 92, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 522, 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.
\(\displaystyle \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))}{\sin (c+d x)^3}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \csc ^3(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^3}d(b \sin (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (a b \csc ^3(c+d x)+b^2 \csc ^2(c+d x)-2 a b \csc (c+d x)-2 b^2+b^2 \sin ^2(c+d x)+a b \sin (c+d x)\right )d(b \sin (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} a b^2 \sin ^2(c+d x)-\frac {1}{2} a b^2 \csc ^2(c+d x)-2 a b^2 \log (b \sin (c+d x))+\frac {1}{3} b^3 \sin ^3(c+d x)-2 b^3 \sin (c+d x)-b^3 \csc (c+d x)}{b^2 d}\) |
Input:
Int[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + b*Sin[c + d*x]),x]
Output:
(-(b^3*Csc[c + d*x]) - (a*b^2*Csc[c + d*x]^2)/2 - 2*a*b^2*Log[b*Sin[c + d* x]] - 2*b^3*Sin[c + d*x] + (a*b^2*Sin[c + d*x]^2)/2 + (b^3*Sin[c + d*x]^3) /3)/(b^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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.98 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(105\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(105\) |
risch | \(2 i a x +\frac {i b \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {7 i b \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {7 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i b \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {4 i a c}{d}-\frac {2 i \left (i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(183\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^3*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(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)))+b*(-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 = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.19 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{4} - 9 \, a \cos \left (d x + c\right )^{2} + 24 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (b \cos \left (d x + c\right )^{4} + 4 \, b \cos \left (d x + c\right )^{2} - 8 \, b\right )} \sin \left (d x + c\right ) - 3 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+b*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*(b*cos(d*x + c)^4 + 4*b*cos(d*x + c)^2 - 8*b)*s in(d*x + c) - 3*a)/(d*cos(d*x + c)^2 - d)
\[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**3*(a+b*sin(d*x+c)),x)
Output:
Integral((a + b*sin(c + d*x))*cos(c + d*x)**2*cot(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {2 \, b \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left (\sin \left (d x + c\right )\right ) - 12 \, b \sin \left (d x + c\right ) - \frac {3 \, {\left (2 \, b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/6*(2*b*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(sin(d*x + c)) - 12 *b*sin(d*x + c) - 3*(2*b*sin(d*x + c) + a)/sin(d*x + c)^2)/d
Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {2 \, b \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 12 \, b \sin \left (d x + c\right ) - \frac {3 \, {\left (2 \, b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*(2*b*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(abs(sin(d*x + c))) - 12*b*sin(d*x + c) - 3*(2*b*sin(d*x + c) + a)/sin(d*x + c)^2)/d
Time = 22.44 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.66 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {2\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {18\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {82\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+22\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {2\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:
int(cos(c + d*x)^2*cot(c + d*x)^3*(a + b*sin(c + d*x)),x)
Output:
(2*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (b*tan(c/2 + (d*x)/2))/(2*d) - (a/ 2 + 2*b*tan(c/2 + (d*x)/2) + (3*a*tan(c/2 + (d*x)/2)^2)/2 - (13*a*tan(c/2 + (d*x)/2)^4)/2 - (15*a*tan(c/2 + (d*x)/2)^6)/2 + 22*b*tan(c/2 + (d*x)/2)^ 3 + (82*b*tan(c/2 + (d*x)/2)^5)/3 + 18*b*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*ta n(c/2 + (d*x)/2)^8)) - (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (2*a*log(tan(c/2 + (d*x)/2)))/d
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.35 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +8 \sin \left (d x +c \right )^{5} b +12 \sin \left (d x +c \right )^{4} a -48 \sin \left (d x +c \right )^{3} b +15 \sin \left (d x +c \right )^{2} a -24 \sin \left (d x +c \right ) b -12 a}{24 \sin \left (d x +c \right )^{2} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^3*(a+b*sin(d*x+c)),x)
Output:
(48*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a - 48*log(tan((c + d*x)/ 2))*sin(c + d*x)**2*a + 8*sin(c + d*x)**5*b + 12*sin(c + d*x)**4*a - 48*si n(c + d*x)**3*b + 15*sin(c + d*x)**2*a - 24*sin(c + d*x)*b - 12*a)/(24*sin (c + d*x)**2*d)