Integrand size = 27, antiderivative size = 120 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \csc (c+d x)}{d}-\frac {a b \csc ^2(c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {4 a b \log (\sin (c+d x))}{d}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{d}+\frac {a b \sin ^2(c+d x)}{d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \] Output:
(2*a^2-b^2)*csc(d*x+c)/d-a*b*csc(d*x+c)^2/d-1/3*a^2*csc(d*x+c)^3/d-4*a*b*l n(sin(d*x+c))/d+(a^2-2*b^2)*sin(d*x+c)/d+a*b*sin(d*x+c)^2/d+1/3*b^2*sin(d* x+c)^3/d
Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (6 a^2-3 b^2\right ) \csc (c+d x)-3 a b \csc ^2(c+d x)-a^2 \csc ^3(c+d x)-12 a b \log (\sin (c+d x))+3 \left (a^2-2 b^2\right ) \sin (c+d x)+3 a b \sin ^2(c+d x)+b^2 \sin ^3(c+d x)}{3 d} \] Input:
Integrate[Cos[c + d*x]*Cot[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]
Output:
((6*a^2 - 3*b^2)*Csc[c + d*x] - 3*a*b*Csc[c + d*x]^2 - a^2*Csc[c + d*x]^3 - 12*a*b*Log[Sin[c + d*x]] + 3*(a^2 - 2*b^2)*Sin[c + d*x] + 3*a*b*Sin[c + d*x]^2 + b^2*Sin[c + d*x]^3)/(3*d)
Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98, 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.
| \(\displaystyle \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))^2}{\sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \csc ^4(c+d x) (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^4(c+d x) (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^4}d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (a^2 \csc ^4(c+d x)+2 a b \csc ^3(c+d x)+\frac {\left (b^4-2 a^2 b^2\right ) \csc ^2(c+d x)}{b^2}-4 a b \csc (c+d x)+b^2 \sin ^2(c+d x)+a^2 \left (1-\frac {2 b^2}{a^2}\right )+2 a b \sin (c+d x)\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (a^2-2 b^2\right ) \sin (c+d x)+b \left (2 a^2-b^2\right ) \csc (c+d x)-\frac {1}{3} a^2 b \csc ^3(c+d x)+a b^2 \sin ^2(c+d x)-a b^2 \csc ^2(c+d x)-4 a b^2 \log (b \sin (c+d x))+\frac {1}{3} b^3 \sin ^3(c+d x)}{b d}\) |
Input:
Int[Cos[c + d*x]*Cot[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]
Output:
(b*(2*a^2 - b^2)*Csc[c + d*x] - a*b^2*Csc[c + d*x]^2 - (a^2*b*Csc[c + d*x] ^3)/3 - 4*a*b^2*Log[b*Sin[c + d*x]] + b*(a^2 - 2*b^2)*Sin[c + d*x] + a*b^2 *Sin[c + d*x]^2 + (b^3*Sin[c + d*x]^3)/3)/(b*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 = 2.73 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.47
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(176\) |
| default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{6}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{4}}{2}-\cos \left (d x +c \right )^{2}-2 \ln \left (\sin \left (d x +c \right )\right )\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(176\) |
| risch | \(4 i x a b +\frac {i b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d}-\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 d}-\frac {a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}-\frac {i b^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {8 i a b c}{d}+\frac {2 i \left (6 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-8 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(308\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/3/sin(d*x+c)^3*cos(d*x+c)^6+1/sin(d*x+c)*cos(d*x+c)^6+(8/3+co s(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+2*a*b*(-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^2*(-1/sin(d*x+c)*co s(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 = 158, normalized size of antiderivative = 1.32 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {2 \, b^{2} \cos \left (d x + c\right )^{6} - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 24 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 24 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 16 \, a^{2} + 16 \, b^{2} - 3 \, {\left (2 \, a b \cos \left (d x + c\right )^{4} - 3 \, a b \cos \left (d x + c\right )^{2} - a b\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/6*(2*b^2*cos(d*x + c)^6 - 6*(a^2 - b^2)*cos(d*x + c)^4 + 24*(a^2 - b^2)* cos(d*x + c)^2 - 24*(a*b*cos(d*x + c)^2 - a*b)*log(1/2*sin(d*x + c))*sin(d *x + c) - 16*a^2 + 16*b^2 - 3*(2*a*b*cos(d*x + c)^4 - 3*a*b*cos(d*x + c)^2 - a*b)*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))
\[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cos {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**4*(a+b*sin(d*x+c))**2,x)
Output:
Integral((a + b*sin(c + d*x))**2*cos(c + d*x)*cot(c + d*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {b^{2} \sin \left (d x + c\right )^{3} + 3 \, a b \sin \left (d x + c\right )^{2} - 12 \, a b \log \left (\sin \left (d x + c\right )\right ) + 3 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right ) - \frac {3 \, a b \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/3*(b^2*sin(d*x + c)^3 + 3*a*b*sin(d*x + c)^2 - 12*a*b*log(sin(d*x + c)) + 3*(a^2 - 2*b^2)*sin(d*x + c) - (3*a*b*sin(d*x + c) - 3*(2*a^2 - b^2)*sin (d*x + c)^2 + a^2)/sin(d*x + c)^3)/d
Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.91 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {b^{2} \sin \left (d x + c\right )^{3} + 3 \, a b \sin \left (d x + c\right )^{2} - 12 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 3 \, a^{2} \sin \left (d x + c\right ) - 6 \, b^{2} \sin \left (d x + c\right ) - \frac {3 \, a b \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/3*(b^2*sin(d*x + c)^3 + 3*a*b*sin(d*x + c)^2 - 12*a*b*log(abs(sin(d*x + c))) + 3*a^2*sin(d*x + c) - 6*b^2*sin(d*x + c) - (3*a*b*sin(d*x + c) - 3*( 2*a^2 - b^2)*sin(d*x + c)^2 + a^2)/sin(d*x + c)^3)/d
Time = 22.80 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.62 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a^2-4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (23\,a^2-36\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (36\,a^2-44\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {158\,a^2}{3}-\frac {164\,b^2}{3}\right )-\frac {a^2}{3}-6\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+26\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+30\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {7\,a^2}{8}-\frac {b^2}{2}\right )}{d}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {4\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {4\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int(cos(c + d*x)*cot(c + d*x)^4*(a + b*sin(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)^2*(6*a^2 - 4*b^2) + tan(c/2 + (d*x)/2)^8*(23*a^2 - 36* b^2) + tan(c/2 + (d*x)/2)^4*(36*a^2 - 44*b^2) + tan(c/2 + (d*x)/2)^6*((158 *a^2)/3 - (164*b^2)/3) - a^2/3 - 6*a*b*tan(c/2 + (d*x)/2)^3 + 26*a*b*tan(c /2 + (d*x)/2)^5 + 30*a*b*tan(c/2 + (d*x)/2)^7 - 2*a*b*tan(c/2 + (d*x)/2))/ (d*(8*tan(c/2 + (d*x)/2)^3 + 24*tan(c/2 + (d*x)/2)^5 + 24*tan(c/2 + (d*x)/ 2)^7 + 8*tan(c/2 + (d*x)/2)^9)) - (a^2*tan(c/2 + (d*x)/2)^3)/(24*d) + (tan (c/2 + (d*x)/2)*((7*a^2)/8 - b^2/2))/d - (a*b*tan(c/2 + (d*x)/2)^2)/(4*d) - (4*a*b*log(tan(c/2 + (d*x)/2)))/d + (4*a*b*log(tan(c/2 + (d*x)/2)^2 + 1) )/d
Time = 0.18 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.38 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a b -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a b +4 \sin \left (d x +c \right )^{6} b^{2}+12 \sin \left (d x +c \right )^{5} a b +12 \sin \left (d x +c \right )^{4} a^{2}-24 \sin \left (d x +c \right )^{4} b^{2}+15 \sin \left (d x +c \right )^{3} a b +24 \sin \left (d x +c \right )^{2} a^{2}-12 \sin \left (d x +c \right )^{2} b^{2}-12 \sin \left (d x +c \right ) a b -4 a^{2}}{12 \sin \left (d x +c \right )^{3} d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^4*(a+b*sin(d*x+c))^2,x)
Output:
(48*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*a*b - 48*log(tan((c + d*x )/2))*sin(c + d*x)**3*a*b + 4*sin(c + d*x)**6*b**2 + 12*sin(c + d*x)**5*a* b + 12*sin(c + d*x)**4*a**2 - 24*sin(c + d*x)**4*b**2 + 15*sin(c + d*x)**3 *a*b + 24*sin(c + d*x)**2*a**2 - 12*sin(c + d*x)**2*b**2 - 12*sin(c + d*x) *a*b - 4*a**2)/(12*sin(c + d*x)**3*d)