Integrand size = 21, antiderivative size = 102 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-a^3 x+\frac {3}{2} a b^2 x-\frac {3 a^2 b \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \cos (c+d x)}{d}-\frac {b^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
-a^3*x+3/2*a*b^2*x-3*a^2*b*arctanh(cos(d*x+c))/d+3*a^2*b*cos(d*x+c)/d-1/3* b^3*cos(d*x+c)^3/d-a^3*cot(d*x+c)/d+3/2*a*b^2*cos(d*x+c)*sin(d*x+c)/d
Time = 1.39 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.40 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\left (36 a^2 b-3 b^3\right ) \cos (c+d x)-b^3 \cos (3 (c+d x))-6 a^3 \cot \left (\frac {1}{2} (c+d x)\right )+9 a b^2 \sin (2 (c+d x))+6 a \left (-2 a^2 c+3 b^2 c-2 a^2 d x+3 b^2 d x-6 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{12 d} \] Input:
Integrate[Cot[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]
Output:
((36*a^2*b - 3*b^3)*Cos[c + d*x] - b^3*Cos[3*(c + d*x)] - 6*a^3*Cot[(c + d *x)/2] + 9*a*b^2*Sin[2*(c + d*x)] + 6*a*(-2*a^2*c + 3*b^2*c - 2*a^2*d*x + 3*b^2*d*x - 6*a*b*Log[Cos[(c + d*x)/2]] + 6*a*b*Log[Sin[(c + d*x)/2]] + a^ 2*Tan[(c + d*x)/2]))/(12*d)
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3201, 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.
| \(\displaystyle \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\tan (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3201 |
| \(\displaystyle \int \left (a^3 \cot ^2(c+d x)+3 a^2 b \cos (c+d x) \cot (c+d x)+3 a b^2 \cos ^2(c+d x)+b^3 \sin (c+d x) \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 \cot (c+d x)}{d}+a^3 (-x)-\frac {3 a^2 b \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \cos (c+d x)}{d}+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} a b^2 x-\frac {b^3 \cos ^3(c+d x)}{3 d}\) |
Input:
Int[Cot[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]
Output:
-(a^3*x) + (3*a*b^2*x)/2 - (3*a^2*b*ArcTanh[Cos[c + d*x]])/d + (3*a^2*b*Co s[c + d*x])/d - (b^3*Cos[c + d*x]^3)/(3*d) - (a^3*Cot[c + d*x])/d + (3*a*b ^2*Cos[c + d*x]*Sin[c + d*x])/(2*d)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Int[ExpandIntegrand[(g*Tan[e + f*x])^p, (a + b*Si n[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Time = 1.72 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 a^{2} b \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {b^{3} \cos \left (d x +c \right )^{3}}{3}}{d}\) | \(96\) |
| default | \(\frac {a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 a^{2} b \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {b^{3} \cos \left (d x +c \right )^{3}}{3}}{d}\) | \(96\) |
| risch | \(-a^{3} x +\frac {3 a \,b^{2} x}{2}-\frac {3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}-\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {3 b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}-\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{12 d}\) | \(204\) |
Input:
int(cot(d*x+c)^2*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(-cot(d*x+c)-d*x-c)+3*a^2*b*(cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)) )+3*a*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)-1/3*b^3*cos(d*x+c)^3)
Time = 0.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.40 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {9 \, a b^{2} \cos \left (d x + c\right )^{3} + 9 \, a^{2} b \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, a^{2} b \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) + {\left (2 \, b^{3} \cos \left (d x + c\right )^{3} - 18 \, a^{2} b \cos \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x\right )} \sin \left (d x + c\right )}{6 \, d \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-1/6*(9*a*b^2*cos(d*x + c)^3 + 9*a^2*b*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*a^2*b*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(2*a^3 - 3*a *b^2)*cos(d*x + c) + (2*b^3*cos(d*x + c)^3 - 18*a^2*b*cos(d*x + c) + 3*(2* a^3 - 3*a*b^2)*d*x)*sin(d*x + c))/(d*sin(d*x + c))
\[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \cot ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**2*(a+b*sin(d*x+c))**3,x)
Output:
Integral((a + b*sin(c + d*x))**3*cot(c + d*x)**2, x)
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {4 \, b^{3} \cos \left (d x + c\right )^{3} + 12 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} - 18 \, a^{2} b {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \] Input:
integrate(cot(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/12*(4*b^3*cos(d*x + c)^3 + 12*(d*x + c + 1/tan(d*x + c))*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^2 - 18*a^2*b*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (96) = 192\).
Time = 0.16 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.95 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {18 \, a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {3 \, {\left (6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b + 2 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cot(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
1/6*(18*a^2*b*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^3*tan(1/2*d*x + 1/2*c) - 3*(2*a^3 - 3*a*b^2)*(d*x + c) - 3*(6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/t an(1/2*d*x + 1/2*c) - 2*(9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1/2 *d*x + 1/2*c)^4 + 6*b^3*tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/ 2*c)^2 - 9*a*b^2*tan(1/2*d*x + 1/2*c) - 18*a^2*b + 2*b^3)/(tan(1/2*d*x + 1 /2*c)^2 + 1)^3)/d
Time = 18.01 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.83 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )\,\left (\frac {a\,b^2\,3{}\mathrm {i}}{2}-a^3\,1{}\mathrm {i}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2\,b-\frac {4\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3+6\,a\,b^2\right )-3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a\,b^2-3\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (12\,a^2\,b-4\,b^3\right )-a^3+24\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {3\,a^2\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,d} \] Input:
int(cot(c + d*x)^2*(a + b*sin(c + d*x))^3,x)
Output:
(a^3*tan(c/2 + (d*x)/2))/(2*d) - (log(tan(c/2 + (d*x)/2) - 1i)*((a*b^2*3i) /2 - a^3*1i))/d + (tan(c/2 + (d*x)/2)*(12*a^2*b - (4*b^3)/3) - tan(c/2 + ( d*x)/2)^6*(6*a*b^2 + a^3) - 3*a^3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2 )^2*(6*a*b^2 - 3*a^3) + tan(c/2 + (d*x)/2)^5*(12*a^2*b - 4*b^3) - a^3 + 24 *a^2*b*tan(c/2 + (d*x)/2)^3)/(d*(2*tan(c/2 + (d*x)/2) + 6*tan(c/2 + (d*x)/ 2)^3 + 6*tan(c/2 + (d*x)/2)^5 + 2*tan(c/2 + (d*x)/2)^7)) + (3*a^2*b*log(ta n(c/2 + (d*x)/2)))/d - (a*log(tan(c/2 + (d*x)/2) + 1i)*(2*a^2 - 3*b^2)*1i) /(2*d)
Time = 0.20 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.68 \[ \int \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{3}+9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+18 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{3}-6 \cos \left (d x +c \right ) a^{3}+18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} b -6 \sin \left (d x +c \right ) a^{3} d x -18 \sin \left (d x +c \right ) a^{2} b +9 \sin \left (d x +c \right ) a \,b^{2} d x +2 \sin \left (d x +c \right ) b^{3}}{6 \sin \left (d x +c \right ) d} \] Input:
int(cot(d*x+c)^2*(a+b*sin(d*x+c))^3,x)
Output:
(2*cos(c + d*x)*sin(c + d*x)**3*b**3 + 9*cos(c + d*x)*sin(c + d*x)**2*a*b* *2 + 18*cos(c + d*x)*sin(c + d*x)*a**2*b - 2*cos(c + d*x)*sin(c + d*x)*b** 3 - 6*cos(c + d*x)*a**3 + 18*log(tan((c + d*x)/2))*sin(c + d*x)*a**2*b - 6 *sin(c + d*x)*a**3*d*x - 18*sin(c + d*x)*a**2*b + 9*sin(c + d*x)*a*b**2*d* x + 2*sin(c + d*x)*b**3)/(6*sin(c + d*x)*d)