Integrand size = 27, antiderivative size = 137 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 a x}{b^3}+\frac {2 \sqrt {a^2-b^2} \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^3 d}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{b^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))} \] Output:
-2*a*x/b^3+2*(a^2-b^2)^(1/2)*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/( a^2-b^2)^(1/2))/a^2/b^3/d-arctanh(cos(d*x+c))/a^2/d-cos(d*x+c)/b^2/d-(a^2- b^2)*cos(d*x+c)/a/b^2/d/(a+b*sin(d*x+c))
Time = 1.19 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {2 a (c+d x)}{b^3}+\frac {2 \left (2 a^4-a^2 b^2-b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^3 \sqrt {a^2-b^2}}-\frac {\cos (c+d x)}{b^2}-\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {\left (-a^2+b^2\right ) \cos (c+d x)}{a b^2 (a+b \sin (c+d x))}}{d} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^2,x]
Output:
((-2*a*(c + d*x))/b^3 + (2*(2*a^4 - a^2*b^2 - b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*b^3*Sqrt[a^2 - b^2]) - Cos[c + d*x]/b^2 - Log[Cos[(c + d*x)/2]]/a^2 + Log[Sin[(c + d*x)/2]]/a^2 + ((-a^2 + b^2)*Cos[ c + d*x])/(a*b^2*(a + b*Sin[c + d*x])))/d
Time = 0.76 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.21, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3371, 3042, 3536, 3042, 3139, 1083, 217, 4257}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x) (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3371 |
| \(\displaystyle \frac {\int \frac {\csc (c+d x) \left (b^2-a \sin (c+d x) b-2 a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {b^2-a \sin (c+d x) b-2 a^2 \sin (c+d x)^2}{\sin (c+d x) (a+b \sin (c+d x))}dx}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 3536 |
| \(\displaystyle \frac {\frac {\left (2 a^4-a^2 b^2-b^4\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a b}+\frac {b^2 \int \csc (c+d x)dx}{a}-\frac {2 a^2 x}{b}}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^4-a^2 b^2-b^4\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a b}+\frac {b^2 \int \csc (c+d x)dx}{a}-\frac {2 a^2 x}{b}}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {2 \left (2 a^4-a^2 b^2-b^4\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}+\frac {b^2 \int \csc (c+d x)dx}{a}-\frac {2 a^2 x}{b}}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {-\frac {4 \left (2 a^4-a^2 b^2-b^4\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a b d}+\frac {b^2 \int \csc (c+d x)dx}{a}-\frac {2 a^2 x}{b}}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {b^2 \int \csc (c+d x)dx}{a}-\frac {2 a^2 x}{b}+\frac {2 \left (2 a^4-a^2 b^2-b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a b d \sqrt {a^2-b^2}}}{a b^2}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a b^2 d (a+b \sin (c+d x))}+\frac {-\frac {2 a^2 x}{b}+\frac {2 \left (2 a^4-a^2 b^2-b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a b d \sqrt {a^2-b^2}}-\frac {b^2 \text {arctanh}(\cos (c+d x))}{a d}}{a b^2}-\frac {\cos (c+d x)}{b^2 d}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x])^2,x]
Output:
((-2*a^2*x)/b + (2*(2*a^4 - a^2*b^2 - b^4)*ArcTan[(2*b + 2*a*Tan[(c + d*x) /2])/(2*Sqrt[a^2 - b^2])])/(a*b*Sqrt[a^2 - b^2]*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a*d))/(a*b^2) - Cos[c + d*x]/(b^2*d) - ((a^2 - b^2)*Cos[c + d*x])/ (a*b^2*d*(a + b*Sin[c + d*x]))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*b^2*d*f*(m + 1))), x] + (-Simp[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2)*((d*Sin[e + f* x])^(n + 1)/(b^2*d*f*(m + n + 4))), x] - Simp[1/(a*b^2*(m + 1)*(m + n + 4)) Int[(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^n*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 2 )*(n + 3) - b^2*(m + n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; Fre eQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && Lt Q[m, -1] && !LtQ[n, -1] && NeQ[m + n + 4, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Simp[(A*b^2 - a*b*B + a^2*C) /(b*(b*c - a*d)) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)) Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a , b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.83 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.41
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {4 \left (-\frac {b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{3} b}{2}+\frac {a \,b^{3}}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (2 a^{4}-a^{2} b^{2}-b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{2} b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {4 \left (\frac {b}{2+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{3}}}{d}\) | \(193\) |
| default | \(\frac {\frac {\frac {4 \left (-\frac {b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{3} b}{2}+\frac {a \,b^{3}}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (2 a^{4}-a^{2} b^{2}-b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{2} b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {4 \left (\frac {b}{2+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{3}}}{d}\) | \(193\) |
| risch | \(-\frac {2 a x}{b^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {2 i \left (-a^{2}+b^{2}\right ) \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \,b^{3} a \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {2 i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{3}}+\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d b \,a^{2}}-\frac {2 i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{3}}-\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d b \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) | \(358\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(4/b^3/a^2*((-1/2*b^2*(a^2-b^2)*tan(1/2*d*x+1/2*c)-1/2*a^3*b+1/2*a*b^3 )/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+1/2*(2*a^4-a^2*b^2-b^4 )/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)) )+1/a^2*ln(tan(1/2*d*x+1/2*c))-4/b^3*(1/2*b/(1+tan(1/2*d*x+1/2*c)^2)+a*arc tan(tan(1/2*d*x+1/2*c))))
Time = 0.17 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.77 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [-\frac {4 \, a^{4} d x - {\left (2 \, a^{3} + a b^{2} + {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + {\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (2 \, a^{3} b d x + a^{2} b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{4} d \sin \left (d x + c\right ) + a^{3} b^{3} d\right )}}, -\frac {4 \, a^{4} d x + 2 \, {\left (2 \, a^{3} + a b^{2} + {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + {\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (b^{4} \sin \left (d x + c\right ) + a b^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (2 \, a^{3} b d x + a^{2} b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{4} d \sin \left (d x + c\right ) + a^{3} b^{3} d\right )}}\right ] \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
[-1/2*(4*a^4*d*x - (2*a^3 + a*b^2 + (2*a^2*b + b^3)*sin(d*x + c))*sqrt(-a^ 2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b ^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b ^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2*(2*a^3*b - a*b^3) *cos(d*x + c) + (b^4*sin(d*x + c) + a*b^3)*log(1/2*cos(d*x + c) + 1/2) - ( b^4*sin(d*x + c) + a*b^3)*log(-1/2*cos(d*x + c) + 1/2) + 2*(2*a^3*b*d*x + a^2*b^2*cos(d*x + c))*sin(d*x + c))/(a^2*b^4*d*sin(d*x + c) + a^3*b^3*d), -1/2*(4*a^4*d*x + 2*(2*a^3 + a*b^2 + (2*a^2*b + b^3)*sin(d*x + c))*sqrt(a^ 2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 2* (2*a^3*b - a*b^3)*cos(d*x + c) + (b^4*sin(d*x + c) + a*b^3)*log(1/2*cos(d* x + c) + 1/2) - (b^4*sin(d*x + c) + a*b^3)*log(-1/2*cos(d*x + c) + 1/2) + 2*(2*a^3*b*d*x + a^2*b^2*cos(d*x + c))*sin(d*x + c))/(a^2*b^4*d*sin(d*x + c) + a^3*b^3*d)]
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)/(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (132) = 264\).
Time = 0.15 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.09 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (d x + c\right )} a}{b^{3}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2} b^{3}} + \frac {2 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{2} b^{2}}}{d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
-(2*(d*x + c)*a/b^3 - log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 2*(2*a^4 - a^2* b^2 - b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d* x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^2*b^3) + 2*(a^2*b*tan (1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 + 2*a^3*tan(1/2*d*x + 1/2 *c)^2 - a*b^2*tan(1/2*d*x + 1/2*c)^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - b^3* tan(1/2*d*x + 1/2*c) + 2*a^3 - a*b^2)/((a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan (1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c ) + a)*a^2*b^2))/d
Time = 34.83 (sec) , antiderivative size = 2773, normalized size of antiderivative = 20.24 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)^3*cot(c + d*x))/(a + b*sin(c + d*x))^2,x)
Output:
log(tan(c/2 + (d*x)/2))/(a^2*d) + (atan((((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^(1/2)/(a^2*b))*(((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^(1/2)/(a^ 2*b))*(((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^(1/2)/(a^2*b))*(((32*(4*a^ 4*b^10 - 3*a^6*b^8))/(a^2*b^5) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^14 - 17* a^6*b^12 + 2*a^8*b^10))/(a^3*b^8))*((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2 )^(1/2)/(a^2*b)) + (32*(8*a^2*b^10 - 7*a^6*b^6))/(a^2*b^5) + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^14 - 5*a^4*b^12 - 18*a^6*b^10 + 8*a^8*b^8))/(a^3*b^8)) + (32*(4*b^10 + 3*a^2*b^8 - 7*a^4*b^6 + 16*a^6*b^4 - 12*a^8*b^2))/(a^2*b^ 5) + (32*tan(c/2 + (d*x)/2)*(6*a^4*b^10 - 11*a^2*b^12 + 101*a^6*b^8 - 100* a^8*b^6 + 8*a^10*b^4))/(a^3*b^8)) - (32*(28*a^8 + 2*a^2*b^6 - 28*a^4*b^4 - 6*a^6*b^2))/(a^2*b^5) + (32*tan(c/2 + (d*x)/2)*(b^12 + 2*a^2*b^10 + 61*a^ 4*b^8 - 20*a^6*b^6 - 72*a^8*b^4 + 32*a^10*b^2))/(a^3*b^8))*1i - ((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^(1/2)/(a^2*b))*(((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^(1/2)/(a^2*b))*((32*(4*b^10 + 3*a^2*b^8 - 7*a^4*b^6 + 16*a^6 *b^4 - 12*a^8*b^2))/(a^2*b^5) - ((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^( 1/2)/(a^2*b))*((32*(8*a^2*b^10 - 7*a^6*b^6))/(a^2*b^5) - ((32*(4*a^4*b^10 - 3*a^6*b^8))/(a^2*b^5) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^14 - 17*a^6*b^1 2 + 2*a^8*b^10))/(a^3*b^8))*((2*(b^2 - a^2)^(1/2))/b^3 + (b^2 - a^2)^(1/2) /(a^2*b)) + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^14 - 5*a^4*b^12 - 18*a^6*b^10 + 8*a^8*b^8))/(a^3*b^8)) + (32*tan(c/2 + (d*x)/2)*(6*a^4*b^10 - 11*a^2...
Time = 0.18 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.36 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{2} b +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) b^{3}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3}+2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a \,b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}-2 \cos \left (d x +c \right ) a^{3} b +\cos \left (d x +c \right ) a \,b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{3}-2 \sin \left (d x +c \right ) a^{3} b d x -\sin \left (d x +c \right ) a^{2} b^{2}-2 a^{4} d x -a^{3} b}{a^{2} b^{3} d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c))^2,x)
Output:
(4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( c + d*x)*a**2*b + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a **2 - b**2))*sin(c + d*x)*b**3 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2 )*a + b)/sqrt(a**2 - b**2))*a**3 + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x) /2)*a + b)/sqrt(a**2 - b**2))*a*b**2 - cos(c + d*x)*sin(c + d*x)*a**2*b**2 - 2*cos(c + d*x)*a**3*b + cos(c + d*x)*a*b**3 + log(tan((c + d*x)/2))*sin (c + d*x)*b**4 + log(tan((c + d*x)/2))*a*b**3 - 2*sin(c + d*x)*a**3*b*d*x - sin(c + d*x)*a**2*b**2 - 2*a**4*d*x - a**3*b)/(a**2*b**3*d*(sin(c + d*x) *b + a))